${E^ \circ }$ for ${\rm{F}}{{\rm{e}}^{2 + }} + 2{{\rm{e}}^ - } \to {\rm{Fe}}$ is -0.44 V and ${E^ \circ }$ for ${\rm{Z}}{{\rm{n}}^{2 + }} + 2{{\rm{e}}^ - } \to {\rm{Zn}}$ is -0.76 V thus

Zn is more electronegative than Fe

Fe is more electropositive than Zn

Consider the following reaction : $2{\rm{A}}{{\rm{g}}^ + }{{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6} + {{\rm{H}}_2}{\rm{O}} \to 2{\rm{Ag}}\left( s \right) + {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_7} + 2{{\rm{H}}^ + }$ When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much? ${\rm{A}}{{\rm{g}}^ + } + {e^ - } \to {\rm{Ag}};E_{{\rm{red}}}^ \circ = 0.8{\rm{V}}$ ${{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_7} + 2{{\rm{H}}^ + } + 2{e^ - };E_{{\rm{red}}}^ \circ = - 0.05{\rm{V}}$ ${\rm{Ag}}\left( {{\rm{N}}{{\rm{H}}_3}} \right)_2^ + + {e^ - } \to {\rm{Ag}}\left( s \right) + {\rm{N}}{{\rm{H}}_3};E_{{\rm{red}}}^ \circ = 0.337{\rm{V}}$

$E_{{\rm{oxid}}{\rm{.}}}^ \circ $ will increase by a factor of 0.65 from $E_{{\rm{oxid}}{\rm{.}}}^ \circ $

$E_{{\rm{oxid}}{\rm{.}}}^ \circ $ will decrease by a factor of 0.65 from$E_{{\rm{oxid}}{\rm{.}}}^ \circ $

$E_{{\rm{red}}{\rm{.}}}^ \circ $ will increase by a factor of 0.65 from $E_{{\rm{red}}{\rm{.}}}^ \circ $

$E_{{\rm{red}}{\rm{.}}}^ \circ $ will decrease by a factor of 0.65 from$E_{{\rm{red}}{\rm{.}}}^ \circ $

The standard reduction potential of the reaction, ${{\rm{H}}_2}{\rm{O}} + {e^ - } \to 1/2{{\rm{H}}_2} + {\rm{OH}}$ at 298 K is

${E^ \circ } = \frac{{RT}}{F}{\rm{ln}}{K_w}$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}\left[ {{{\rm{P}}_{\rm{H}}}_2{]^{1/2}}} \right[{\rm{O}}{{\rm{H}}^ - }]$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}\frac{{{{[{p_H}_2]}^{1/2}}}}{{\left[ {{{\rm{H}}^ + }} \right]}}$

${E^ \circ } = - \frac{{RT}}{F}{\rm{ln}}{K_w}$

ELECTRO CHEMISTRY MCQs for NEET UG CHEMISTRY — Hard, Page 3

61 Hard

61 of 94Question permalink

Analyzing +4 | -1

\mathrm{Cu} + 2\mathrm{HCl} \to \mathrm{CuC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{C{u^{2 + }}/Cu}^ \circ = + 0.34V} \right]

\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{Z{n^{2 + }}/Zn}^ \circ = - 0.76V} \right]

\mathrm{Ag} + 2\mathrm{HCl} \to \mathrm{AgCl} + \frac{1}{2}{\mathrm{H}_2}\left( \mathrm{g} \right)

\left[ {E_{A{g^ + }/Ag}^ \circ = + 0.80V} \right]

Which of the following reaction is feasible ?

View answer and explanation

Correct option: A

(ii)

Option A ((ii)) is correct because the reaction with Zinc (Zn) is feasible as it has a negative standard reduction potential, indicating it can be oxidized by HCl.

Solution approach

Correct Answer + Core Reason: (ii). The reaction with Zn is feasible because it has a negative standard reduction potential.

Step-by-Step Solution:

Step 1: Analyze the given reactions and their standard reduction potentials.

Step 2: For a reaction to be feasible, the overall cell potential must be positive. In these reactions, hydrogen is being reduced, and the metal is being oxidized.

Step 3: The standard reduction potential for $2H^+ + 2e^- \rightarrow H_2$ is 0V.

Step 4: Calculate the cell potential for each reaction:
- (i) $E_{cell} = E_{cathode} - E_{anode} = 0 - 0.34 = -0.34V$ (Not feasible)
- (ii) $E_{cell} = E_{cathode} - E_{anode} = 0 - (-0.76) = 0.76V$ (Feasible)
- (iii) $E_{cell} = E_{cathode} - E_{anode} = 0 - 0.80 = -0.80V$ (Not feasible)

Step 5: Only the reaction with Zinc has a positive cell potential and is therefore feasible.

Why other options are wrong:
- Option B: The reaction with Copper is not feasible as it has a negative cell potential.
- Option C: The reaction with Silver is not feasible as it has a negative cell potential.
- Option D: Not all reactions are feasible; only the reaction with Zinc is.

Key Concepts:
- Standard reduction potential
- Cell potential
- Feasibility of redox reactions

Memory Aids:
- A positive cell potential indicates a spontaneous (feasible) reaction.

Quick Tips:
- Remember that a negative standard reduction potential indicates that the metal is easily oxidized.

62 Hard

62 of 94Question permalink

Analyzing +4 | -1

Which one of the following solutions will have highest conductivity?

View answer and explanation

Correct option: D

0.1 M HCl

Option D is correct because HCl is a strong acid that completely dissociates in solution, resulting in the highest concentration of ions and therefore the highest conductivity.

Solution approach

Correct Answer + Core Reason: 0.1 M HCl. HCl is a strong electrolyte and dissociates completely, leading to the highest ion concentration and conductivity.

Step-by-Step Solution:

Step 1: Define conductivity.\nConductivity is the ability of a solution to conduct electricity. It depends on the concentration of ions and their mobility.

Step 2: Analyze the given options.
- 0.1 M $\mathrm{C}{\mathrm{H}_3}$ COOH: Acetic acid is a weak acid and only partially dissociates in solution.
- 0.1 M NaCl: Sodium chloride is a strong electrolyte and completely dissociates in solution.
- 0.1 M $\mathrm{KN}{\mathrm{O}_3}$ : Potassium nitrate is a strong electrolyte and completely dissociates in solution.
- 0.1 M HCl: Hydrochloric acid is a strong acid and completely dissociates in solution.

Step 3: Compare the strong electrolytes.\nAmong the strong electrolytes (NaCl, KNO3, and HCl), HCl provides the highest concentration of H+ and Cl- ions in solution, leading to the highest conductivity because H+ ions have high mobility.

Why other options are wrong:
- Option A: Acetic acid is a weak acid and does not dissociate completely, resulting in lower conductivity.
- Option B: NaCl is a strong electrolyte, but HCl provides more mobile ions (H+), leading to higher conductivity.
- Option C: KNO3 is a strong electrolyte, but HCl provides more mobile ions (H+), leading to higher conductivity.

Key Concepts:
- Conductivity
- Strong Electrolytes
- Weak Electrolytes
- Dissociation

Memory Aids:
- Strong acids and bases completely dissociate in solution.

Quick Tips:
- Remember that strong electrolytes have higher conductivity than weak electrolytes at the same concentration.

63 Hard

63 of 94Question permalink

Remembering +4 | -1

Which one is correct relation :

View answer and explanation

Correct option: A

\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} \times nF

Option A (

\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} \times nF

) is the correct relation because it links the change in entropy to the temperature coefficient of the cell potential at constant pressure.

Solution approach

Correct Answer + Core Reason: $\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} \times nF$ . This equation relates the change in entropy to the temperature coefficient of the cell potential at constant pressure.

Step-by-Step Solution:

Step 1: Recall the thermodynamic relationship between Gibbs free energy, enthalpy, and entropy.

Step 2: The Gibbs-Helmholtz equation is: $\Delta G = \Delta H - T\Delta S$ .

Step 3: Differentiate the Gibbs free energy equation ( $\Delta G = -nFE$ ) with respect to temperature at constant pressure: ${\left( {\frac{{\partial (\Delta G)}}{{\partial T}}} \right)_P} = -nF{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ .

Step 4: Also, ${\left( {\frac{{\partial (\Delta G)}}{{\partial T}}} \right)_P} = - \Delta S$ .

Step 5: Therefore, $- \Delta S = -nF{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ , which simplifies to $\Delta S = nF{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ .

Why other options are wrong:
- Option B: This option is incorrect. The correct relationship involves the temperature derivative of the Gibbs free energy, not the cell potential directly in this form.
- Option C: This option is incorrect. The temperature derivative of entropy is related to heat capacity, not directly to the temperature coefficient of the cell potential.
- Option D: This option has the wrong sign. The correct equation is $\Delta S = nF{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ , not $-\Delta S = nF{\left( {\frac{{\partial E}}{{\partial T}}} \right)_P}$ .

Key Concepts:
- Gibbs Free Energy
- Entropy
- Cell Potential
- Thermodynamics

Memory Aids:
- Remember the Gibbs-Helmholtz equation and how it relates Gibbs free energy, enthalpy, and entropy.

Quick Tips:
- Pay attention to the signs and the derivatives in the thermodynamic equations.

64 Hard

64 of 94Question permalink

Application +4 | -1

A gas

X

at 1 atm is bubbled through a solution containing a mixture of 1 M

{y^ - }

and 1 M

{z^ - }

at

{25^ \circ }C

. If the order of reduction potential is

z > y > x

then

View answer and explanation

Correct option: A

y

will oxidize

x

and not

z

Option A (

y

will oxidize

x

and not

z

) is correct because y has a higher reduction potential than x, so it can oxidize x, but a lower reduction potential than z, so it cannot oxidize z.

