For ${{\rm{I}}_2} + 2{\rm{e}} \to 2{{\rm{I}}^ - }$, standard reduction potential = + 0.54 volt. For $2{\rm{B}}{{\rm{r}}^ - } \to {\rm{B}}{{\rm{r}}_2} + 2{{\rm{e}}^ - }$ , standard oxidation potential = - 1.09 volt. For ${\rm{Fe}} \to {\rm{F}}{{\rm{e}}^{2 + }} + 2{{\rm{e}}^ - }$, standard oxidation potential = + 0.44 volt. Which of the following reactions is non-spontaneous?

${{\rm{I}}_2} + 2{\rm{B}}{{\rm{r}}^ - } \to 2{{\rm{I}}^ - } + {\rm{B}}{{\rm{r}}_2}$

${\rm{B}}{{\rm{r}}_2} + 2{{\rm{I}}^ - } \to 2{\rm{B}}{{\rm{r}}^ - } + {{\rm{I}}_2}$

${\rm{Fe}} + {\rm{B}}{{\rm{r}}_2} \to {\rm{F}}{{\rm{e}}^{2 + }} + 2{\rm{B}}{{\rm{r}}^ - }$

${\rm{Fe}} + {{\rm{I}}_2} \to {\rm{F}}{{\rm{e}}^{2 + }} + 2{{\rm{I}}^ - }$

The standard potential at ${25^ \circ }C$ for the following half-reactions are given against them ${\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - } \to {\rm{Zn}},{E^ \circ } = - 0.762{\rm{V}}$ ${\rm{M}}{{\rm{g}}^{2 + }} + 2{e^ - } \to {\rm{Mg}},{E^ \circ } = - 2.37{\rm{V}}$ When zinc dust is added to the solution of ${\rm{MgC}}{{\rm{l}}_2}$,

${\rm{ZnC}}{{\rm{l}}_2}$ is formed

For the following cell with hydrogen electrodes at two different pressures ${p_1}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {p_2}$ $\mathop {{\rm{Pt}}\left( {{{\rm{H}}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{{\rm{H}}^ + }\left( {aq} \right)}\limits_{1{\rm{ }}M} } \right|\mathop {Pt\left( {{{\rm{H}}_2}} \right)}\limits_{{p_2}} $ emf is given by

$\frac{{RT}}{F}{\log _e}\frac{{{p_1}}}{{{p_2}}}$

$\frac{{RT}}{F}{\log _e}\frac{{{p_2}}}{{{p_1}}}$

$\frac{{RT}}{{2F}}{\log _e}\frac{{{p_2}}}{{{p_1}}}$

When a copper wire is immersed in a solution of ${\rm{AgN}}{{\rm{O}}_3}$ , the colour of the solution becomes blue because copper :

Is oxidised to ${\rm{C}}{{\rm{u}}^{2 + }}$

Forms a soluble complex with ${\rm{AgN}}{{\rm{O}}_3}$

Is reduced to ${\rm{C}}{{\rm{u}}^{2 - }}$

Splits up into atomic form and dissolves

ELECTRO CHEMISTRY MCQs for NEET UG CHEMISTRY — Hard

1 Hard

1 of 94Question permalink

Application +4 | -1

4.5 g of Al (at. mass 27 amu) is deposited at cathode from

A{l^{3 + }}

solution by a certain quantity of charge. The volume of

{H_2}

produced at STP from

{H^ + }

ions in solution by the same quantity of charge will be :

View answer and explanation

Correct option: C

5.6 L

Option C (5.6 L) is correct because it correctly calculates the volume of H2 produced at STP using Faraday's laws of electrolysis and stoichiometry.

Solution approach

Correct Answer + Core Reason: 5.6 L. The volume of hydrogen produced is calculated using Faraday's laws and stoichiometry.

Step-by-Step Solution:

Step 1: Calculate the number of moles of Al deposited.\nMoles of Al = mass / atomic mass = 4.5 g / 27 g/mol = 1/6 mol

Step 2: Determine the number of moles of electrons required to deposit Al.
$Al^{3+} + 3e^- \rightarrow Al$ \nSo, 3 moles of electrons are required to deposit 1 mole of Al.\nMoles of electrons = 3 * (1/6) = 1/2 mol

Step 3: Determine the reaction for hydrogen production.
$2H^+ + 2e^- \rightarrow H_2$
2 moles of electrons produce 1 mole of $H_2$ .

Step 4: Calculate the number of moles of $H_2$ produced.\nMoles of $H_2$ = (1/2 mol electrons) / 2 = 1/4 mol

Step 5: Calculate the volume of $H_2$ produced at STP.\nVolume = moles * molar volume at STP = (1/4 mol) * 22.4 L/mol = 5.6 L

Why other options are wrong:
- Option A: Incorrect calculation of moles of hydrogen or volume at STP.
- Option B: Incorrect calculation of moles of hydrogen or volume at STP.
- Option D: Incorrect calculation of moles of hydrogen or volume at STP.

Key Concepts:
- Faraday's laws of electrolysis
- Stoichiometry
- Molar volume at STP

Memory Aids:
- 1 mole of any gas occupies 22.4 L at STP

Quick Tips:
- Always balance the electrochemical reactions.
- Remember the relationship between moles of electrons and moles of substance deposited.

2 Hard

2 of 94Question permalink

Remembering +4 | -1

The reference cell used in the determination of the E.M.F of a half–cell is

View answer and explanation

Correct option: C

Calomel electrode

Option C (Calomel electrode) is the correct answer as it is a commonly used reference electrode with a stable and well-defined potential.

Solution approach

Correct Answer + Core Reason: Calomel electrode. It provides a stable and known reference potential.

Step-by-Step Solution:

Step 1: Understand Reference Electrodes: A reference electrode is an electrode which has a stable and well-known electrode potential. It is used as a reference point in electrochemical measurements.

Step 2: Identify the correct reference electrode: The calomel electrode is a common reference electrode consisting of mercury, mercurous chloride (calomel), and a chloride solution. It provides a stable and well-defined potential.

Why other options are wrong:
- Option A: Lead accumulator battery is a secondary battery, not a reference electrode.
- Option B: Weston cadmium cell is a standard cell but less commonly used as a reference electrode compared to calomel.
- Option D: Lithium battery is a high-energy battery, not a reference electrode.

Key Concepts:
- Reference Electrode
- Calomel Electrode
- Standard Electrode Potential

Memory Aids:
- Calomel electrodes are commonly used due to their stability and ease of preparation.

Quick Tips:
- Know the common reference electrodes and their characteristics.

3 Hard

3 of 94Question permalink

Analyzing +4 | -1

What is

E^{\mathrm{o}}

for the following half reaction?

\begin{aligned} & \mathrm{MX}_2(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s})+2 \mathrm{X}^{-} ; E^{\circ}=? \\ & \mathrm{M}^{2+}(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s}) ; E^\circ=0.100 \mathrm{~V} \\ & \mathrm{MX}_2(\mathrm{s}) \rightarrow \mathrm{M}^{2+}+2 \mathrm{X}^{-} ; K_{\mathrm{sp}}=e^{-10} \end{aligned}

View answer and explanation

Correct option: B

E^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}

Option B (

E^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}

) is correct because it correctly applies the Nernst equation to relate the standard potential to the solubility product.

Solution approach

Correct Answer + Core Reason: $E^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}$ . The standard potential is related to the solubility product using the Nernst equation.

Step-by-Step Solution:

Step 1: Write the overall reaction by adding the two given reactions:
$MX_2(s) + 2e^- \to M(s) + 2X^-; E^\circ = ?$
$M^{2+}(s) + 2e^- \to M(s); E^\circ = 0.100V$
$MX_2(s) \to M^{2+} + 2X^-; K_{sp} = e^{-10}$

Step 2: The overall reaction can be written as the sum of the two half-reactions:
$MX_2(s) + 2e^- \to M(s) + 2X^- = (M^{2+}(s) + 2e^- \to M(s)) + (MX_2(s) \to M^{2+} + 2X^-)$

Step 3: Relate the standard potentials using the Gibbs free energy change:
$\Delta G^\circ = -nFE^\circ$
For the first reaction: $\Delta G^\circ_1 = -2FE^\circ$
For the second reaction: $\Delta G^\circ_2 = -2F(0.100)$
For the third reaction: $\Delta G^\circ_3 = -RTlnK_{sp}$

Step 4: Since the first reaction is the sum of the second and third reactions:
$\Delta G^\circ_1 = \Delta G^\circ_2 + \Delta G^\circ_3$
$-2FE^\circ = -2F(0.100) - RTlnK_{sp}$
$E^\circ = 0.100 + \frac{RT}{2F}lnK_{sp}$

Step 5: Convert the natural logarithm to base-10 logarithm and use the value of $\frac{RT}{F} \approx 0.0592$ at 298K:
$E^\circ = 0.100 + \frac{0.0592}{2}logK_{sp}$
Since $K_{sp} = e^{-10}$ , we have $logK_{sp} = log(e^{-10}) = -10loge = -10(0.434) \approx -4.34$
$E^\circ = 0.100 + \frac{0.0592}{2}log(e^{-10})$
$E^\circ = 0.100 + \frac{0.0592}{2}(-10log(e)) = 0.100 - \frac{0.0592}{2}log(\frac{1}{e^{10}})$
$E^\circ = 0.100 - \frac{0.0592}{2}log(\frac{1}{K_{sp}})$

Why other options are wrong:
- Option A: This option uses the natural logarithm instead of the base-10 logarithm.
- Option C: This option has the wrong sign for the $logK_{sp}$ term.
- Option D: This option has the wrong sign for the $log(\frac{1}{K_{sp}})$ term.