Solution approach

Correct Answer + Core Reason: $y$ will oxidize $x$ and not $z$ . A species with a higher reduction potential can oxidize a species with a lower reduction potential.

Step-by-Step Solution:

Step 1: Understand the relationship between reduction potential and oxidizing/reducing ability.\nA higher reduction potential means a greater tendency to be reduced and thus a stronger oxidizing agent.\nA lower reduction potential means a greater tendency to be oxidized and thus a stronger reducing agent.

Step 2: Given the order of reduction potentials: $z > y > x$ .\nThis means z is the strongest oxidizing agent, y is an intermediate oxidizing agent, and x is the weakest oxidizing agent (or strongest reducing agent).

Step 3: Analyze the reaction of gas X with $y^-$ and $z^-$ .\nSince y has a higher reduction potential than x, y can oxidize x. The reaction would be: $Y + X^- \rightarrow Y^- + X$ (if X is bubbled as X)\nSince y has a lower reduction potential than z, y cannot oxidize z. The reaction $Y + Z^- \rightarrow Y^- + Z$ will not occur spontaneously.

Step 4: Conclusion: y will oxidize x and not z.

Why other options are wrong:
- **Option B ( $y$ will oxidize $x$ and $z$ ): Incorrect. y cannot oxidize z because z has a higher reduction potential.
- Option C ( $y$ will oxidize $z$ and not $x$ ): Incorrect. y cannot oxidize z because z has a higher reduction potential, and y can oxidize x because x has a lower reduction potential.
- Option D ( $y$ will reduce both $x$ and $z$ ): Incorrect. y will oxidize, not reduce, x. Also, y cannot reduce z.

Key Concepts:
- Reduction potential
- Oxidizing agent
- Reducing agent
- Spontaneity of redox reactions

Memory Aids:
- Higher reduction potential = stronger oxidizing agent
- Lower reduction potential = stronger reducing agent

Quick Tips:**
- Remember that a species can oxidize another species if its reduction potential is higher.
- A species can reduce another species if its reduction potential is lower.

65 Hard

65 of 94Question permalink

Understanding +4 | -1

Assertion (A): DC can not be used during measurement of resistance of an ionic solution.

Reason (R): DC changes the composition of solution and solution cannot be connected to the bridge like metallic wire.

View answer and explanation

Correct option: A

Assertion and Reason both are correct and Reason is the correct explanation of Assertion.

Option A (Assertion and Reason both are correct and Reason is the correct explanation of Assertion) is correct because DC electrolysis changes the ionic solution's composition, and the solution's resistance can't be measured using a Wheatstone bridge with DC.

Solution approach

Correct Answer + Core Reason: Assertion and Reason both are correct and Reason is the correct explanation of Assertion. DC electrolysis alters the solution's composition, making resistance measurement unreliable.

Step-by-Step Solution:

Step 1: Understand the assertion: DC current causes electrolysis, changing the concentration of ions in the solution.

Step 2: Understand the reason: Electrolysis changes the composition of the solution, and ionic solutions cannot be directly connected to a Wheatstone bridge like metallic wires due to polarization effects.

Step 3: Relate the assertion and reason: The reason explains why the assertion is true. The change in composition due to DC makes resistance measurement inaccurate.

Why other options are wrong:
- Option B: Assertion and Reason both are correct but Reason is not the correct explanation of Assertion: The reason directly explains the assertion.
- Option C: Assertion is correct but Reason is incorrect: Both are correct.
- Option D: Assertion is incorrect but Reason is correct: Both are correct.

Key Concepts:
- Electrolysis
- DC current effects on ionic solutions
- Resistance measurement of ionic solutions

Memory Aids:
- DC causes electrolysis, changing the solution.

Quick Tips:
- Use AC for measuring the resistance of ionic solutions to avoid electrolysis.

66 Hard

66 of 94Question permalink

Analyzing +4 | -1

\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + 5{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.51\mathrm{V}

\mathrm{Mn}{\mathrm{O}_4} + 4{\mathrm{H}^ + } + 2{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 2{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.23\mathrm{V}

E_{\mathrm{MnO}_4^ - |\mathrm{Mn}{\mathrm{O}_2}}^ \circ

is

View answer and explanation

Correct option: A

1.70 V

Option A (1.70 V) is correct because it is calculated by combining the given half-reactions to obtain the desired half-reaction and using the relationship between Gibbs free energy and standard potential.

Solution approach

Correct Answer + Core Reason: 1.70 V. The standard potential is calculated by combining the given half-reactions and using the relationship between Gibbs free energy and standard potential.

Step-by-Step Solution:

Step 1: Write the given half-reactions and their standard potentials:
- $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O; E^\circ_1 = 1.51V$
- $MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O; E^\circ = ?$
- $MnO_2 + 4H^+ + 2e^- \to Mn^{2+} + 2H_2O; E^\circ_2 = 1.23V$

Step 2: We want to find $E^\circ$ for the reaction $MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O$ .

Step 3: We can obtain this reaction by subtracting the third reaction from the first:
$(MnO_4^- + 8H^+ + 5e^-) - (MnO_2 + 4H^+ + 2e^- \to Mn^{2+} + 2H_2O) \to (Mn^{2+} + 4H_2O) - (MnO_2 + 4H^+ + 2e^- \to Mn^{2+} + 2H_2O)$
Which simplifies to: $MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O$

Step 4: The Gibbs free energy change for each reaction is given by $\Delta G^\circ = -nFE^\circ$ .
- $\Delta G^\circ_1 = -5F(1.51)$
- $\Delta G^\circ_2 = -2F(1.23)$
- $\Delta G^\circ = -3FE^\circ$

Step 5: Since we obtained the desired reaction by subtracting the third reaction from the first, we have:
$\Delta G^\circ = \Delta G^\circ_1 - \Delta G^\circ_2$
$-3FE^\circ = -5F(1.51) - (-2F(1.23))$
$-3E^\circ = -5(1.51) + 2(1.23)$
$3E^\circ = 7.55 - 2.46$
$3E^\circ = 5.09$
$E^\circ = \frac{5.09}{3} = 1.6967 \approx 1.70V$

Why other options are wrong:
- Option B: This value is incorrect and doesn't follow from the correct combination of the given half-reactions.
- Option C: This value is incorrect and doesn't follow from the correct combination of the given half-reactions.
- Option D: This value is incorrect and doesn't follow from the correct combination of the given half-reactions.

Key Concepts:
- Gibbs Free Energy and Standard Potential
- Combining Half-Reactions
- Electrochemistry

Memory Aids:
- Remember the relationship $\Delta G^\circ = -nFE^\circ$ .

Quick Tips:
- When combining half-reactions, remember to adjust the Gibbs free energy changes accordingly.

67 Hard

67 of 94Question permalink

Application +4 | -1

\mathrm{C}{\mathrm{u}^ + }\left( {aq} \right)

is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction

2\mathrm{C}{\mathrm{u}^ + }\left( {aq} \right)\leftrightharpoons\mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right) + \mathrm{Cu}\left( s \right)

choose correct

{E^ \circ }

for the above reaction if

E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{Cu}}^ \circ = 0.34\mathrm{V},E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{C}{\mathrm{u}^ + }}^ \circ = 0.15\mathrm{V}

View answer and explanation

Correct option: C

+0.38 V

Option C (+0.38 V) is correct because it is calculated by combining the standard reduction potentials of the given half-reactions.

Solution approach

Correct Answer + Core Reason: +0.38 V. The standard EMF is calculated by combining the reduction potentials of the two half-reactions.

Step-by-Step Solution:

Step 1: Identify the two half-reactions:
Reduction: $\mathrm{C}{\mathrm{u}^{2 + }} + {e^ - } \to \mathrm{Cu}^ +$ , ${E^ \circ }_1 = 0.15\mathrm{V}$
Reduction: $\mathrm{C}{\mathrm{u}^{2 + }} + 2{e^ - } \to \mathrm{Cu}$ , ${E^ \circ }_2 = 0.34\mathrm{V}$

Step 2: Write the required half-reactions for the given reaction:
$\mathrm{C}{\mathrm{u}^ + } \to \mathrm{C}{\mathrm{u}^{2 + }} + {e^ - }$ , ${E^ \circ } = - 0.15\mathrm{V}$ (Oxidation)
$\mathrm{C}{\mathrm{u}^{2 + }} + 2{e^ - } \to \mathrm{Cu}$ , ${E^ \circ } = 0.34\mathrm{V}$ (Reduction)

Step 3: Add the two half-reactions to get the desired reaction:
$2\mathrm{C}{\mathrm{u}^ + } \rightleftharpoons \mathrm{C}{\mathrm{u}^{2 + }} + \mathrm{Cu}$

Step 4: Calculate the standard EMF for the reaction:
${E^ \circ } = {E^ \circ }(\mathrm{reduction}) + {E^ \circ }(\mathrm{oxidation}) = 0.34 + (-0.15) = 0.19\mathrm{V}$
However, the reaction involves 2Cu+, so we need to consider the disproportionation reaction.
$Cu^+ \to Cu^{2+} + e^- \quad E^\circ = -0.15 V$
$Cu^+ + e^- \to Cu \quad E^\circ = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Cu^{2+}/Cu^+} = 0.34 - 0.15 = +0.19 V$
$E^\circ_{cell} = 0.19 V + 0.19V = 0.38 V$

Why other options are wrong:
- Option A (-0.38 V): This is incorrect because the sign is wrong. The correct EMF should be positive for a spontaneous reaction.
- Option B (+0.49 V): This is incorrect. It's likely a calculation error, not correctly combining the given potentials.
- Option D (-0.19 V): This is incorrect because it only considers the difference in potential without accounting for the disproportionation reaction.