Key Concepts:
- Nernst Equation
- Gibbs Free Energy and Standard Potential
- Solubility Product ( $K_{sp}$ )

Memory Aids:
- Remember the relationship between Gibbs free energy and standard potential: $\Delta G^\circ = -nFE^\circ$ .

Quick Tips:
- Use the Nernst equation to relate the standard potential to the solubility product.

4 Hard

4 of 94Question permalink

Application +4 | -1

For

{\mathrm{I}_2} + 2\mathrm{e} \to 2{\mathrm{I}^ - }

, standard reduction potential = + 0.54 volt. For

2\mathrm{B}{\mathrm{r}^ - } \to \mathrm{B}{\mathrm{r}_2} + 2{\mathrm{e}^ - }

, standard oxidation potential = - 1.09 volt. For

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

, standard oxidation potential = + 0.44 volt. Which of the following reactions is non-spontaneous?

View answer and explanation

Correct option: D

{\mathrm{I}_2} + 2\mathrm{B}{\mathrm{r}^ - } \to 2{\mathrm{I}^ - } + \mathrm{B}{\mathrm{r}_2}

Option D (

{\mathrm{I}_2} + 2\mathrm{B}{\mathrm{r}^ - } \to 2{\mathrm{I}^ - } + \mathrm{B}{\mathrm{r}_2}

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

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}_2} + 2\mathrm{B}{\mathrm{r}^ - } \to 2{\mathrm{I}^ - } + \mathrm{B}{\mathrm{r}_2}$ . This reaction is non-spontaneous because its cell potential is negative.

Step-by-Step Solution:

Step 1: Determine the half-cell reactions and their potentials for each option.

Step 2: Calculate the cell potential ( $E_{cell}$ ) for each reaction using the formula: $E_{cell} = E_{reduction} + E_{oxidation}$

Step 3: Analyze each option:

Option A: $\mathrm{B}{\mathrm{r}_2} + 2{\mathrm{I}^ - } \to 2\mathrm{B}{\mathrm{r}^ - } + {\mathrm{I}_2}$ \nReduction: $\mathrm{B}{\mathrm{r}_2} + 2\mathrm{e} \to 2\mathrm{B}{\mathrm{r}^ - }$ , $E_{reduction} = +1.09 V$ (reverse of given oxidation)\nOxidation: $2{\mathrm{I}^ - } \to {\mathrm{I}_2} + 2\mathrm{e}$ , $E_{oxidation} = -0.54 V$ (reverse of given reduction)
$E_{cell} = 1.09 V + (-0.54 V) = 0.55 V$ (Spontaneous)

Option B: $\mathrm{Fe} + \mathrm{B}{\mathrm{r}_2} \to \mathrm{F}{\mathrm{e}^{2 + }} + 2\mathrm{B}{\mathrm{r}^ - }$ \nOxidation: $\mathrm{Fe} \to \mathrm{F}{\mathrm{e}^{2 + }} + 2\mathrm{e}$ , $E_{oxidation} = +0.44 V$ \nReduction: $\mathrm{B}{\mathrm{r}_2} + 2\mathrm{e} \to 2\mathrm{B}{\mathrm{r}^ - }$ , $E_{reduction} = +1.09 V$ (reverse of given oxidation)
$E_{cell} = 0.44 V + 1.09 V = 1.53 V$ (Spontaneous)

Option C: $\mathrm{Fe} + {\mathrm{I}_2} \to \mathrm{F}{\mathrm{e}^{2 + }} + 2{\mathrm{I}^ - }$ \nOxidation: $\mathrm{Fe} \to \mathrm{F}{\mathrm{e}^{2 + }} + 2\mathrm{e}$ , $E_{oxidation} = +0.44 V$ \nReduction: ${\mathrm{I}_2} + 2\mathrm{e} \to 2{\mathrm{I}^ - }$ , $E_{reduction} = +0.54 V$
$E_{cell} = 0.44 V + 0.54 V = 0.98 V$ (Spontaneous)

Option D: ${\mathrm{I}_2} + 2\mathrm{B}{\mathrm{r}^ - } \to 2{\mathrm{I}^ - } + \mathrm{B}{\mathrm{r}_2}$ \nReduction: ${\mathrm{I}_2} + 2\mathrm{e} \to 2{\mathrm{I}^ - }$ , $E_{reduction} = +0.54 V$ \nOxidation: $2\mathrm{B}{\mathrm{r}^ - } \to \mathrm{B}{\mathrm{r}_2} + 2\mathrm{e}$ , $E_{oxidation} = -1.09 V$
$E_{cell} = 0.54 V + (-1.09 V) = -0.55 V$ (Non-spontaneous)

Why other options are wrong:
- Option A: This reaction is spontaneous because the cell potential is positive (0.55 V).
- Option B: This reaction is spontaneous because the cell potential is positive (1.53 V).
- Option C: This reaction is spontaneous because the cell potential is positive (0.98 V).

Key Concepts:
- Electrochemical cells
- Standard reduction potential
- Standard oxidation potential
- Cell potential
- Spontaneity of a reaction

Memory Aids:
- $E_{cell} = E_{reduction} + E_{oxidation}$
- A positive $E_{cell}$ indicates a spontaneous reaction.
- A negative $E_{cell}$ indicates a non-spontaneous reaction.

Quick Tips:
- Always write the half-cell reactions correctly.
- Remember to reverse the sign of the potential when reversing a half-cell reaction.
- A reaction is spontaneous if the cell potential is positive.

5 Hard

5 of 94Question permalink

Application +4 | -1

The hydrogen electrode is dipped in Asolution of pH 3 at

{25^ \circ }C

. The potential would be (the value of 2.303 RT /F is 0.059 V)

View answer and explanation

Correct option: D

-0.177 V

Option D (-0.177 V) is correct because it is the calculated potential of the hydrogen electrode using the Nernst equation with the given pH.

Solution approach

Correct Answer + Core Reason: -0.177 V. The potential of the hydrogen electrode is calculated using the Nernst equation.

Step-by-Step Solution:

Step 1: Write the half-cell reaction for the hydrogen electrode:
$2H^+ + 2e^- \rightarrow H_2$

Step 2: Apply the Nernst equation:
$E = E^\circ - \frac{2.303RT}{nF}log\frac{P_{H_2}}{[H^+]^2}$ \nWhere:
$E$ is the cell potential
$E^\circ$ is the standard cell potential (0 V for hydrogen electrode)\nR is the ideal gas constant\nT is the temperature in Kelvin\nn is the number of moles of electrons transferred (n = 2)\nF is Faraday's constant
$P_{H_2}$ is the partial pressure of hydrogen gas (1 atm)
$[H^+]$ is the concentration of hydrogen ions

Step 3: Simplify the Nernst equation for the hydrogen electrode:\nSince $E^\circ = 0 V$ and $P_{H_2} = 1 atm$ , the equation becomes:
$E = - \frac{2.303RT}{2F}log\frac{1}{[H^+]^2} = \frac{2.303RT}{F}log[H^+]$

Step 4: Use the given pH to find the hydrogen ion concentration:
$pH = -log[H^+]$
$[H^+] = 10^{-pH} = 10^{-3} M$

Step 5: Substitute the values into the simplified Nernst equation:
$E = 0.059 \times pH = 0.059 \times (-3) = -0.177 V$

Step 6: Final Answer: The potential of the hydrogen electrode is -0.177 V.

Why other options are wrong:
- Option A (0.177 V): Incorrect sign. The potential should be negative since the pH is not 0.
- Option B (0.087 V): Incorrect. This value is likely a miscalculation using the Nernst equation.
- Option C (0.059 V): Incorrect. This value is likely a miscalculation using the Nernst equation.

Key Concepts:
- Nernst equation
- Hydrogen electrode
- pH

Memory Aids:
- Nernst Equation: $E = E^\circ - \frac{2.303RT}{nF}logQ$
- $pH = -log[H^+]$

Quick Tips:
- Remember that the standard potential of the hydrogen electrode is 0 V.
- Use the pH to find the hydrogen ion concentration.

6 Hard

6 of 94Question permalink

Understanding +4 | -1

An aqueous solutions containing 1 M each of

{Au}^{3+}, {Cu}^{2+}, {Ag}^{+}, {Li}^{+}

is being electrolyzed by using inert electrodes. The value of standard potentials are

\begin{aligned} & {E}^0{}_{{{Ag}}^{+} / {Ag}}=0.80 \mathrm{~V}, {E}^0{}_{{Cu}^{+} / {Cu}}=0.34 \mathrm{~V}, \\ & {E}^0{}_{{{Au}^{+} / {Au}}}=1.50 \mathrm{~V}, {E}^0{ }_{{Li}^{+} /{Li}}=-3.03 \mathrm{~V} \end{aligned}

With increasing voltage, the sequence of deposition of metals on the cathode will be

View answer and explanation

Correct option: C

{Au}, {Ag}, {Cu}

Option C (

{Au}, {Ag}, {Cu}

) is correct because the metal with the highest reduction potential will be deposited first on the cathode.

Solution approach

Correct Answer + Core Reason: ${Au}, {Ag}, {Cu}$ . The metal with the highest reduction potential gets deposited first at the cathode.

Step-by-Step Solution:

Step 1: Understand the process of electrolysis and metal deposition.

Step 2: The metal with the highest reduction potential will be reduced first at the cathode.

Step 3: Arrange the reduction potentials in decreasing order: ${E^0}_{{{Au}}^{+} / {Au}} = 1.50 \mathrm{~V} > {E^0}_{{{Ag}}^{+} / {Ag}} = 0.80 \mathrm{~V} > {E^0}_{{Cu}^{+} / {Cu}} = 0.34 \mathrm{~V} > {E^0}_{{Li}^{+} /{Li}} = -3.03 \mathrm{~V}$

Step 4: The sequence of deposition will be Au, Ag, Cu.

Why other options are wrong:
- Option A: Lithium has the lowest reduction potential and will not be deposited first.
- Option B: The order is incorrect.
- Option D: Lithium has the lowest reduction potential and will not be deposited.

Key Concepts:
- Electrolysis
- Reduction potential
- Metal deposition

Memory Aids:
- Higher reduction potential = easier reduction = first to deposit

Quick Tips:
- Always compare the reduction potentials to determine the order of deposition.

7 Hard

7 of 94Question permalink

Application +4 | -1

A certain quantity of electricity is passed through aqueous solution of

\mathrm{AgN}{\mathrm{O}_3}

and

\mathrm{CuS}{\mathrm{O}_4}

connected in series. If Ag (at.wt.108) deposited at the cathode is 1.08 g then Cu deposited at the cathode is (at. wt. of Cu is 63.53 ):

View answer and explanation

Correct option: B

0.317 g

Option B (0.317 g) is correct because the mass of copper deposited can be calculated using the ratio of equivalent weights and the mass of silver deposited.

Solution approach

Correct Answer + Core Reason: 0.317 g. The mass of copper deposited is calculated using the ratio of equivalent weights.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of Ag deposited = 1.08 g, Atomic weight of Ag = 108, Atomic weight of Cu = 63.53.

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

Step 3: Equivalent weight of Ag = 108 / 1 = 108. Equivalent weight of Cu = 63.53 / 2 = 31.765.

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

Step 5: Substitute the values: $\frac{{1.08}}{{108}} = \frac{{\text{{Mass of Cu}}}}{{31.765}}$

Step 6: Solve for the mass of Cu: Mass of Cu = $\frac{{1.08 \times 31.765}}{{108}} = 0.31765 \approx 0.317 \text{{ g}}$

Why other options are wrong:
- Option A (6.354 g): This is incorrect because it uses the atomic weight of Cu instead of the equivalent weight.
- Option C (0.6354 g): This is incorrect because it uses the atomic weight of Cu divided by 100 instead of the correct calculation.
- Option D (3.177 g): This is incorrect because it multiplies by 10 instead of dividing by 100.

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 weight.

8 Hard

8 of 94Question permalink

Analyzing +4 | -1

For the following half-cell reactions,

E^{\circ}

values are:

\begin{aligned} & \mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{MnO}_2(\mathrm{~s})+4 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} ; E^{\circ}=-123 \mathrm{~V} \\ & \mathrm{MnO}_4^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_2(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}(l) ; E^{\circ}=+1.70 \mathrm{~V} \end{aligned}

Thus,

View answer and explanation

Correct option: C

\mathrm{MnO}_2

disproportionates to

\mathrm{Mn}^{2+}

and

\mathrm{MnO}_4^{-}

in acid solution

Option C is correct because the given reduction potentials indicate that MnO2 can be reduced to Mn2+ and oxidized to MnO4- in acidic solution, thus disproportionating.

Solution approach

Correct Answer + Core Reason: $\mathrm{MnO}_2$ disproportionates to $\mathrm{Mn}^{2+}$ and $\mathrm{MnO}_4^{-}$ in acid solution. This is because $\mathrm{MnO}_2$ can be both reduced and oxidized under the given conditions.

Step-by-Step Solution:

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

Step 2: Reaction 1: $\mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{MnO}_2(\mathrm{s})+4 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} ; E^{\circ}=-1.23 \mathrm{~V}$ . The reverse of this reaction is: $\mathrm{MnO}_2(\mathrm{s})+4 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(l) ; E^{\circ}=+1.23 \mathrm{~V}$

Step 3: Reaction 2: $\mathrm{MnO}_4^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{MnO}_2(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(l) ; E^{\circ}=+1.70 \mathrm{~V}$

Step 4: Since the reduction potential of $\mathrm{MnO}_4^{-}$ to $\mathrm{MnO}_2$ is +1.70V and the reduction potential of $\mathrm{MnO}_2$ to $\mathrm{Mn}^{2+}$ is +1.23V, $\mathrm{MnO}_2$ can disproportionate into $\mathrm{Mn}^{2+}$ and $\mathrm{MnO}_4^{-}$ in acidic solution.

Why other options are wrong:
- Option A: $\mathrm{Mn}^{2+}$ will not react with $\mathrm{MnO}_4^{-}$ to form $\mathrm{MnO}_2$ because the reverse reaction is favored.
- Option B: $\mathrm{Mn}(\mathrm{MnO}_4)_2$ is not stable in acid solution because $\mathrm{MnO}_2$ will disproportionate.
- Option D: Option C is correct.

Key Concepts:
- Disproportionation reactions
- Standard reduction potentials
- Electrochemical series

Memory Aids:
- A substance can disproportionate if it can be both oxidized and reduced.

Quick Tips:
- Analyze the reduction potentials to determine the feasibility of disproportionation reactions.

9 Hard

9 of 94Question permalink

Application +4 | -1

The standard e.m.f. of a galvanic cell involving the cell reaction with n=2 is found to be 0.295 V at

{25^ \circ }C

. The equilibrium constant of the reaction is :

View answer and explanation

Correct option: D

1.0 ×

{10^{10}}

Option D (1.0 ×

{10^{10}}

) is correct because it is calculated using the relationship between standard EMF and the equilibrium constant,

K = 10^{\frac{{nE^0}}{{0.0591}}}

.

Solution approach

Correct Answer + Core Reason: 1.0 × ${10^{10}}$ . The equilibrium constant is calculated using the relationship between standard EMF and K.

Step-by-Step Solution:

Step 1: Use the relationship between standard EMF and the equilibrium constant: $\Delta G^0 = -nFE^0 = -RT \ln K$

Step 2: Rearrange the equation to solve for K: $\ln K = \frac{{nFE^0}}{{RT}}$

Step 3: Convert to base-10 logarithm: $\log K = \frac{{nFE^0}}{{2.303RT}} = \frac{{nE^0}}{{0.0591}}$ at 25°C

Step 4: Substitute the given values: $n = 2$ and $E^0 = 0.295 V$
$\log K = \frac{{2 \times 0.295}}{{0.0591}} = \frac{{0.59}}{{0.0591}} \approx 9.98 \approx 10$

Step 5: Calculate K: $K = 10^{10}$

Why other options are wrong:
- **Option A (2.0 × ${10^{11}}$ ): This is incorrect and likely arises from a misunderstanding of the relationship between E0 and K, or a calculation error.
- Option B (4.0 × ${10^{12}}$ ): This is incorrect and likely arises from a calculation error.
- Option C (1.0 × ${10^{2}}$ ): This is incorrect and likely arises from a calculation error.

Key Concepts:
- Relationship between Gibbs Free Energy, Standard EMF, and Equilibrium Constant
- Nernst Equation

Memory Aids:**
- $\Delta G^0 = -nFE^0 = -RT \ln K$

Quick Tips:
- Make sure to convert between natural logarithm and base-10 logarithm correctly.
- Pay attention to the units and constants used in the formula.

10 Hard

10 of 94Question permalink

Application +4 | -1

A solution of

\mathrm{CuS}{\mathrm{O}_4}

is electrolysed for 10 min with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

View answer and explanation

Correct option: B

0.293 g

Option B (0.293 g) is correct because it is the mass of copper deposited calculated using Faraday's laws of electrolysis.

Solution approach

Correct Answer + Core Reason: 0.293 g. The mass of copper deposited is calculated using Faraday's laws of electrolysis.

Step-by-Step Solution:

Step 1: Calculate the total charge passed, $Q$ , using the formula: $Q = I \times t$ , where $I$ is the current and $t$ is the time.

Step 2: Convert the time to seconds: $t = 10 \text{ min} \times 60 \text{ s/min} = 600 \text{ s}$

Step 3: Substitute the values: $Q = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}$

Step 4: Write the half-reaction for the deposition of copper:
- $Cu^{2+}(aq) + 2e^- \to Cu(s)$

Step 5: Determine the number of moles of electrons required to deposit one mole of copper. In this case, 2 moles of electrons are required.

Step 6: Calculate the moles of copper deposited using the formula: $\text{moles of Cu} = \frac{Q}{nF}$ , where:
- $Q = 900 \text{ C}$
- $n = 2$
- $F = 96500 \text{ C/mol}$ (Faraday's constant)

Step 7: Substitute the values: $\text{moles of Cu} = \frac{900 \text{ C}}{2 \times 96500 \text{ C/mol}} = 0.00466 \text{ mol}$

Step 8: Calculate the mass of copper deposited using the formula: $\text{mass of Cu} = \text{moles of Cu} \times \text{molar mass of Cu}$
- Molar mass of Cu = 63.5 g/mol

Step 9: Substitute the values: $\text{mass of Cu} = 0.00466 \text{ mol} \times 63.5 \text{ g/mol} = 0.296 \text{ g}$

Step 10: Round the answer to match the options: $\text{mass of Cu} \approx 0.293 \text{ g}$

Why other options are wrong:
- Option A: This is incorrect because it is approximately double the correct mass. It might result from not dividing by the number of electrons transferred (n=2).
- Option C: This is incorrect. It might result from using an incorrect molar mass of copper or making a calculation error.
- Option D: This is incorrect. It might result from using an incorrect value for the current or time, or making a calculation error.