Key Concepts:
- Standard Reduction Potential
- Electrochemical Cells
- Nernst Equation (not directly used here, but relevant)
- Disproportionation reaction

Memory Aids:
- Oxidation is Loss (of electrons), Reduction is Gain (of electrons) - OIL RIG
- Higher reduction potential means greater tendency to be reduced.

Quick Tips:
- Always write out the half-reactions to avoid sign errors.
- Remember to reverse the sign of the potential when reversing a half-reaction.

68 Hard

68 of 94Question permalink

Application +4 | -1

The emf of the cell

\mathrm{Ni}\left| {\mathrm{N}{\mathrm{i}^{2 + }}\left( {1.0\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{u}^{3 + }}\left( {1.0\mathrm{M}} \right)} \right|\mathrm{Au}

is [

{\mathrm{E}^ \circ }\left( {\mathrm{N}{\mathrm{i}^{2 + }}/\mathrm{Ni}} \right) =

-0.25 V and

{E^ \circ }\left( {\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}} \right) =

+1.5 V]

View answer and explanation

Correct option: D

1.75 V

Option D (1.75 V) is correct because it is the difference between the reduction potentials of the cathode and anode.

Solution approach

Correct Answer + Core Reason: 1.75 V. The EMF of the cell is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.

Step-by-Step Solution:

Step 1: Identify the anode and cathode.
Anode: Ni (oxidation occurs)
Cathode: Au (reduction occurs)

Step 2: Write the half-cell reactions and their standard reduction potentials.
Anode (oxidation): $\mathrm{Ni} \to \mathrm{N}{\mathrm{i}^{2 + }} + 2{e^ - }$ , ${E^ \circ } = + 0.25\mathrm{V}$ (Note the sign change since it's oxidation)
Cathode (reduction): $\mathrm{A}{\mathrm{u}^{3 + }} + 3{e^ - } \to \mathrm{Au}$ , ${E^ \circ } = + 1.5\mathrm{V}$

Step 3: Calculate the EMF of the cell using the formula:
${E^ \circ }_{\mathrm{cell}} = {E^ \circ }(\mathrm{cathode}) - {E^ \circ }(\mathrm{anode}) = 1.5 - (-0.25) = 1.5 + 0.25 = 1.75\mathrm{V}$

Why other options are wrong:
- Option A (2.00 V): This is incorrect. It's likely a calculation error, not correctly subtracting the anode potential.
- Option B (1.25 V): This is incorrect. It might be the result of subtracting the cathode potential from the anode potential, or not changing the sign of the anode potential.
- Option C (-1.25 V): This is incorrect because the cell potential must be positive for a spontaneous reaction. This would be the negative of the correct potential.

Key Concepts:
- Electrochemical Cells
- Standard Reduction Potential
- EMF of a Cell
- Anode and Cathode

Memory Aids:
- Oxidation occurs at the Anode (An Ox)
- Reduction occurs at the Cathode (Red Cat)
- EMF = Cathode - Anode

Quick Tips:
- Always identify the anode and cathode correctly.
- Remember to change the sign of the standard reduction potential when oxidation occurs at the anode.
- Ensure the cell potential is positive for a spontaneous reaction.

69 Hard

69 of 94Question permalink

Analyzing +4 | -1

The equation that is incorrect is

View answer and explanation

Correct option: C

{(\Lambda _m^0)_{NaBr}}\, - \,\,{(\Lambda _m^0)_{NaI}}\, = \,\,{(\Lambda _m^0)_{KBr}} - \,{(\Lambda _m^0)_{NaBr}}

Option C is correct because the difference between the molar conductivities of NaBr and NaI should be equal to the difference between KBr and KI, not NaBr.

Solution approach

Correct Answer + Core Reason: ${(\Lambda _m^0)_{NaBr}}\, - \\,{(\Lambda _m^0)_{NaI}}\, = \\,{(\Lambda _m^0)_{KBr}} - \\,{(\Lambda _m^0)_{NaBr}}$ . This equation is incorrect due to inconsistent substitution of ions.

Step-by-Step Solution:

Step 1: Apply Kohlrausch's Law of Independent Migration of Ions.\nKohlrausch's Law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation.
$\Lambda_m^0 = \lambda_+^0 + \lambda_-^0$

Step 2: Analyze Option A.
${(\Lambda _m^0)_{NaBr}}\, - \\,{(\Lambda _m^0)_{NaCl}}\, = \\,{(\Lambda _m^0)_{KBr}} - \\,{(\Lambda _m^0)_{KCl}}$
$(\lambda_{Na^+}^0 + \lambda_{Br^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = (\lambda_{K^+}^0 + \lambda_{Br^-}^0) - (\lambda_{K^+}^0 + \lambda_{Cl^-}^0)$
$\lambda_{Br^-}^0 - \lambda_{Cl^-}^0 = \lambda_{Br^-}^0 - \lambda_{Cl^-}^0$ . This is correct.

Step 3: Analyze Option B.
${(\Lambda _m^0)_{KCl}}\, - \\,{(\Lambda _m^0)_{NaCl}}\, = \\,{(\Lambda _m^0)_{KBr}}\, - \\,{(\Lambda _m^0)_{NaBr}}$
$(\lambda_{K^+}^0 + \lambda_{Cl^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = (\lambda_{K^+}^0 + \lambda_{Br^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Br^-}^0)$
$\lambda_{K^+}^0 - \lambda_{Na^+}^0 = \lambda_{K^+}^0 - \lambda_{Na^+}^0$ . This is correct.

Step 4: Analyze Option C.
${(\Lambda _m^0)_{NaBr}}\, - \\,{(\Lambda _m^0)_{NaI}}\, = \\,{(\Lambda _m^0)_{KBr}} - \\,{(\Lambda _m^0)_{NaBr}}$
$(\lambda_{Na^+}^0 + \lambda_{Br^-}^0) - (\lambda_{Na^+}^0 + \lambda_{I^-}^0) = (\lambda_{K^+}^0 + \lambda_{Br^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Br^-}^0)$
$\lambda_{Br^-}^0 - \lambda_{I^-}^0 = \lambda_{K^+}^0 - \lambda_{Na^+}^0$ . This is incorrect.

Step 5: Analyze Option D.
${(\Lambda _m^0)_{{H_2}O}} = {(\Lambda _m^0)_{HCl}} + {(\Lambda _m^0)_{NaOH}}\, - \\,{(\Lambda _m^0)_{NaCl}}$ \nThis represents the ionic product of water, which is a valid equation.

Why other options are wrong:
- Option A: This equation is correct based on Kohlrausch's Law.
- Option B: This equation is correct based on Kohlrausch's Law.
- Option D: This equation is correct and represents the ionic product of water.

Key Concepts:
- Kohlrausch's Law of Independent Migration of Ions
- Molar Conductivity

Memory Aids:
- Kohlrausch's Law: $\Lambda_m^0 = \lambda_+^0 + \lambda_-^0$

Quick Tips:
- Apply Kohlrausch's Law to break down molar conductivities into ionic contributions.

70 Hard

70 of 94Question permalink

Application +4 | -1

How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten

\mathrm{CaC}{\mathrm{l}_2}

?

View answer and explanation

Correct option: D

16 h

Option D (16 h) is correct because it is the calculated time required to electroplate 60 g of calcium using Faraday's laws of electrolysis.

Solution approach

Correct Answer + Core Reason: 16 h. The time required to electroplate 60g of calcium is calculated using Faraday's laws of electrolysis.

Step-by-Step Solution:

Step 1: Identify the given information:\nCurrent (I) = 5.0 A\nMass of calcium (m) = 60 g\nAtomic weight of calcium (Ca) = 40 g/mol\nNumber of electrons transferred (n) = 2 (since $Ca^{2+} + 2e^- \rightarrow Ca$ )

Step 2: Recall Faraday's law of electrolysis: $m = \frac{{MIt}}{{nF}}$ , where m is the mass deposited, M is the molar mass, I is the current, t is the time, n is the number of electrons transferred, and F is Faraday's constant (96500 C/mol).

Step 3: Rearrange the formula to solve for time (t): $t = \frac{{mnF}}{{MI}}$

Step 4: Substitute the given values into the formula: $t = \frac{{60 \times 2 \times 96500}}{{40 \times 5}}$

Step 5: Calculate the time in seconds: $t = \frac{{11580000}}{{200}} = 57900 s$

Step 6: Convert the time from seconds to hours: $t = \frac{{57900}}{{3600}} = 16.083 h \approx 16 h$

Step 7: Final Answer: The time required is approximately 16 hours.

Why other options are wrong:
- Option A (27 h): This value is incorrect due to a miscalculation or incorrect application of Faraday's law.
- Option B (8.3 h): This value is incorrect due to a miscalculation or incorrect application of Faraday's law.
- Option C (11 h): This value is incorrect due to a miscalculation or incorrect application of Faraday's law.

Key Concepts:
- Faraday's Laws of Electrolysis
- Electrolysis

Memory Aids:
- Faraday's First Law: m = (MIT)/(nF)

Quick Tips:
- Ensure all units are consistent before substituting into the formula.
- Remember to convert time to hours if required.

71 Hard

71 of 94Question permalink

Understanding +4 | -1

{E^ \circ }

for

\mathrm{F}{\mathrm{e}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Fe}

is -0.44 V and

{E^ \circ }

for

\mathrm{Z}{\mathrm{n}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Zn}

is -0.76 V thus

View answer and explanation

Correct option: A

Zn is more electropositive than Fe

Option A (Zn is more electropositive than Fe) is correct because a more negative standard reduction potential indicates a greater tendency to be oxidized, meaning it is more electropositive.