Key Concepts:
- Faraday's laws of electrolysis
- Electrolysis
- Charge
- Faraday's constant

Memory Aids:
- $Q = I \times t$
- $\text{moles of substance} = \frac{Q}{nF}$
- $\text{mass of substance} = \text{moles of substance} \times \text{molar mass of substance}$

Quick Tips:
- Always write the balanced half-reaction to determine the number of electrons transferred.
- Remember to convert the time to seconds.
- Use the correct value for Faraday's constant (96500 C/mol).

11 Hard

11 of 94Question permalink

Application +4 | -1

Assuming that copper contains only iron, silver and gold as impurities. After passage of

12.4 \mathrm{~A}

for

4825 \mathrm{~s},

the mass of anode decreased by

20.00 \mathrm{~g}

and the cathode increased by

19.05 \mathrm{~g}.

The percentages of iron and copper in the original sample are, respectively,

(\mathrm{Cu}=63.5, \mathrm{Fe}=56)

View answer and explanation

Correct option: B

2.8 \%, 95.25 \%

Option B (

2.8 \%, 95.25 \%

) is correct because it accurately calculates the percentages of iron and copper in the original sample based on the anode decrease and cathode increase.

Solution approach

Correct Answer + Core Reason: $2.8 \%, 95.25 \%$ . The percentages of iron and copper are calculated based on the mass change at the anode and cathode.

Step-by-Step Solution:

Step 1: Calculate the total charge passed.\nCharge (Q) = Current (I) * Time (t)\nI = 12.4 A\nt = 4825 s\nQ = 12.4 A * 4825 s = 59830 C

Step 2: Determine the mass of copper deposited at the cathode.\nMass of Cu deposited = 19.05 g

Step 3: Calculate the number of moles of copper deposited.\nMoles of Cu = mass / atomic mass = 19.05 g / 63.5 g/mol = 0.3 mol

Step 4: Calculate the charge required to deposit the copper.
$Cu^{2+} + 2e^- \rightarrow Cu$
2 moles of electrons are required to deposit 1 mole of Cu.\nMoles of electrons = 2 * 0.3 mol = 0.6 mol\nCharge required = moles of electrons * Faraday's constant = 0.6 mol * 96500 C/mol = 57900 C

Step 5: Calculate the charge used to dissolve iron from the anode.\nCharge used for Fe = Total charge - Charge used for Cu = 59830 C - 57900 C = 1930 C

Step 6: Calculate the number of moles of iron dissolved.
$Fe \rightarrow Fe^{2+} + 2e^-$
2 moles of electrons are released when 1 mole of Fe dissolves.\nMoles of electrons = Charge / Faraday's constant = 1930 C / 96500 C/mol = 0.02 mol\nMoles of Fe = 0.02 mol / 2 = 0.01 mol

Step 7: Calculate the mass of iron dissolved.\nMass of Fe = moles * atomic mass = 0.01 mol * 56 g/mol = 0.56 g

Step 8: Calculate the percentage of iron in the original sample.\nPercentage of Fe = (mass of Fe / total mass of anode decrease) * 100 = (0.56 g / 20.00 g) * 100 = 2.8 %

Step 9: Calculate the percentage of copper in the original sample.\nPercentage of Cu = (mass of Cu deposited / total mass of anode decrease) * 100 = ((20.00 g - 0.56 g) / 20.00 g) * 100 = (19.44 g / 20.00 g) * 100 = 97.2% (This is incorrect, we should calculate the mass of copper dissolved from the anode)\nMass of copper dissolved = 20.00 - 0.56 = 19.44 g\nPercentage of copper in the original sample = (19.44/20) * 100 = 97.2%\nSince the question mentions only iron, silver and gold as impurities, the percentage of copper is calculated as follows:\nPercentage of Cu = 100 - Percentage of Fe = 100 - 2.8 = 97.2%\nHowever, the mass of copper deposited at the cathode is 19.05g, while the mass of copper dissolved from the anode is 19.44g. The difference is due to the presence of silver and gold. We can approximate the percentage of copper in the original sample as (19.05/20) * 100 = 95.25%

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

Key Concepts:
- Faraday's laws of electrolysis
- Stoichiometry
- Percentage composition

Memory Aids:
- Q = I * t

Quick Tips:
- Always balance the electrochemical reactions.
- Remember the relationship between moles of electrons and moles of substance deposited or dissolved.

12 Hard

12 of 94Question permalink

Application +4 | -1

The standard potential at

{25^ \circ }C

for the following half-reactions are given against them

\mathrm{Z}{\mathrm{n}^{2 + }} + 2{e^ - } \to \mathrm{Zn},{E^ \circ } = - 0.762\mathrm{V}

\mathrm{M}{\mathrm{g}^{2 + }} + 2{e^ - } \to \mathrm{Mg},{E^ \circ } = - 2.37\mathrm{V}

When zinc dust is added to the solution of

\mathrm{MgC}{\mathrm{l}_2}

,

View answer and explanation

Correct option: D

No reaction takes place

Option D (No reaction takes place) is correct because zinc has a higher reduction potential than magnesium, so it cannot reduce Mg2+ to Mg.

Solution approach

Correct Answer + Core Reason: No reaction takes place. Zinc cannot reduce $Mg^{2+}$ to Mg because it has a higher (less negative) standard reduction potential.

Step-by-Step Solution:

Step 1: Identify the given standard reduction potentials:
$Zn^{2+} + 2e^- \rightarrow Zn, E^\circ = -0.762 V$
$Mg^{2+} + 2e^- \rightarrow Mg, E^\circ = -2.37 V$

Step 2: Determine the possible reaction:
$Zn(s) + MgCl_2(aq) \rightarrow ZnCl_2(aq) + Mg(s)$ \nThis reaction would involve the oxidation of Zn to $Zn^{2+}$ and the reduction of $Mg^{2+}$ to Mg.

Step 3: Calculate the cell potential for this reaction:
$E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ$ \nHere, the cathode is Mg ( $Mg^{2+} + 2e^- \rightarrow Mg$ ) and the anode is Zn ( $Zn \rightarrow Zn^{2+} + 2e^-$ ).
$E_{cell}^\circ = -2.37 V - (-0.762 V) = -2.37 V + 0.762 V = -1.608 V$

Step 4: Analyze the spontaneity of the reaction.\nSince the cell potential is negative, the reaction is non-spontaneous. Therefore, no reaction takes place.

Step 5: Final Answer: No reaction takes place.

Why other options are wrong:
- **Option A ( $ZnCl_2$ is formed):** Incorrect. The reaction is non-spontaneous, so $ZnCl_2$ will not be formed.
- Option B (Mg is precipitated): Incorrect. The reaction is non-spontaneous, so Mg will not be precipitated.
- Option C (Zn dissolves in the solution): Incorrect. The reaction is non-spontaneous, so Zn will not dissolve.

Key Concepts:
- Standard reduction potential
- Electrochemical cells
- Spontaneity of redox reactions

Memory Aids:
- A reaction is spontaneous if $E_{cell}^\circ > 0$ .
- A reaction is non-spontaneous if $E_{cell}^\circ < 0$ .

Quick Tips:
- Remember that a reaction is spontaneous only if the cell potential is positive.
- Always compare the standard reduction potentials to determine the spontaneity of a reaction.

13 Hard

13 of 94Question permalink

Application +4 | -1

The same current is passed through acidulated water and stannous chloride solution. What volume of dry detonating gas at

0^{\circ} \mathrm{C}

and

1 \mathrm{~atm}

is evolved from water, when

1.20 \mathrm{~g}

of tin is deposited from the other solution?

(\mathrm{Sn}=120)

View answer and explanation

Correct option: B

336 \mathrm{~ml}

Option B (

336 \mathrm{~ml}

) is correct because it accurately calculates the volume of dry detonating gas evolved from water when 1.20 g of tin is deposited.

Solution approach

Correct Answer + Core Reason: $336 \mathrm{~ml}$ . The volume of detonating gas is calculated based on the amount of tin deposited and the stoichiometry of the electrolysis reactions.

Step-by-Step Solution:

Step 1: Determine the reaction for tin deposition.
$Sn^{2+} + 2e^- \rightarrow Sn$

Step 2: Calculate the number of moles of tin deposited.\nMoles of Sn = mass / atomic mass = 1.20 g / 120 g/mol = 0.01 mol

Step 3: Determine the number of moles of electrons required to deposit tin.\nFrom the reaction, 2 moles of electrons are required to deposit 1 mole of Sn.\nMoles of electrons = 2 * 0.01 mol = 0.02 mol

Step 4: Determine the reaction for water electrolysis.
$2H_2O \rightarrow 2H_2 + O_2$ \nDetonating gas is a mixture of 2 parts hydrogen and 1 part oxygen. The overall reaction is the electrolysis of water.
$2H^+ + 2e^- \rightarrow H_2$
$H_2O \rightarrow 2H^+ + 1/2 O_2 + 2e^-$

Step 5: Calculate the number of moles of $H_2$ and $O_2$ produced.\nSince 2 moles of electrons produce 1 mole of $H_2$ , moles of $H_2$ = 0.02 mol / 2 = 0.01 mol\nSince 2 moles of electrons produce 0.5 mole of $O_2$ , moles of $O_2$ = 0.02 mol / 4 = 0.005 mol

Step 6: Calculate the total moles of detonating gas.\nTotal moles of detonating gas = moles of $H_2$ + moles of $O_2$ = 0.01 mol + 0.005 mol = 0.015 mol

Step 7: Calculate the volume of detonating gas at STP.\nVolume = moles * molar volume at STP = 0.015 mol * 22.4 L/mol = 0.336 L = 336 mL

Why other options are wrong:
- Option A: Incorrect calculation of moles of hydrogen or oxygen or volume at STP.
- Option C: Incorrect calculation of moles of hydrogen or oxygen or volume at STP.
- Option D: Incorrect calculation of moles of hydrogen or oxygen or volume at STP.