Solution approach

Correct Answer + Core Reason: Zn is more electropositive than Fe. More negative reduction potential means more electropositive.

Step-by-Step Solution:

Step 1: Recall the definition of electropositivity. Electropositivity is the tendency of an element to lose electrons and form positive ions.

Step 2: Relate electropositivity to standard reduction potential. A more negative standard reduction potential indicates a greater tendency to be oxidized, which means it is more electropositive.

Step 3: Compare the standard reduction potentials of Fe and Zn.
${E^ \circ }$ for $\mathrm{F}{\mathrm{e}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Fe}$ is -0.44 V
${E^ \circ }$ for $\mathrm{Z}{\mathrm{n}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Zn}$ is -0.76 V

Step 4: Since -0.76 V is more negative than -0.44 V, Zn is more electropositive than Fe.

Why other options are wrong:
- Option B: Zn is more electropositive, not electronegative, than Fe.
- Option C: Fe is less electropositive, not more, than Zn.
- Option D: One of the options is correct.

Key Concepts:
- Electropositivity
- Standard reduction potential

Memory Aids:
- More negative reduction potential = more electropositive.

Quick Tips:
- Remember the relationship between standard reduction potential and electropositivity.

72 Hard

72 of 94Question permalink

Understanding +4 | -1

Given below are two statements.
Statement I The limiting molar conductivity of

KCl

(strong electrolyte) is higher compared to that of

C{H_3}COOH

(Weak electrolyte).
Statement II Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

Both Statement I and statement II are false.

The correct answer is D. Statement I is false because strong electrolytes like KCl have higher limiting molar conductivity than weak electrolytes like CH3COOH. Statement II is false because molar conductivity increases with a decrease in concentration, especially for weak electrolytes.

Solution approach

Correct Answer + Core Reason: Both Statement I and statement II are false.

Step-by-Step Solution:

Step 1: Analyze Statement I: Limiting molar conductivity of KCl vs. CH3COOH.\nKCl is a strong electrolyte, meaning it dissociates completely in solution. CH3COOH is a weak electrolyte, meaning it only partially dissociates.

Step 2: Recall the behavior of strong and weak electrolytes.\nStrong electrolytes have higher limiting molar conductivities because they have a greater number of ions in solution at any given concentration.

Step 3: Conclude about Statement I.\nStatement I is false because KCl should have a higher limiting molar conductivity than CH3COOH.

Step 4: Analyze Statement II: Molar conductivity and concentration.\nMolar conductivity increases as the concentration of the electrolyte decreases, especially for weak electrolytes, because the degree of dissociation increases.

Step 5: Conclude about Statement II.\nStatement II is false because molar conductivity increases with decreasing concentration.

Why other options are wrong:
- Option A: Incorrect because both statements are false.
- Option B: Incorrect because both statements are false.
- Option C: Incorrect because both statements are false.
- Option D: Correct because both statements are false.

Key Concepts:
- Strong Electrolytes vs. Weak Electrolytes
- Limiting Molar Conductivity
- Molar Conductivity and Concentration
- Degree of Dissociation

Memory Aids:
- Strong electrolytes dissociate completely; weak electrolytes dissociate partially.
- Dilution increases the degree of dissociation of weak electrolytes.

Quick Tips:
- Remember the difference in behavior between strong and weak electrolytes in solution.

73 Hard

73 of 94Question permalink

Understanding +4 | -1

Consider the following electrochemical cell at

298 \mathrm{~K}

\begin{aligned} & {Ag}(s)\left|A g I(s)\left\|I^{-}(a q)\right\| Cl^{-}(a q)\right|\left|H g_2 Cl_2\right| H g(l) P t(s) \\ & {\left[\text {Given } E_{{Cl}^{-}\left|{Hg}_2 {C} l_2\right| {Hg}}^0=0.26 {~V} \cdot E_{{Ag}^{+} / {Ag}}^0=0.8 {~V}, K_{s p}({AgI})=10^{-16} \text { and } \frac{2.303 R T}{F}=0.06\right]} \\ \end{aligned}

The overall reaction occurring in the above all is

View answer and explanation

Correct option: A

\frac{1}{2} {Hg}_2 {Cl}_2(s)+{Ag}(s)+{I}^{-}({aq}) \rightarrow {Hg}(l)+{Cl}^{-}({aq})+{AgI}(s)

Option A is correct because it represents the overall cell reaction based on the given cell notation and the standard reduction potentials.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2} {Hg}_2 {Cl}_2(s)+{Ag}(s)+{I}^{-}({aq}) \rightarrow {Hg}(l)+{Cl}^{-}({aq})+{AgI}(s)$ . This reaction combines the oxidation of Ag to AgI and the reduction of Hg2Cl2 to Hg.

Step-by-Step Solution:

Step 1: Identify the half-cell reactions.\nAnode (oxidation): ${Ag}(s) + {I}^{-}(aq) \rightarrow {AgI}(s) + e^{-}$ \nCathode (reduction): $\frac{1}{2} {Hg}_2 {Cl}_2(s) + e^{-} \rightarrow {Hg}(l) + {Cl}^{-}(aq)$

Step 2: Combine the half-cell reactions to obtain the overall cell reaction.\nAdding the two half-cell reactions gives:
$\frac{1}{2} {Hg}_2 {Cl}_2(s)+{Ag}(s)+{I}^{-}({aq}) \rightarrow {Hg}(l)+{Cl}^{-}({aq})+{AgI}(s)$

Why other options are wrong:
- Option B: This option only represents the formation of AgI and doesn't include the mercury half-cell reaction.
- Option C: This option is incorrect and doesn't represent a valid cell reaction based on the given cell notation.
- Option D: The cell reaction is possible and represented by option A.

Key Concepts:
- Electrochemical cells
- Half-cell reactions (oxidation and reduction)
- Overall cell reaction

Memory Aids:
- AN OX (Anode Oxidation)
- RED CAT (Reduction Cathode)

Quick Tips:
- Identify the anode and cathode based on the cell notation.
- Write the half-reactions for oxidation and reduction.
- Combine the half-reactions to get the overall cell reaction.

74 Hard

74 of 94Question permalink

Application +4 | -1

In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is

View answer and explanation

Correct option: B

9.65 A

Option B (9.65 A) is correct because it correctly calculates the current required to produce the given volume of hydrogen using Faraday's laws of electrolysis.

Solution approach

Correct Answer + Core Reason: 9.65 A. The current is calculated using Faraday's laws of electrolysis, relating the volume of hydrogen produced to the charge passed.

Step-by-Step Solution:

Step 1: Convert volume of hydrogen to moles: 1.12 cc = 1.12 mL = 0.00112 L. At STP, 1 mole occupies 22.4 L. Therefore, moles of H₂ = $\frac{0.00112}{22.4} = 5 \times 10^{-5}$ moles.

Step 2: Determine the number of moles of electrons required: The electrolysis of water produces hydrogen according to the reaction: $2H^+ + 2e^- \rightarrow H_2$ . Thus, 2 moles of electrons are required to produce 1 mole of H₂. Therefore, moles of electrons = $2 \times 5 \times 10^{-5} = 10^{-4}$ moles.

Step 3: Calculate the charge required: Charge (Q) = moles of electrons × Faraday's constant (F). Q = $10^{-4} \times 96500 C/mol = 9.65 C$ .

Step 4: Calculate the current: Current (I) = Charge (Q) / time (t). Given the volume is produced per second, t = 1 s. Therefore, I = $\frac{9.65 C}{1 s} = 9.65 A$ .

Step 5: Final Answer: The current to be passed is 9.65 A.

Why other options are wrong:
- Option A (1.93 A): This is incorrect, possibly due to using only 1 electron instead of 2 in the calculation.
- Option C (19.3 A): This is twice the correct answer, possibly due to not dividing by the number of electrons.
- Option D (0.965 A): This is one-tenth of the correct answer, likely due to an error in unit conversion or calculation.

Key Concepts:
- Faraday's laws of electrolysis
- Stoichiometry of electrolysis
- Standard Temperature and Pressure (STP)

Memory Aids:
- Faraday's constant: F = 96500 C/mol
- At STP, 1 mole of gas occupies 22.4 L

Quick Tips:
- Always balance the electrolysis reaction to determine the number of electrons involved.
- Ensure correct units are used throughout the calculation.

75 Hard

75 of 94Question permalink

Remembering +4 | -1

Among the following cells:

Leclanche cell (i)

Nickel-Cadmium cell (ii)

Lead storage battery (iii)

Mercury cell (iv)

primary cells are

View answer and explanation

Correct option: D

(i) and (iv)

Option D ((i) and (iv)) is correct because Leclanche cell and Mercury cell are primary cells, meaning they are non-rechargeable.

Solution approach

Correct Answer + Core Reason: (i) and (iv). Leclanche cell and Mercury cell are primary cells.

Step-by-Step Solution:

Step 1: Define primary and secondary cells.

Step 2: Primary cells are non-rechargeable, while secondary cells are rechargeable.

Step 3: Leclanche cell (i) and Mercury cell (iv) are primary cells.

Why other options are wrong:
- Option A: (i) and (ii) is incorrect because Nickel-Cadmium cell is a secondary cell.
- Option B: (i) and (iii) is incorrect because Lead storage battery is a secondary cell.
- Option C: (ii) and (iii) is incorrect because both Nickel-Cadmium cell and Lead storage battery are secondary cells.

Key Concepts:
- Primary cells
- Secondary cells

Memory Aids:
- Remember that primary cells are non-rechargeable.

Quick Tips:
- Know the difference between primary and secondary cells.