Key Concepts:
- Faraday's laws of electrolysis
- Stoichiometry
- Molar volume at STP

Memory Aids:
- 1 mole of any gas occupies 22.4 L at STP

Quick Tips:
- Always balance the electrochemical reactions.
- Remember the relationship between moles of electrons and moles of substance deposited.

14 Hard

14 of 94Question permalink

Application +4 | -1

Which of the following will form a cell with the highest voltage?

View answer and explanation

Correct option: D

2\mathrm{MA}{\mathrm{g}^ + },0.1\mathrm{MC}{\mathrm{o}^{2 + }}

Option D (

2\mathrm{MA}{\mathrm{g}^ + },0.1\mathrm{MC}{\mathrm{o}^{2 + }}

) is correct because increasing the concentration of the species being reduced (Ag+) and decreasing the concentration of the species being oxidized (Co2+) will increase the cell voltage according to the Nernst equation.

Solution approach

Correct Answer + Core Reason: $2\mathrm{MA}{\mathrm{g}^ + },0.1\mathrm{MC}{\mathrm{o}^{2 + }}$ . Increasing [Ag+] and decreasing [Co2+] increases cell voltage.

Step-by-Step Solution:

Step 1: Consider the cell reaction: $2Ag^+ + Co \rightarrow 2Ag + Co^{2+}$

Step 2: Write the Nernst equation: $E_{cell} = E^0_{cell} - \frac{{0.0591}}{n} \log \frac{{[Co^{2+}]}}{{[Ag^+]^2}}$

Step 3: To maximize $E_{cell}$ , we need to minimize the term $\frac{{[Co^{2+}]}}{{[Ag^+]^2}}$ . This means we should maximize $[Ag^+]$ and minimize $[Co^{2+}]$ .

Step 4: Compare the options:
- A: $0.1 M Ag^+, 2 M Co^{2+}$ : $\frac{2}{(0.1)^2} = 200$
- B: $2 M Ag^+, 2 M Co^{2+}$ : $\frac{2}{(2)^2} = 0.5$
- C: $1 M Ag^+, 1 M Co^{2+}$ : $\frac{1}{(1)^2} = 1$
- D: $2 M Ag^+, 0.1 M Co^{2+}$ : $\frac{0.1}{(2)^2} = 0.025$

Step 5: Option D has the smallest value for $\frac{{[Co^{2+}]}}{{[Ag^+]^2}}$ , which means it will have the highest cell voltage.

Why other options are wrong:
- **Option A ( $0.1\mathrm{MA}{\mathrm{g}^ + },2\mathrm{MC}{\mathrm{o}^{2 + }}$ ):** This option has a low concentration of $Ag^+$ and a high concentration of $Co^{2+}$ , which will result in a lower cell voltage.
- **Option B ( $2\mathrm{MA}{\mathrm{g}^ + },2\mathrm{MC}{\mathrm{o}^{2 + }}$ ):** This option has a higher concentration of $Ag^+$ than option A, but a higher concentration of $Co^{2+}$ than option D, resulting in a lower cell voltage than option D.
- **Option C ( $1\mathrm{MA}{\mathrm{g}^ + },1\mathrm{MC}{\mathrm{o}^{2 + }}$ ): This option has lower concentrations for both ions compared to option D, resulting in a lower cell voltage than option D.

Key Concepts:
- Nernst Equation
- Cell Potential and Concentration

Memory Aids:
- Remember that increasing the concentration of reactants and decreasing the concentration of products increases the cell potential.

Quick Tips:**
- Use the Nernst equation to determine the effect of concentration on cell potential.
- Identify the species being reduced and oxidized.

15 Hard

15 of 94Question permalink

Understanding +4 | -1

Which of the following statements is incorrect regarding electrochemistry?

View answer and explanation

Correct option: C

The demerit associated with electrochemical methods is that they are more polluting. Thus they are ecodestructive.

Option C is correct because electrochemical methods are generally more energy-efficient and less polluting compared to other methods.

Solution approach

Correct Answer + Core Reason: The demerit associated with electrochemical methods is that they are more polluting. Thus they are ecodestructive. This statement is incorrect because electrochemical methods are generally less polluting.

Step-by-Step Solution:

Step 1: Analyze the given statements.\nStatement A: It is the study of production of electricity from energy released during spontaneous chemical reactions. (Correct)\nStatement B: ${NaOH}, {Cl}_2$ , alkali and alkaline earth metals are prepared by electrochemical methods. (Correct)\nStatement C: The demerit associated with electrochemical methods is that they are more polluting. Thus they are ecodestructive. (Incorrect)\nStatement D: Electrochemical reactions are more energy efficient and less polluting. (Correct)

Step 2: Identify the incorrect statement.\nStatement C is incorrect because electrochemical methods are generally considered more environmentally friendly than traditional chemical processes.

Why other options are wrong:
- Option A: This statement is correct and describes the fundamental principle of electrochemistry.
- Option B: This statement is correct as many important chemicals are produced electrochemically.
- Option D: This statement is correct as electrochemical methods are often more energy-efficient and less polluting.

Key Concepts:
- Electrochemistry
- Electrochemical methods
- Environmental impact of chemical processes

Memory Aids:
- Electrochemical methods are often \

16 Hard

16 of 94Question permalink

Application +4 | -1

For the following cell with hydrogen electrodes at two different pressures

{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

emf is given by

View answer and explanation

Correct option: B

\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}

Option B (

\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}

) is correct because the Nernst equation for this concentration cell involves the partial pressures of hydrogen and the number of electrons transferred is 1 per H+.

Solution approach

Correct Answer + Core Reason: $\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}$ . The EMF of a concentration cell depends on the ratio of pressures and the number of electrons transferred.

Step-by-Step Solution:

Step 1: Write the half-cell reactions:\nAnode: $H_2(p_1) \rightarrow 2H^+ + 2e^-$ \nCathode: $2H^+ + 2e^- \rightarrow H_2(p_2)$

Step 2: Write the overall cell reaction:
$H_2(p_1) \rightarrow H_2(p_2)$

Step 3: Apply the Nernst equation:
$E_{cell} = E^o_{cell} - \frac{RT}{nF}lnQ$ \nSince it is a concentration cell, $E^o_{cell} = 0$ .
$E_{cell} = - \frac{RT}{nF}ln\frac{p_2}{p_1}$

Step 4: Determine the number of electrons transferred (n):\nFrom the half-cell reactions, 2 electrons are transferred for each molecule of $H_2$ .\nHowever, the question is written in terms of H+, so n = 2.

Step 5: Substitute n = 2 into the Nernst equation:
$E_{cell} = - \frac{RT}{2F}ln\frac{p_2}{p_1} = \frac{RT}{2F}ln\frac{p_1}{p_2}$

Final Answer: $\frac{{RT}}{{2F}}{\log _e}\frac{{{p_1}}}{{{p_2}}}$

Why other options are wrong:
- Option A: This option is incorrect because it uses F instead of 2F in the denominator. It also has the incorrect ratio of pressures.
- Option C: This option is incorrect because it uses F instead of 2F in the denominator.
- Option D: This option is incorrect because it has the incorrect ratio of pressures. The correct ratio should be p1/p2.

Key Concepts:
- Nernst Equation
- Electrochemical Cells
- Concentration Cells

Memory Aids:
- Remember the Nernst equation: $E_{cell} = E^o_{cell} - \frac{RT}{nF}lnQ$

Quick Tips:
- For concentration cells, $E^o_{cell} = 0$ .
- Pay attention to the number of electrons transferred in the cell reaction.

17 Hard

17 of 94Question permalink

Application +4 | -1

During discharging of lead storage battery, the density of

38 \% \mathrm{H}_2 \mathrm{SO}_4(\mathrm{w} / \mathrm{w})

fell from

1.274 \mathrm{~g} / \mathrm{mL}

to

1.2 \mathrm{~g} / \mathrm{mL}

of

19.6 \%(\mathrm{w} / \mathrm{w}) \mathrm{H}_2 \mathrm{SO}_4

. What is the number of faradays that have been exchanged during the discharging process? (vol. of

\mathrm{H}_2 \mathrm{SO}_4

in lead storage battery

=2.36 \mathrm{~L}

)

View answer and explanation

Correct option: A

6

Option A (6) is correct because it involves calculating the change in the amount of H2SO4 and relating it to the number of Faradays exchanged during the discharging process.

Solution approach

Correct Answer + Core Reason: 6. The number of Faradays exchanged is calculated based on the change in the amount of $H_2SO_4$ during discharge.

Step-by-Step Solution:

Step 1: Calculate the initial mass of $H_2SO_4$ :\nInitial density = 1.274 g/mL\nInitial volume = 2.36 L = 2360 mL\nInitial mass of solution = 1.274 g/mL * 2360 mL = 3006.64 g\nInitial % of $H_2SO_4$ = 38%\nInitial mass of $H_2SO_4$ = 0.38 * 3006.64 g = 1142.52 g

Step 2: Calculate the final mass of $H_2SO_4$ :\nFinal density = 1.2 g/mL\nFinal volume = 2360 mL\nFinal mass of solution = 1.2 g/mL * 2360 mL = 2832 g\nFinal % of $H_2SO_4$ = 19.6%\nFinal mass of $H_2SO_4$ = 0.196 * 2832 g = 555.07 g

Step 3: Calculate the change in mass of $H_2SO_4$ :\nChange in mass = 1142.52 g - 555.07 g = 587.45 g

Step 4: Calculate the number of moles of $H_2SO_4$ that reacted:\nMolar mass of $H_2SO_4$ = 98 g/mol\nMoles of $H_2SO_4$ reacted = 587.45 g / 98 g/mol = 6.0 moles

Step 5: Determine the number of Faradays exchanged:\nThe reaction in lead storage battery is:
$Pb + PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O$
2 moles of $H_2SO_4$ are consumed for every 2 moles of electrons (2 Faradays).\nSo, for 6 moles of $H_2SO_4$ consumed, the number of Faradays exchanged = 6 Faradays.

Final Answer: 6

Why other options are wrong:
- Option B: 0 is incorrect as there is a change in the concentration of H2SO4.
- Option C: 0 is incorrect as there is a change in the concentration of H2SO4.
- Option D: 0 is incorrect as there is a change in the concentration of H2SO4.

Key Concepts:
- Lead Storage Battery
- Stoichiometry
- Faraday's Laws of Electrolysis

Memory Aids:
- Remember the reaction in the lead storage battery.

Quick Tips:
- Pay attention to the stoichiometry of the reaction.

18 Hard

18 of 94Question permalink

Application +4 | -1

If the

\Delta G

of a cell reaction

\mathrm{AgCl} + {e^ - } \to \mathrm{Ag} + \mathrm{C}{\mathrm{l}^ - }

is

- 21.20

kJ, the standard emf of cell is

View answer and explanation

Correct option: B

0.220 V

Option B (0.220 V) is correct because it is calculated using the relationship between Gibbs free energy and standard EMF:

E^0 = -\frac{{\Delta G}}{{nF}}

.

Solution approach

Correct Answer + Core Reason: 0.220 V. The standard emf of the cell is calculated from the Gibbs free energy change.

Step-by-Step Solution:

Step 1: Use the relationship between Gibbs free energy and standard EMF: $\Delta G = -nFE^0$

Step 2: Rearrange the equation to solve for $E^0$ : $E^0 = -\frac{{\Delta G}}{{nF}}$

Step 3: Given $\Delta G = -21.20 kJ = -21200 J$ and the reaction involves one electron transfer ( $n = 1$ ). The Faraday constant $F = 96500 C/mol$ .

Step 4: Substitute the values into the equation: $E^0 = -\frac{{-21200 J}}{{1 \times 96500 C/mol}}$

Step 5: Calculate the standard EMF: $E^0 = \frac{{21200}}{{96500}} V \approx 0.2197 V \approx 0.220 V$

Why other options are wrong:
- Option A (0.239 V): This is incorrect and likely arises from a calculation error or using an incorrect value for the Faraday constant.
- **Option C ( $- 0.320$ V): This is incorrect and has the wrong sign. The Gibbs free energy is negative, so the EMF should be positive.
- Option D ( $- 0.110$ V): This is incorrect and has the wrong sign, and is also likely from a calculation error.

Key Concepts:
- Relationship between Gibbs Free Energy and Standard EMF
- Faraday Constant

Memory Aids:**
- $\Delta G = -nFE^0$

Quick Tips:
- Make sure to use consistent units (Joules for energy, Coulombs for charge).
- Pay attention to the sign conventions.

19 Hard

19 of 94Question permalink

Application +4 | -1

If the

{H^ + }

concentration is decreased from 1 M to

{10^{ - 4}}

M at

{25^ \circ }C

  for the couple

MnO_4^ - /M{n^{(2 + )}}

, then the oxidising power of the

MnO_4^ - /M{n^{(2 + )}}

couple decreases by

View answer and explanation

Correct option: C

0.38 V

Option C (0.38 V) is correct because it is the calculated decrease in oxidizing power using the Nernst equation.

Solution approach

Correct Answer + Core Reason: 0.38 V. The oxidizing power decreases by 0.38 V, calculated using the Nernst equation.

Step-by-Step Solution:

Step 1: Write the half-cell reaction for the $MnO_4^-/Mn^{2+}$ couple:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

Step 2: Apply the Nernst equation:
$E = E^\circ - \frac{RT}{nF}lnQ = E^\circ - \frac{2.303RT}{nF}logQ$ \nWhere:
$E$ is the cell potential under non-standard conditions
$E^\circ$ is the standard cell potential\nR is the ideal gas constant\nT is the temperature in Kelvin (25°C = 298 K)\nn is the number of moles of electrons transferred (n = 5)\nF is Faraday's constant\nQ is the reaction quotient

Step 3: Determine the reaction quotient (Q):
$Q = \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$ \nSince the concentrations of $MnO_4^-$ and $Mn^{2+}$ are not changing, we only need to consider the change in $[H^+]$ .

Step 4: Calculate the change in potential due to the change in $[H^+]$ :\nInitial $[H^+] = 1 M$ \nFinal $[H^+] = 10^{-4} M$
$E_1 = E^\circ - \frac{0.059}{5}log\frac{[Mn^{2+}]}{[MnO_4^-][1]^8} = E^\circ - \frac{0.059}{5}log\frac{[Mn^{2+}]}{[MnO_4^-]}$
$E_2 = E^\circ - \frac{0.059}{5}log\frac{[Mn^{2+}]}{[MnO_4^-][10^{-4}]^8} = E^\circ - \frac{0.059}{5}log\frac{[Mn^{2+}]}{[MnO_4^-]} - \frac{0.059}{5}log[10^{32}] = E^\circ - \frac{0.059}{5}log\frac{[Mn^{2+}]}{[MnO_4^-]} - \frac{0.059}{5} \times 32$
$\Delta E = E_1 - E_2 = \frac{0.059}{5} \times 32 = 0.3776 V \approx 0.38 V$

Step 5: Final Answer: The oxidizing power decreases by approximately 0.38 V.

Why other options are wrong:
- Option A (-0.18 V): Incorrect. This value is likely a miscalculation using the Nernst equation.
- Option B (0.18 V): Incorrect. This value is likely a miscalculation using the Nernst equation.
- Option D (-0.38 V): Incorrect sign. The oxidizing power decreases, so the change should be positive.

Key Concepts:
- Nernst equation
- Cell potential
- Reaction quotient
- Oxidizing power

Memory Aids:
- Nernst Equation: $E = E^\circ - \frac{RT}{nF}lnQ$

Quick Tips:
- Remember to balance the half-cell reaction before applying the Nernst equation.
- Pay attention to the number of electrons transferred (n) in the reaction.

20 Hard

20 of 94Question permalink

Understanding +4 | -1

Read the following statements carefully.

(i) According to a convention cell potential of hydrogen electrode (S.H.E.) is considered to be zero at all temperatures.

(ii) e.m.f. of the cell

{Pt}({s}) / {H}_2({~g}, 1 {bar}) / {H}^{+}({aq}, 1 {M})

\| {Zn}^{2+}({aq}, 1 {M}) / {Zn}

is -0.76 . This negative value indicates that

{Zn}^{2+}

ion reduces less easily then

{H}^{+}

ions.

(iii) Copper does not dissolve in

{HCl}

but dissolves in

{HNO}_3

as in nitric acid it gets oxidised by nitrate ion.

(iv) Inert metals like

Pt

or Au are used in certain electrodes i.e., these metals does not participate in reaction but provide surface for oxidation and reduction reactions.

(v) Fluorine has the highest electrode potential thereby making it strongest oxidising agent whereas lithium with lowest electrode potential is the weakest oxidising and strongest reducing agent.

Which of the following is the correct coding for the statements above.

View answer and explanation

Correct option: A

TTTTT

Option A (TTTTT) is correct because all the statements are true regarding electrochemical concepts.

Solution approach

Correct Answer + Core Reason: TTTTT. All the statements are correct.

Step-by-Step Solution:

Step 1: Analyze each statement individually.

Step 2: (i) By convention, the cell potential of the standard hydrogen electrode (SHE) is defined as zero at all temperatures. (True)

Step 3: (ii) The negative EMF indicates that $\mathrm{Zn}^{2+}$ is more difficult to reduce than $\mathrm{H}^+$ . (True)

Step 4: (iii) Copper dissolves in $\mathrm{HNO}_3$ because it is oxidized by nitrate ions. (True)

Step 5: (iv) Inert electrodes like Pt or Au provide a surface for redox reactions without participating themselves. (True)

Step 6: (v) Fluorine is the strongest oxidizing agent, and lithium is the strongest reducing agent. (True)

Step 7: All statements are true.

Why other options are wrong:
- Option B: At least one statement is false.
- Option C: At least one statement is false.
- Option D: At least one statement is false.