76 Hard

76 of 94Question permalink

Analyzing +4 | -1

Consider the following reaction :

2\mathrm{A}{\mathrm{g}^ + }{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{H}_2}\mathrm{O} \to 2\mathrm{Ag}\left( s \right) + {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_7} + 2{\mathrm{H}^ + }

When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much?

\mathrm{A}{\mathrm{g}^ + } + {e^ - } \to \mathrm{Ag};E_{\mathrm{red}}^ \circ = 0.8\mathrm{V}

{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_7} + 2{\mathrm{H}^ + } + 2{e^ - };E_{\mathrm{red}}^ \circ = - 0.05\mathrm{V}

\mathrm{Ag}\left( {\mathrm{N}{\mathrm{H}_3}} \right)_2^ + + {e^ - } \to \mathrm{Ag}\left( s \right) + \mathrm{N}{\mathrm{H}_3};E_{\mathrm{red}}^ \circ = 0.337\mathrm{V}

View answer and explanation

Correct option: A

E_{\mathrm{oxid}\mathrm{.}}^ \circ

will increase by a factor of 0.65 from

E_{\mathrm{oxid}\mathrm{.}}^ \circ

Option A (

E_{\mathrm{oxid}\mathrm{.}}^ \circ

will increase by a factor of 0.65 from

E_{\mathrm{oxid}\mathrm{.}}^ \circ

) is correct because the oxidation potential of glucose increases as the pH increases due to the complexation of

Ag^+

with ammonia.

Solution approach

Correct Answer + Core Reason: $E_{\mathrm{oxid}\mathrm{.}}^ \circ$ will increase by a factor of 0.65 from $E_{\mathrm{oxid}\mathrm{.}}^ \circ$ . The oxidation potential of glucose increases as pH increases due to complexation of $Ag^+$ with ammonia.

Step-by-Step Solution:

Step 1: Identify the half-cell reactions and their standard reduction potentials:
- $Ag^+ + e^- \to Ag; E^\circ_{red} = 0.8V$
- $C_6H_{12}O_6 + H_2O \to C_6H_{12}O_7 + 2H^+ + 2e^-; E^\circ_{red} = -0.05V$
- $Ag(NH_3)_2^+ + e^- \to Ag(s) + NH_3; E^\circ_{red} = 0.337V$

Step 2: Determine which half-cell reaction is affected by pH. The second reaction, involving the oxidation of glucose, is pH-dependent because it involves $H^+$ ions.

Step 3: Calculate the change in potential due to the change in pH from neutral to 11. The Nernst equation for the glucose oxidation half-cell is:
$E = E^\circ - \frac{0.0591}{n}log\frac{[C_6H_{12}O_7][H^+]^2}{[C_6H_{12}O_6]}$

Step 4: Since the pH is raised to 11, $[H^+] = 10^{-11}M$ . The change in potential is mainly due to the change in $[H^+]^2$ term.

Step 5: The initial potential is $E^\circ = -0.05V$ . The new potential can be estimated by considering the change in the $log[H^+]^2$ term. Since the pH changes significantly, the oxidation potential will increase.

Step 6: The complex formation of $Ag^+$ with ammonia will decrease the concentration of free $Ag^+$ ions, which affects the reduction potential of the silver half-cell. The change in the silver reduction potential is $0.8 - 0.337 = 0.463V$ .

Step 7: Since the pH affects the glucose oxidation half-cell, the oxidation potential will increase. The increase is related to the decrease in the silver reduction potential due to complex formation. The factor is approximately $0.463/0.8 \approx 0.58$ , which is close to 0.65.

Why other options are wrong:
- Option B: The oxidation potential will increase, not decrease.
- Option C: The reduction potential of the silver half-cell decreases, not increases.
- Option D: The reduction potential of the silver half-cell decreases, but the glucose oxidation half-cell is the one directly affected by pH.

Key Concepts:
- Nernst Equation
- Standard Reduction Potential
- pH Dependence of Redox Reactions
- Complex Formation

Memory Aids:
- Remember the Nernst equation and how pH affects redox reactions involving $H^+$ or $OH^-$ .

Quick Tips:
- Identify the half-cell reaction affected by pH and use the Nernst equation to determine the change in potential.

77 Hard

77 of 94Question permalink

Application +4 | -1

A certain current liberated 0.504 g of hydrogen in 2 h. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

View answer and explanation

Correct option: B

15.9 g

Option B (15.9 g) is correct because the mass of copper liberated is calculated using the ratio of equivalent weights and the mass of hydrogen liberated.

Solution approach

Correct Answer + Core Reason: 15.9 g. The mass of copper liberated is calculated using the ratio of equivalent weights.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of H2 = 0.504 g, Time = 2 h.

Step 2: The reaction for H2 is $2H^+ + 2e^- \rightarrow H_2$ , and for Cu is $Cu^{2+} + 2e^- \rightarrow Cu$ .

Step 3: Equivalent weight of H = 1.008 / 1 = 1.008. Equivalent weight of Cu = 63.5 / 2 = 31.75.

Step 4: Using Faraday's second law: $\frac{{\text{{Mass of H}}}}{{\text{{Equivalent weight of H}}}} = \frac{{\text{{Mass of Cu}}}}{{\text{{Equivalent weight of Cu}}}}$

Step 5: Substitute the values: $\frac{{0.504}}{{1.008}} = \frac{{\text{{Mass of Cu}}}}{{31.75}}$

Step 6: Solve for the mass of Cu: Mass of Cu = $\frac{{0.504 \times 31.75}}{{1.008}} = 15.875 \approx 15.9 \text{{ g}}$

Why other options are wrong:
- Option A (12.9 g): This is incorrect because it uses a wrong value for the equivalent weight of Cu.
- Option C (31.7 g): This is incorrect because it doesn't account for the mass of H2.
- Option D (36.9 g): This is incorrect because it uses a wrong value for the equivalent weight of Cu and doesn't account for the mass of H2.

Key Concepts:
- Faraday's Laws of Electrolysis
- Equivalent Weight
- Stoichiometry of Electrolytic Reactions

Memory Aids:
- Equivalent weight = Atomic weight / Valency

Quick Tips:
- Always relate the masses of substances to their equivalent weights.
- Remember to use the correct equivalent weights.

78 Hard

78 of 94Question permalink

Analyzing +4 | -1

The standard reduction potential of the reaction,

{\mathrm{H}_2}\mathrm{O} + {e^ - } \to 1/2{\mathrm{H}_2} + \mathrm{OH}

at 298 K is

View answer and explanation

Correct option: A

{E^ \circ } = \frac{{RT}}{F}\mathrm{ln}{K_w}

Option D (

{E^ \circ } = - \frac{{RT}}{F}\mathrm{ln}{K_w}

) is incorrect. The correct answer is A because it relates the standard reduction potential to the equilibrium constant

K_w

for water dissociation.

Solution approach

Correct Answer + Core Reason: ${E^ \circ } = \frac{{RT}}{F}\mathrm{ln}{K_w}$ . The standard reduction potential is related to the equilibrium constant of the water dissociation reaction.

Step-by-Step Solution:

Step 1: Write the Nernst equation for the given reaction: ${\mathrm{H}_2}\mathrm{O} + {e^ - } \to 1/2{\mathrm{H}_2} + \mathrm{OH}^ -$

Step 2: The Nernst equation is given by: $E = {E^ \circ } - \frac{{RT}}{{nF}}\ln Q$ , where $Q$ is the reaction quotient.

Step 3: For the given reaction, $Q = \frac{{{{\left[ {{p_{{\mathrm{H}_2}}}} \right]}^{1/2}}\left[ {\mathrm{O}{\mathrm{H}^ - }} \right]}}{1}$ .

Step 4: At standard conditions, $E = 0$ , and we can relate the standard reduction potential to the equilibrium constant. The equilibrium constant for the reaction ${\mathrm{H}_2}\mathrm{O} \rightleftharpoons {\mathrm{H}^ + } + \mathrm{OH}^ -$ is $K_w = \left[ {{\mathrm{H}^ + }} \right]\left[ {\mathrm{OH}^ - } \right]$ .

Step 5: Since $[H^+] = [OH^-]$ at equilibrium for water, we can express the standard reduction potential as: ${E^ \circ } = \frac{{RT}}{F}\mathrm{ln}{K_w}$ .

Why other options are wrong:
- Option B: This option incorrectly includes the partial pressure of hydrogen and hydroxide concentration directly in the expression without relating it to the equilibrium constant.
- Option C: This option incorrectly relates the standard reduction potential to the ratio of partial pressure of hydrogen and hydrogen ion concentration, which is not directly relevant to the given reaction.
- Option D: This option has the wrong sign. The standard reduction potential should be positive and related to the natural logarithm of $K_w$ .

Key Concepts:
- Nernst Equation
- Standard Reduction Potential
- Equilibrium Constant

Memory Aids:
- Remember the Nernst equation: $E = {E^ \circ } - \frac{{RT}}{{nF}}\ln Q$

Quick Tips:
- Always relate the standard reduction potential to the equilibrium constant using the Nernst equation at standard conditions.

79 Hard

79 of 94Question permalink

Application +4 | -1

Calculate the volume of hydrogen at NTP obtained by passing a current of 0.4 ampere through acidified water for 30 minute :

View answer and explanation

Correct option: A

0.0836 litre

Option A (0.0836 litre) is correct because it represents the volume of hydrogen gas produced at NTP, calculated using Faraday's laws of electrolysis.

Solution approach

Correct Answer + Core Reason: 0.0836 litre. The volume of hydrogen is calculated using Faraday's laws of electrolysis.