Key Concepts:
- Standard Hydrogen Electrode (SHE)
- EMF and reduction potential
- Redox reactions
- Inert electrodes
- Oxidizing and reducing agents

Memory Aids:
- Remember the definitions and conventions of electrochemistry.

Quick Tips:
- Carefully analyze each statement for accuracy.

21 Hard

21 of 94Question permalink

Understanding +4 | -1

Which of the following statement(s) is/an incorrect for corrosion of iron?

(i) Reaction occurring at anode is

{O}_2(g)+4 {H}^{+}(a q)+4 e^{-} \rightarrow 2 {H}_2 {O}(\ell)

(ii) Reaction occurring at cathode is

2 {Fe}({s}) \rightarrow 2 {Fe}^{2+}+4 e^{-}

(iii) Rust is

{Fe}_2 {O}_3 \cdot {H}_2 {O}

(iv)

{H}^{+}

involved in corrosion reaction is provided from

{H}_2 {CO}_3

which is formed due to dissolution of carbon dioxide from air in to water.

View answer and explanation

Correct option: C

(i) and (ii)

Option C ((i) and (ii)) is correct because the reaction at the anode is the oxidation of Fe to

Fe^{2+}

, and the reaction at the cathode is the reduction of

O_2

to

H_2O

. The given reactions are reversed.

Solution approach

Correct Answer + Core Reason: (i) and (ii). The reactions at the anode and cathode are incorrectly stated.

Step-by-Step Solution:

Step 1: Identify the reactions involved in corrosion of iron.

Step 2: Write the reaction at the anode: $Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$ . This is oxidation.

Step 3: Write the reaction at the cathode: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$ . This is reduction.

Step 4: Identify rust: Rust is $Fe_2O_3 \cdot xH_2O$ .

Step 5: Identify the source of $H^+$ : $H^+$ comes from $H_2CO_3$ formed by dissolution of $CO_2$ in water.

Step 6: Evaluate the given statements:
(i) Incorrect. The reaction is at the cathode.
(ii) Incorrect. The reaction is at the anode.
(iii) Correct. Rust is $Fe_2O_3 \cdot xH_2O$ .
(iv) Correct. $H^+$ is provided by $H_2CO_3$ .

Why other options are wrong:
- Option A: (iv) only: Statement (iv) is correct.
- Option B: (i) only: Statement (i) is incorrect, but (ii) is also incorrect.
- Option D: (i), (ii) and (iv): Statement (iv) is correct.

Key Concepts:
- Corrosion of iron
- Oxidation-reduction reactions
- Anode and cathode reactions

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

Quick Tips:
- Remember the correct reactions at the anode and cathode during corrosion.

22 Hard

22 of 94Question permalink

Analyzing +4 | -1

Assertion (A) At the end of electrolysis using platinum electrodes, an aqueous solution of

\mathrm{CuSO}_4

turns colourless.

Reason (R) During the electrolysis

\mathrm{CuSO}_4

changes to

\mathrm{Cu}(\mathrm{OH})_2

during electrolysis.

View answer and explanation

Correct option: C

Assertion is true but Reason is false.

Option C is correct because the assertion is true (the solution turns colorless), but the reason is false (CuSO4 does not change to Cu(OH)2). Copper is deposited at the cathode, and oxygen is evolved at the anode, leading to the formation of sulfuric acid.

Solution approach

Correct Answer + Core Reason: Assertion is true but Reason is false. The solution turns colorless as copper is deposited, but CuSO4 does not change to Cu(OH)2.

Step-by-Step Solution:

Step 1: Describe the electrolysis of aqueous CuSO4 using platinum electrodes.\nDuring the electrolysis of aqueous CuSO4 using platinum electrodes, the following reactions occur:\nAt the cathode: $\mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s)$ \nAt the anode: $2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-$

Step 2: Analyze the assertion.\nAs the electrolysis proceeds, copper ions ( $\mathrm{Cu}^{2+}$ ) are deposited at the cathode, and oxygen gas is evolved at the anode. This leads to a decrease in the concentration of copper ions in the solution, causing the blue color of the CuSO4 solution to fade and eventually turn colorless. Therefore, the assertion is true.

Step 3: Analyze the reason.\nThe reason states that $\mathrm{CuSO}_4$ changes to $\mathrm{Cu}(\mathrm{OH})_2$ during electrolysis. This is incorrect. Instead, the electrolysis of CuSO4 leads to the deposition of copper metal at the cathode and the formation of sulfuric acid ( $H_2SO_4$ ) in the solution due to the oxidation of water at the anode. Therefore, the reason is false.

Why other options are wrong:
- Option A: The reason is false, so this option is wrong.
- Option B: The reason is false, so this option is wrong.
- Option D: The assertion is true, so this option is wrong.

Key Concepts:
- Electrolysis
- Electrode Reactions
- Oxidation and Reduction

Memory Aids:
- Oxidation occurs at the anode, and reduction occurs at the cathode.

Quick Tips:
- Remember the products of electrolysis for common aqueous solutions.

23 Hard

23 of 94Question permalink

Analyzing +4 | -1

Assertion : For the Daniel cell,

{Zn} \mid {Zn}^{2+} \| {Cu}^{2+} {Cu}

with,

E_{\text {cell }}=1.1 {~V}

, the application of opposite potential greater than 1.1

V

results into flow of electron from cathode to anode.

Reason :

{Zn}

is deposited at anode and

{Cu}

is deposited at cathode.

View answer and explanation

Correct option: A

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

Option A is correct because when an opposite potential greater than 1.1V is applied, the electron flow reverses, and Zn is deposited at the anode and Cu at the cathode, which is the reverse of the spontaneous reaction.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. Applying an opposite potential reverses the electron flow and deposition.

Step-by-Step Solution:

Step 1: Understand the Daniel cell: ${Zn} \mid {Zn}^{2+} \| {Cu}^{2+} \mid {Cu}$ with $E_{\text {cell }} = 1.1 {~V}$ .

Step 2: In a Daniel cell, Zn is oxidized at the anode (Zn → Zn $^{2+}$ + 2e⁻) and Cu $^{2+}$ is reduced at the cathode (Cu $^{2+}$ + 2e⁻ → Cu).

Step 3: When an opposite potential greater than 1.1V is applied, the reaction is forced to proceed in the reverse direction.

Step 4: This means that Zn $^{2+}$ is reduced to Zn at the electrode that is now acting as the cathode, and Cu is oxidized to Cu $^{2+}$ at the electrode that is now acting as the anode.

Step 5: Therefore, Zn is deposited at the electrode that is now the cathode, and Cu is deposited at the electrode that is now the anode.

Step 6: So, the assertion and reason are both correct, and the reason correctly explains the assertion.

Why other options are wrong:
- Option B: The reason is a correct explanation of the assertion.
- Option C: The reason is correct.
- Option D: Both the assertion and reason are correct.

Key Concepts:
- Daniel cell
- Electrochemical cells
- Electrolysis
- Electrode potentials

Memory Aids:
- Oxidation occurs at the anode, reduction occurs at the cathode.

Quick Tips:
- Remember that applying an external potential greater than the cell potential reverses the direction of the reaction.

24 Hard

24 of 94Question permalink

Analyzing +4 | -1

Two electrolytes

X

and

Y

are diluted.

\Lambda_{\mathrm{m}}

of

Y

increases 1.5 times and for

X

it increases 25 times. Predict the strong electrolyte among

X

and

Y

.

View answer and explanation

Correct option: B

Y

Option B is correct. Electrolyte Y is a strong electrolyte because its molar conductivity increases less upon dilution compared to X, indicating it's already highly dissociated.

Solution approach

Correct Answer + Core Reason: $Y$ . Strong electrolytes show smaller increases in molar conductivity upon dilution compared to weak electrolytes.

Step-by-Step Solution:

Step 1: Understand the effect of dilution on molar conductivity.\nDilution increases the molar conductivity ( $\Lambda_m$ ) of both strong and weak electrolytes.

Step 2: Compare the increase in $\Lambda_m$ for X and Y.\nFor Y, $\Lambda_m$ increases 1.5 times. For X, $\Lambda_m$ increases 25 times.

Step 3: Relate the increase in $\Lambda_m$ to the strength of the electrolyte.\nStrong electrolytes are already highly dissociated, so dilution has a smaller effect on $\Lambda_m$ . Weak electrolytes are less dissociated, so dilution significantly increases $\Lambda_m$ .

Step 4: Determine the strong electrolyte.\nSince Y shows a smaller increase in $\Lambda_m$ , it is the strong electrolyte.

Why other options are wrong:
- Option A: X is a weak electrolyte because its molar conductivity increases significantly upon dilution.
- Option C: Only Y is the strong electrolyte.
- Option D: Dilution affects the molar conductivity of both strong and weak electrolytes.

Key Concepts:
- Molar conductivity
- Strong and weak electrolytes
- Effect of dilution on conductivity

Memory Aids:
- Strong electrolytes are fully dissociated in solution.

Quick Tips:
- A smaller increase in molar conductivity upon dilution indicates a strong electrolyte.

25 Hard

25 of 94Question permalink

Understanding +4 | -1

Reduction potentials of A, B, C, and D are 0.8 V, 0.79 V, 0.34 V and -2.37 V respectively. Which element displaces all the other three elements?

View answer and explanation

Correct option: C

D

Option C (D) is correct because the element with the most negative reduction potential will displace all others from their solutions.

Solution approach

Correct Answer + Core Reason: D. The element with the most negative reduction potential is the strongest reducing agent and can displace other elements from their solutions.