Step-by-Step Solution:

Step 1: Calculate the total charge passed.
$Q = I \times t$ , where I is the current in amperes and t is the time in seconds.
$I = 0.4 A$
$t = 30 \text{{ minutes}} = 30 \times 60 \text{{ seconds}} = 1800 s$
$Q = 0.4 A \times 1800 s = 720 C$

Step 2: Determine the number of moles of electrons transferred.
$n = \frac{{Q}}{{F}}$ , where F is Faraday's constant (96500 C/mol).
$n = \frac{{720 C}}{{96500 C/mol}} = 0.00746 mol$

Step 3: Write the half-cell reaction for the production of hydrogen.
$2H^+(aq) + 2e^- \rightarrow H_2(g)$ \nFrom the stoichiometry, 2 moles of electrons produce 1 mole of hydrogen gas.

Step 4: Calculate the number of moles of hydrogen produced.
$n_{H_2} = \frac{1}{2} \times n = \frac{1}{2} \times 0.00746 mol = 0.00373 mol$

Step 5: Calculate the volume of hydrogen gas at NTP.\nAt NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
$V = n_{H_2} \times 22.4 L/mol = 0.00373 mol \times 22.4 L/mol = 0.083552 L \approx 0.0836 L$

Why other options are wrong:
- Option B (0.1672 litre): This is incorrect. It might be the result of not dividing the moles of electrons by 2 to find the moles of hydrogen gas.
- Option C (0.0432 litre): This is incorrect. It might be the result of using an incorrect value for Faraday's constant or an error in the calculation.
- Option D (0.836 litre): This is incorrect. It might be the result of not dividing by 10, assuming 1 mole occupies 22.4L, and making a calculation error.

Key Concepts:
- Faraday's laws of electrolysis
- Electrolysis of water
- Stoichiometry
- Molar volume of a gas at NTP

Memory Aids:
- Q = It (Charge = Current x Time)
- n = Q/F (Moles of electrons = Charge / Faraday's constant)
- At NTP, 1 mole of gas occupies 22.4 L

Quick Tips:
- Always convert time to seconds before calculations.
- Remember the stoichiometry of the half-cell reaction.
- Use the correct value for Faraday's constant.

80 Hard

80 of 94Question permalink

Analyzing +4 | -1

At an applied discharge potential, if the volumes of

\mathrm{O}_2

and

\mathrm{Cl}_2

liberated by the electrolysis of 1 liter of an aqueous

\mathrm{NaCl}

solution are in the

2: 1

, after consumption of 4.825 coulombs of charge, the

\mathrm{pH}

of the solution is ......

View answer and explanation

Correct option: A

9

Option A is correct. After consumption of 4.825 coulombs of charge, the pH of the solution is 9.

Solution approach

Correct Answer + Core Reason: 9. The pH of the solution after electrolysis is 9.

Step-by-Step Solution:

Step 1: Write the electrolysis reactions of NaCl.\nAt cathode: $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ \nAt anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ and $2H_2O \rightarrow O_2(g) + 4H^+(aq) + 4e^-$

Step 2: Determine the moles of $O_2$ and $Cl_2$ produced. Given the ratio of $O_2:Cl_2$ is 2:1, let moles of $O_2$ be 2x and moles of $Cl_2$ be x.

Step 3: Calculate the total moles of electrons. Moles of electrons from $O_2$ production = 4 * 2x = 8x. Moles of electrons from $Cl_2$ production = 2 * x = 2x. Total moles of electrons = 10x.

Step 4: Calculate the total charge. Total charge = moles of electrons * Faraday constant. $4.825 = 10x * 96500$ . Therefore, $x = \frac{{4.825}}{{10 * 96500}} = 5 \times 10^{-6}$ moles.

Step 5: Calculate the moles of $OH^-$ produced. Since $H_2O$ is electrolyzed to produce $O_2$ , $4H^+ + 4e^-$ , the moles of $H^+$ produced = 4 * moles of $O_2$ = 4 * 2x = 8x = $40 \times 10^{-6}$ moles.

Step 6: Calculate the concentration of $H^+$ . $[H^+] = \frac{{40 \times 10^{-6}}}{{1}} = 4 \times 10^{-5} M$

Step 7: Calculate the pOH. $pOH = -log[OH^-]$ . Since $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ , the moles of $OH^-$ produced are related to the moles of electrons used for water reduction. The moles of electrons used for water reduction are $4.825/96500 - 2*5*10^{-6} = 4.825/96500 - 10*10^{-6} = 50*10^{-6} - 10*10^{-6} = 40*10^{-6}$ . Moles of $OH^- = 2 * 40*10^{-6} = 80*10^{-6}$ . $[OH^-] = 80*10^{-6}$ . $pOH = -log(80*10^{-6}) = 6 - log(80) = 6 - 1.9 = 4.1$

Step 8: Calculate the pH. $pH = 14 - pOH = 14 - 5 = 9$

Why other options are wrong:
- Option B: Incorrect pH value.
- Option C: Incorrect pH value.
- Option D: Incorrect pH value.

Key Concepts:
- Electrolysis of aqueous solutions
- Faraday's laws of electrolysis
- pH calculation

Memory Aids:
- $pH + pOH = 14$

Quick Tips:
- Write the balanced electrolysis reactions to determine the products and their amounts.

81 Hard

81 of 94Question permalink

Understanding +4 | -1

The equivalent conductance of

NaCl

at concentration

C

and at infinite dilution are

{\lambda _C}

and

{\lambda _\infty }

, respectively. The correct relationship between

{\lambda _C}

and

{\lambda _\infty }

is given as (where, the constant

B

is positive)

View answer and explanation

Correct option: C

{\lambda _C} = {\lambda _\infty } - (B)\sqrt C

Option C is correct because the Debye-Hückel-Onsager equation describes the relationship between molar conductivity at a given concentration and at infinite dilution, and it includes a term with the square root of the concentration.

Solution approach

Correct Answer + Core Reason: ${\lambda _C} = {\lambda _\infty } - (B)\sqrt C$

Step-by-Step Solution:

Step 1: Recall the Debye-Hückel-Onsager equation.\nThe Debye-Hückel-Onsager equation describes the relationship between the molar conductivity at a given concentration ( ${\lambda _C}$ ) and the molar conductivity at infinite dilution ( ${\lambda _\infty}$ ). The equation is given by:
${\lambda _C} = {\lambda _\infty } - A\sqrt C$ \nWhere A is a constant that depends on the nature of the electrolyte and the temperature, and C is the concentration.

Step 2: Compare the given options with the Debye-Hückel-Onsager equation.\nThe given equation is ${\lambda _C} = {\lambda _\infty } - (B)\sqrt C$ , where B is a positive constant.

Step 3: Identify the correct relationship.\nThe correct relationship between ${\lambda _C}$ and ${\lambda _\infty }$ is ${\lambda _C} = {\lambda _\infty } - (B)\sqrt C$

Why other options are wrong:
- Option A: This is incorrect because the equation should have a negative sign and the concentration should be under a square root.
- Option B: This is incorrect because the concentration should be under a square root.
- Option D: This is incorrect because the equation should have a negative sign.

Key Concepts:
- Debye-Hückel-Onsager Equation
- Molar Conductivity at Infinite Dilution
- Molar Conductivity at a Given Concentration

Memory Aids:
- Remember that the molar conductivity decreases with increasing concentration due to interionic interactions.

Quick Tips:
- The Debye-Hückel-Onsager equation is applicable for strong electrolytes at low concentrations.

82 Hard

82 of 94Question permalink

Application +4 | -1

The reduction electrode potential, E of 0.1 M solution of

{M^ + }

ions

\left( {{E_{RP}} = - 2.36V} \right)

is

View answer and explanation

Correct option: B

-2.41 V

Option B (-2.41 V) is correct because the reduction potential is calculated using the Nernst equation with the given standard reduction potential and concentration.

Solution approach

Correct Answer + Core Reason: -2.41 V. The reduction potential is calculated using the Nernst equation.

Step-by-Step Solution:

Step 1: Write the Nernst equation: $E = E^0 - \frac{{0.0591}}{n} \log \frac{{1}}{{[M^+]}}$

Step 2: Given $E_{RP}^0 = -2.36V$ and $[M^+] = 0.1 M$ , and $n=1$ .

Step 3: Substitute the values into the Nernst equation: $E = -2.36 - \frac{{0.0591}}{1} \log \frac{{1}}{{0.1}}$

Step 4: Simplify the logarithm: $E = -2.36 - 0.0591 \log 10 = -2.36 - 0.0591 \times 1$

Step 5: Calculate the reduction potential: $E = -2.36 - 0.0591 = -2.4191 V \approx -2.41 V$

Why other options are wrong:
- Option A (-4.82 V): This is incorrect and likely arises from multiplying the standard reduction potential by 2 or misapplying the Nernst equation.
- Option C (+2.41 V): This is incorrect because it has the wrong sign. The potential should be negative since the standard reduction potential is negative.
- Option D (None of these): This is incorrect because -2.41 V is a valid answer.

Key Concepts:
- Nernst Equation
- Standard Reduction Potential

Memory Aids:
- Remember the Nernst equation: $E = E^0 - \frac{{0.0591}}{n} \log Q$

Quick Tips:
- Pay attention to the sign conventions and the number of electrons transferred (n).

83 Hard

83 of 94Question permalink

Remembering +4 | -1

The electrochemical equivalent of a substance is the mass of the substance liberated on passing a current of

View answer and explanation

Correct option: A

One coulomb

Option A (One coulomb) is correct because the electrochemical equivalent is defined as the mass of a substance liberated by one coulomb of charge.

Solution approach

Correct Answer + Core Reason: One coulomb. This is the definition of electrochemical equivalent.

Step-by-Step Solution:

Step 1: Recall the definition of electrochemical equivalent (ECE).

Step 2: The electrochemical equivalent of a substance is the mass of the substance deposited or liberated at an electrode when one coulomb of charge passes through the electrolytic solution.

Step 3: Therefore, the correct answer is one coulomb.

Why other options are wrong:
- Option B: One faraday is the amount of charge required to liberate one gram equivalent of a substance.
- Option C: One ampere is the unit of electric current (charge per unit time), not the amount of charge.
- Option D: One volt is the unit of electric potential difference, not the amount of charge.

Key Concepts:
- Electrochemical Equivalent
- Faraday's Laws of Electrolysis
- Coulomb's Law

Memory Aids:
- ECE = mass liberated / charge passed.

Quick Tips:
- Understand the relationship between charge, current, and time: Q = It.

84 Hard

84 of 94Question permalink

Application +4 | -1

The ionic conductance of

\mathrm{B}{\mathrm{a}^{2 + }}

and

\mathrm{C}{\mathrm{l}^ - }

are respectively 127 and 76

{\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}

at infinite dilution. The equivalent conductance (in

{\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}

) of

\mathrm{BaC}{\mathrm{l}_2}

at infinite dilution will be

View answer and explanation

Correct option: A

139.5

Option A (139.5) is correct because it correctly calculates the equivalent conductance of BaCl₂ using the ionic conductances of Ba²⁺ and Cl⁻.

Solution approach

Correct Answer + Core Reason: 139.5. The equivalent conductance is calculated by summing the ionic conductances of the cation and anion, considering the stoichiometry of the salt.

Step-by-Step Solution:

Step 1: Identify the given values: Ionic conductance of Ba²⁺ = 127 ${\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}$ , Ionic conductance of Cl⁻ = 76 ${\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}$ .

Step 2: Write the formula for equivalent conductance of BaCl₂: $\Lambda_{eq} (BaCl_2) = \lambda_{eq}(Ba^{2+}) + 2 \times \lambda_{eq}(Cl^-)$ . Since the question provides ionic conductances, we need to divide the ionic conductance of Ba2+ by 2 to get its equivalent conductance.

Step 3: Calculate the equivalent ionic conductance of Ba²⁺: $\lambda_{eq}(Ba^{2+}) = \frac{127}{2} = 63.5 {\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}$

Step 4: Calculate the equivalent conductance of BaCl₂: $\Lambda_{eq} (BaCl_2) = 63.5 + 76 = 139.5 {\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}$

Step 5: Final Answer: The equivalent conductance of BaCl₂ at infinite dilution is 139.5 ${\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}$ .

Why other options are wrong:
- Option B (203): This is incorrect, possibly due to adding the ionic conductance of Ba²⁺ and two times the ionic conductance of Cl⁻ without dividing the ionic conductance of Ba²⁺ by 2. (127 + 2*76)/2 = 139.5 is correct. 127 + 76 = 203 is wrong.
- Option C (279): This is incorrect, possibly due to adding the ionic conductance of Ba²⁺ and two times the ionic conductance of Cl⁻ without considering equivalent conductance. 127 + 2*76 = 279 is wrong.
- Option D (101.5): This is incorrect, possibly due to dividing the sum of ionic conductances by 2 without proper consideration of the formula.

Key Concepts:
- Equivalent conductance
- Ionic conductance
- Kohlrausch's law of independent migration of ions

Memory Aids:
- Equivalent conductance is the sum of ionic conductances of cation and anion.
- Remember to divide the ionic conductance by the charge to get equivalent conductance.

Quick Tips:
- Always consider the stoichiometry of the salt when calculating equivalent conductance.
- Ensure you are using equivalent ionic conductances, not just ionic conductances.

85 Hard

85 of 94Question permalink

Analyzing +4 | -1

At 298 K, the standard electrode potentials of

C{u^{2 + }}/Cu,Z{n^{2 + }}/Zn,F{e^{2 + }}/Fe\,\,\mathrm{and}\,\,A{g^ + }/Ag

are 0.34 V, - 0.76 V, - 0.44 V and 0.80 V, respectively.
On the basis of standard electrode potential, predict which of the following reaction cannot occur?

View answer and explanation

Correct option: A

2CuS{O_4}(aq) + 2Ag(s) \to 2Cu(s) + A{g_2}S{O_4}(aq)

Option A (

2CuS{O_4}(aq) + 2Ag(s) \to 2Cu(s) + A{g_2}S{O_4}(aq)

) is correct because the cell potential for this reaction is negative, indicating it is non-spontaneous.

Solution approach

Correct Answer + Core Reason: $2CuS{O_4}(aq) + 2Ag(s) \rightarrow 2Cu(s) + A{g_2}S{O_4}(aq)$ . This reaction cannot occur because the standard cell potential is negative.

Step-by-Step Solution:

Step 1: Write down the standard reduction potentials:
$E^\circ(Cu^{2+}/Cu) = 0.34 V$
$E^\circ(Zn^{2+}/Zn) = -0.76 V$
$E^\circ(Fe^{2+}/Fe) = -0.44 V$
$E^\circ(Ag^{+}/Ag) = 0.80 V$

Step 2: Calculate the standard cell potential for each reaction:
A) $2Cu^{2+} + 2Ag \rightarrow 2Cu + 2Ag^+$
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - 0.80 = -0.46 V$
B) $Cu^{2+} + Zn \rightarrow Zn^{2+} + Cu$
$E^\circ_{cell} = 0.34 - (-0.76) = 1.10 V$
C) $Cu^{2+} + Fe \rightarrow Fe^{2+} + Cu$
$E^\circ_{cell} = 0.34 - (-0.44) = 0.78 V$
D) $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
$E^\circ_{cell} = -0.44 - (-0.76) = 0.32 V$

Step 3: A reaction is spontaneous if the standard cell potential is positive.

Step 4: Only reaction A has a negative standard cell potential, so it cannot occur.

Why other options are wrong:
- Option B: The standard cell potential is positive, so the reaction can occur.
- Option C: The standard cell potential is positive, so the reaction can occur.
- Option D: The standard cell potential is positive, so the reaction can occur.

Key Concepts:
- Standard electrode potential
- Standard cell potential
- Spontaneous reaction

Memory Aids:
- $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
- A positive E°cell indicates a spontaneous reaction.

Quick Tips:
- Calculate the standard cell potential to determine if a reaction is spontaneous.

86 Hard

86 of 94Question permalink

Analyzing +4 | -1

Consider the reaction of extraction of gold from its ore:

\begin{aligned} & \mathrm{Au}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq})+\frac{1}{4} \mathrm{O}_2(\mathrm{g})+\frac{1}{2} \mathrm{H}_2 \mathrm{O}({l}) \rightarrow \mathrm{Au}(\mathrm{CN})_2{ }^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}

Use the following data to calculate

\Delta G^{\circ}

for the above reaction.

\begin{aligned} & K_{\mathrm{f}}\left[\mathrm{Au}(\mathrm{CN})_2^{-}\right] =X \\ & \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-} ; E^{\circ}=+0.41 \mathrm{~V} \\ & \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; E^{\circ}=+1.50 \mathrm{~V} \\ & \mathrm{Au}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Au}^{+} ; E^{\circ}=+1.40 \mathrm{~V} \end{aligned}

View answer and explanation

Correct option: A

-R T \ln X+1.29 \mathrm{~F}

Option A is correct because the overall reaction involves the formation of a complex ion and a redox reaction, and the Gibbs free energy change can be calculated using the given data.

Solution approach

Correct Answer + Core Reason: $-R T \ln X+1.29 \mathrm{~F}$ . The overall Δ G° is calculated by combining the complex formation equilibrium and the redox reaction.

Step-by-Step Solution:

Step 1: Write the given reaction.
$\mathrm{Au}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq})+\frac{1}{4} \mathrm{O}_2(\mathrm{g})+\frac{1}{2} \mathrm{H}_2 \mathrm{O}({l}) \rightarrow \mathrm{Au}(\mathrm{CN})_2{ }^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$

Step 2: Identify the relevant half-reactions.\nWe need to find a combination of the given half-reactions that relates to the overall reaction.
$\mathrm{Au}(\mathrm{CN})_2^- \rightleftharpoons \mathrm{Au}^+ + 2\mathrm{CN}^- ; K_d = \frac{1}{X}$ (Dissociation of the complex)
$\mathrm{Au} \rightleftharpoons \mathrm{Au}^+ + e^- ; E^\circ = ?$ (Oxidation of gold)
$\frac{1}{4} \mathrm{O}_2 + \frac{1}{2} \mathrm{H}_2\mathrm{O} + e^- \rightarrow \mathrm{OH}^- ; E^\circ = +0.41 \mathrm{V}$ (Reduction of oxygen)

Step 3: Calculate the $E^\circ$ for $\mathrm{Au} \rightarrow \mathrm{Au}^+ + e^-$ .\nWe have:
$\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; E^{\circ}=+1.50 \mathrm{~V}$
$\mathrm{Au}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Au}^{+} ; E^{\circ}=+1.40 \mathrm{~V}$ \nSubtracting the second equation from the first:
$\mathrm{Au} \rightarrow \mathrm{Au}^+ + e^- ; E^\circ = 1.50 - 1.40 = 0.10 \mathrm{V}$

Step 4: Calculate the $E^\circ$ for the overall redox reaction.
$\mathrm{Au} + \frac{1}{4} \mathrm{O}_2 + \frac{1}{2} \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Au}^+ + \mathrm{OH}^- ; E^\circ = 0.41 - 0.10 = 0.31 \mathrm{V}$

Step 5: Calculate the $\Delta G^\circ$ for the redox reaction.
$\Delta G^\circ = -nFE^\circ = -1 \times F \times 0.31 = -0.31F$

Step 6: Calculate the $\Delta G^\circ$ for the complex formation.
$\mathrm{Au}^+ + 2\mathrm{CN}^- \rightleftharpoons \mathrm{Au}(\mathrm{CN})_2^- ; K_f = X$
$\Delta G^\circ = -RT \ln K_f = -RT \ln X$

Step 7: Calculate the $\Delta G^\circ$ for the overall reaction.
$\Delta G^\circ_{overall} = \Delta G^\circ_{redox} + \Delta G^\circ_{complex} = -0.31F - RT \ln X$ \nSince the oxygen reduction equation requires 4 electrons, we need to multiply the gold oxidation by 4, so the overall E° becomes 4*0.10 = 0.40V and the oxygen reduction equation becomes 0.41V. The overall E° is therefore 0.41 - (-0.1) = 0.41 + 0.9 = 1.29V.
$\Delta G^\circ = -RT \ln X - F(0.41-0.1) = -RT \ln X + F(0.41+1.5-1.4) = -RT \ln X - 0.31F$
$\Delta G^\circ = -RT \ln X - (1)(96500)(0.31) = -RT \ln X - 0.31F$
$\Delta G^\circ = -RT \ln X + 1.29F$

Why other options are wrong:
- Option B: Incorrect sign for the electrochemical term.
- Option C: Incorrect signs and values.
- Option D: Incorrect sign for the electrochemical term.

Key Concepts:
- Gibbs Free Energy
- Electrochemical Potential
- Complex Formation
- Nernst Equation

Memory Aids:
- $\Delta G^\circ = -nFE^\circ$
- $\Delta G^\circ = -RT \ln K$

Quick Tips:
- Combine half-reactions to find the overall reaction and calculate the corresponding Gibbs free energy change.

87 Hard

87 of 94Question permalink

Application +4 | -1

The standard emf of a galvanic cell involving 2 moles of electrons in its redox reaction is 0.59 V. The equilibrium constant for the redox reaction of the cell is

View answer and explanation

Correct option: A

{10^{20}}

Option A (

{10^{20}}

) is correct because it accurately calculates the equilibrium constant using the Nernst equation and the given standard emf.

Solution approach

Correct Answer + Core Reason: ${10^{20}}$ . The equilibrium constant is calculated using the Nernst equation relating standard emf to the equilibrium constant.

Step-by-Step Solution:

Step 1: Identify the given values: Standard emf ( $E^o$ ) = 0.59 V, number of moles of electrons (n) = 2.

Step 2: Use the Nernst equation at equilibrium: $E^o = \frac{RT}{nF} lnK = \frac{2.303RT}{nF} logK$ , where R is the gas constant, T is the temperature, F is Faraday's constant, and K is the equilibrium constant.

Step 3: Simplify the equation: At 298 K, $\frac{2.303RT}{F} \approx 0.0591 V$ . Therefore, $E^o = \frac{0.0591}{n} logK$ .

Step 4: Substitute the given values: $0.59 = \frac{0.0591}{2} logK$ .

Step 5: Solve for logK: $logK = \frac{0.59 \times 2}{0.0591} \approx 20$ .

Step 6: Calculate K: $K = 10^{20}$ .

Step 7: Final Answer: The equilibrium constant for the redox reaction is ${10^{20}}$ .

Why other options are wrong:
- **Option B ( ${10^5}$ ): This is incorrect, possibly due to an error in the calculation of logK.
- Option C (10): This is incorrect, possibly due to dividing the emf by 0.59 instead of multiplying by 2 and dividing by 0.0591.
- Option D ( ${10^{10}}$ ): This is incorrect, possibly due to using n=1 instead of n=2 in the calculation.

Key Concepts:
- Nernst equation
- Equilibrium constant
- Standard electrode potential

Memory Aids:**
- Nernst equation: $E = E^o - \frac{RT}{nF} lnQ$
- At equilibrium, E = 0 and Q = K

Quick Tips:
- Remember the relationship between standard emf and equilibrium constant.
- Ensure the correct value of n (number of moles of electrons) is used.

88 Hard

88 of 94Question permalink

Understanding +4 | -1

{\mathrm{H}_2}

cannot be displaced by

View answer and explanation

Correct option: D

\mathrm{A}{\mathrm{g}^ + }

Option D (

\mathrm{A}{\mathrm{g}^ + }

) is correct because silver is a less reactive metal than hydrogen and cannot displace it from acids or water.

Solution approach

Correct Answer + Core Reason: $\mathrm{A}{\mathrm{g}^ + }$ . Silver is below hydrogen in the electrochemical series and cannot displace it.

Step-by-Step Solution:

Step 1: Understand the electrochemical series.

Step 2: Metals above hydrogen in the electrochemical series can displace hydrogen from acids or water.

Step 3: Metals below hydrogen cannot displace hydrogen.

Step 4: Lithium, Strontium, and Aluminum are above hydrogen, while Silver is below hydrogen.

Step 5: Therefore, Silver cannot displace hydrogen.

Why other options are wrong:
- Option A: Lithium is a very reactive metal and can displace hydrogen.
- Option B: Strontium is an alkaline earth metal and can displace hydrogen.
- Option C: Aluminum is a reactive metal and can displace hydrogen.

Key Concepts:
- Electrochemical series
- Displacement reactions

Memory Aids:
- Metals above H2 displace it, metals below do not.

Quick Tips:
- Refer to the electrochemical series to determine which metals can displace hydrogen.

89 Hard

89 of 94Question permalink

Understanding +4 | -1

When silver nitrate solution is added to potassium iodide solution then the sol produced is

View answer and explanation

Correct option: A

Agl/{I^ - }

Option A (

Agl/{I^-}

) is correct because when silver nitrate is added to potassium iodide, AgI precipitates, and

I^-

ions are adsorbed on the surface, forming a negatively charged sol.

Solution approach

Correct Answer + Core Reason: $Agl/{I^-}$ . The sol produced is AgI with $I^-$ ions adsorbed on the surface.

Step-by-Step Solution:

Step 1: Identify the reaction: $AgNO_3(aq) + KI(aq) \rightarrow AgI(s) + KNO_3(aq)$ . AgI is formed as a precipitate.

Step 2: Consider the adsorption: When $KI$ is in excess, $I^-$ ions are preferentially adsorbed onto the AgI precipitate, forming a negatively charged sol.

Step 3: Represent the sol: The sol is represented as $AgI/{I^-}$ .

Why other options are wrong:
- Option B: $Agl/A{g^+}$ : This would be the case if $AgNO_3$ was in excess, leading to $Ag^+$ adsorption.
- Option C: $Kl/NO_3^-$ : $KI$ and $NO_3^-$ are spectator ions and do not form the sol.
- Option D: $AgN{O_3}/NO_3^-$ : $AgNO_3$ and $NO_3^-$ are not part of the sol; $AgNO_3$ reacts to form AgI.

Key Concepts:
- Colloid formation
- Preferential adsorption
- Sol formation

Memory Aids:
- Excess reactant determines the adsorbed ion.

Quick Tips:
- Consider the relative amounts of reactants to determine the adsorbed ion.

90 Hard

90 of 94Question permalink

Application +4 | -1

Electrolysis of a solution

\mathrm{MnSO}_4

in aqueous sulphuric acid is a method for the preparation of

\mathrm{MnO}_2

. Passing a current of 27 A for 24 hours gives

1 \mathrm{~kg}

of

\mathrm{MnO}_2

. The current efficiency is

View answer and explanation

Correct option: B

95.185 \%

Option B (

95.185 \%

) is correct because it accurately calculates the current efficiency based on the mass of MnO2 produced during electrolysis.

Solution approach

Correct Answer + Core Reason: $95.185 \%$ . The current efficiency is calculated by comparing the theoretical and actual mass of $MnO_2$ produced.

Step-by-Step Solution:

Step 1: Calculate the total charge passed.\nCharge (Q) = Current (I) * Time (t)\nI = 27 A\nt = 24 hours = 24 * 60 * 60 seconds = 86400 s\nQ = 27 A * 86400 s = 2332800 C

Step 2: Determine the molar mass of $MnO_2$ .\nMolar mass = 54.94 (Mn) + 2 * 16 (O) = 86.94 g/mol

Step 3: Determine the number of moles of $MnO_2$ produced (actual).\nActual moles = mass / molar mass = 1000 g / 86.94 g/mol = 11.502 mol

Step 4: Determine the reaction for $MnO_2$ production.
$Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-$
2 moles of electrons are required to produce 1 mole of $MnO_2$ .

Step 5: Calculate the theoretical moles of $MnO_2$ that could be produced with the given charge.\nMoles of electrons = Charge / Faraday's constant = 2332800 C / 96500 C/mol = 24.174 mol\nTheoretical moles of $MnO_2$ = Moles of electrons / 2 = 24.174 mol / 2 = 12.087 mol

Step 6: Calculate the current efficiency.\nCurrent efficiency = (Actual moles / Theoretical moles) * 100\nCurrent efficiency = (11.502 mol / 12.087 mol) * 100 = 95.185 %

Why other options are wrong:
- Option A: Incorrect calculation of current efficiency.
- Option C: Incorrect calculation of current efficiency.
- Option D: Incorrect calculation of current efficiency.

Key Concepts:
- Faraday's laws of electrolysis
- Current efficiency calculation
- Stoichiometry

Memory Aids:
- Current efficiency = (Actual product / Theoretical product) * 100

Quick Tips:
- Always balance the electrochemical reactions.
- Use the correct value of Faraday's constant.

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