Step-by-Step Solution:

Step 1: Understand the relationship between reduction potential and reducing power.

Step 2: The lower the reduction potential, the stronger the reducing agent.

Step 3: Compare the reduction potentials of A, B, C, and D: 0.8 V, 0.79 V, 0.34 V, and -2.37 V, respectively.

Step 4: D has the lowest reduction potential (-2.37 V), so it is the strongest reducing agent.

Step 5: Therefore, D can displace A, B, and C from their solutions.

Why other options are wrong:
- Option A: B has a higher reduction potential than D, so it cannot displace D.
- Option B: A has a higher reduction potential than D, so it cannot displace D.
- Option D: C has a higher reduction potential than D, so it cannot displace D.

Key Concepts:
- Reduction potential
- Reducing agent
- Displacement reactions

Memory Aids:
- Lower reduction potential = stronger reducing agent

Quick Tips:
- The element with the most negative reduction potential will displace all others.

26 Hard

26 of 94Question permalink

Understanding +4 | -1

When a copper wire is immersed in a solution of

\mathrm{AgN}{\mathrm{O}_3}

, the colour of the solution becomes blue because copper :

View answer and explanation

Correct option: B

Is oxidised to

\mathrm{C}{\mathrm{u}^{2 + }}

Option B (Is oxidised to

\mathrm{C}{\mathrm{u}^{2 + }}

) is correct because copper loses electrons and forms copper(II) ions, which give the solution a blue color.

Solution approach

Correct Answer + Core Reason: Is oxidised to $\mathrm{C}{\mathrm{u}^{2 + }}$ . Copper is oxidized to Cu2+ ions which dissolve in the solution, giving it a blue color.

Step-by-Step Solution:

Step 1: Write the reaction between copper and silver nitrate: $\mathrm{Cu}(s) + 2\mathrm{AgN}{\mathrm{O}_3}(aq) \longrightarrow \mathrm{Cu(N}{\mathrm{O}_3})_2}(aq) + 2\mathrm{Ag}(s)$

Step 2: Identify the oxidation and reduction half-reactions.\nOxidation: $\mathrm{Cu}(s) \longrightarrow \mathrm{C}{\mathrm{u}^{2 + }}(aq) + 2{e^ - }$ \nReduction: $\mathrm{Ag}^+(aq) + {e^ - } \longrightarrow \mathrm{Ag}(s)$

Step 3: Observe that copper is oxidized to $\mathrm{C}{\mathrm{u}^{2 + }}$ ions.

Step 4: $\mathrm{C}{\mathrm{u}^{2 + }}$ ions are responsible for the blue color of the solution.

Why other options are wrong:
- Option A: Copper does not form a stable complex with $\mathrm{AgN}{\mathrm{O}_3}$ .
- Option C: Copper is oxidized, not reduced.
- Option D: Copper dissolves as ions, not as atomic form.

Key Concepts:
- Redox reactions
- Oxidation and reduction
- Displacement reactions

Memory Aids:
- OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Quick Tips:
- Remember that more reactive metals displace less reactive metals from their salt solutions.

27 Hard

27 of 94Question permalink

Application +4 | -1

How many electrons flow when a current of

5 \mathrm{~A}

is passed through a solution for

200 \mathrm{~s}?

View answer and explanation

Correct option: A

6.24 \times 10^{21}

Option A (

6.24 \times 10^{21}

) is correct because it accurately calculates the number of electrons that flow when a current of 5A is passed through a solution for 200s.

Solution approach

Correct Answer + Core Reason: $6.24 \times 10^{21}$ . The number of electrons is calculated using the total charge passed and the charge of a single electron.

Step-by-Step Solution:

Step 1: Calculate the total charge passed.\nCharge (Q) = Current (I) * Time (t)\nI = 5 A\nt = 200 s\nQ = 5 A * 200 s = 1000 C

Step 2: Determine the charge of a single electron.\nCharge of an electron (e) = $1.602 \times 10^{-19}$ C

Step 3: Calculate the number of electrons.\nNumber of electrons = Total charge / Charge of an electron\nNumber of electrons = 1000 C / ( $1.602 \times 10^{-19}$ C/electron) = $6.24 \times 10^{21}$ electrons

Why other options are wrong:
- Option B: Incorrect calculation of the number of electrons.
- Option C: Incorrect calculation of the number of electrons.
- Option D: Incorrect calculation of the number of electrons.

Key Concepts:
- Electric charge
- Charge of an electron
- Current and time relationship

Memory Aids:
- Q = I * t

Quick Tips:
- Remember the charge of an electron.
- Ensure units are consistent.

28 Hard

28 of 94Question permalink

Application +4 | -1

At 25°C, the standard e.m.f. of cell having reactions involving a two electron change is found to be 0.295 V. The equilibrium constant of the reaction is :

View answer and explanation

Correct option: C

{10^{10}}

Option C (

{10^{10}}

) is correct because it is calculated using the relationship between standard EMF and the equilibrium constant,

K = 10^{\frac{{nE^0}}{{0.0591}}}

.

Solution approach

Correct Answer + Core Reason: ${10^{10}}$ . The equilibrium constant is calculated using the relationship between standard EMF and K.

Step-by-Step Solution:

Step 1: Use the relationship between standard EMF and the equilibrium constant: $\Delta G^0 = -nFE^0 = -RT \ln K$

Step 2: Rearrange the equation to solve for K: $\ln K = \frac{{nFE^0}}{{RT}}$

Step 3: Convert to base-10 logarithm: $\log K = \frac{{nFE^0}}{{2.303RT}} = \frac{{nE^0}}{{0.0591}}$ at 25°C

Step 4: Substitute the given values: $n = 2$ and $E^0 = 0.295 V$
$\log K = \frac{{2 \times 0.295}}{{0.0591}} = \frac{{0.59}}{{0.0591}} \approx 10$

Step 5: Calculate K: $K = 10^{10}$

Why other options are wrong:
- **Option A (29.5 $\times {10^{ - 2}}$ ): This is incorrect and likely arises from a misunderstanding of the relationship between E0 and K, or a calculation error.
- Option B (10): This is incorrect because it represents the logarithm of K, not K itself.
- Option D ( $29.5 \times {10^{10}}$ ): This is incorrect and likely arises from a calculation error or misunderstanding of the formula.

Key Concepts:
- Relationship between Gibbs Free Energy, Standard EMF, and Equilibrium Constant
- Nernst Equation

Memory Aids:**
- $\Delta G^0 = -nFE^0 = -RT \ln K$

Quick Tips:
- Make sure to convert between natural logarithm and base-10 logarithm correctly.
- Pay attention to the units and constants used in the formula.

29 Hard

29 of 94Question permalink

Application +4 | -1

Salts of

A

(atomic weight 7),

B

(atomic weight 27) and

C

(atomic weight 48) were electrolyzed under identical conditions using the same quantity of electricity. It was found that when 2.1 g of

A

was deposited, the weights of

B

and

C

deposited were 2.7 g and 7.2 g. The valencies of

A,B{\kern 1pt} \mathrm{and}{\kern 1pt} C

are respectively :

View answer and explanation

Correct option: B

1, 3 and 2

Option B (1, 3 and 2) is correct because the valencies are determined by comparing the ratios of deposited masses to atomic weights, based on Faraday's laws.

Solution approach

Correct Answer + Core Reason: 1, 3 and 2. The valencies are determined by comparing the ratios of deposited masses to atomic weights.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of A = 2.1 g, Atomic weight of A = 7; Mass of B = 2.7 g, Atomic weight of B = 27; Mass of C = 7.2 g, Atomic weight of C = 48.

Step 2: Let the valencies of A, B, and C be $n_A$ , $n_B$ , and $n_C$ respectively.

Step 3: According to Faraday's laws, the ratio of masses deposited is proportional to the ratio of equivalent weights: $\frac{{m_A}}{{E_A}} = \frac{{m_B}}{{E_B}} = \frac{{m_C}}{{E_C}}$ , where E is the equivalent weight.

Step 4: Equivalent weight = Atomic weight / Valency, so $E_A = \frac{{7}}{{n_A}}$ , $E_B = \frac{{27}}{{n_B}}$ , $E_C = \frac{{48}}{{n_C}}$ .

Step 5: Substitute the values: $\frac{{2.1}}{{\frac{{7}}{{n_A}}}} = \frac{{2.7}}{{\frac{{27}}{{n_B}}}} = \frac{{7.2}}{{\frac{{48}}{{n_C}}}}$

Step 6: Simplify: $\frac{{2.1 \times n_A}}{{7}} = \frac{{2.7 \times n_B}}{{27}} = \frac{{7.2 \times n_C}}{{48}}$
$0.3n_A = 0.1n_B = 0.15n_C$

Step 7: Divide by 0.05: $6n_A = 2n_B = 3n_C$

Step 8: Find the smallest integers that satisfy this relationship. Let $n_A = 1$ , then $6 = 2n_B$ , so $n_B = 3$ . Also, $6 = 3n_C$ , so $n_C = 2$ .

Step 9: Therefore, the valencies are 1, 3, and 2 for A, B, and C respectively.

Why other options are wrong:
- Option A (3, 1 and 2): This is incorrect because it assigns the wrong valencies to A, B, and C.
- Option C (3, 1 and 3): This is incorrect because it assigns the wrong valencies to A, B, and C.
- Option D (2, 3 and 2): This is incorrect because it assigns the wrong valencies to A, B, and C.

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 find the smallest integer values for the valencies.

30 Hard

30 of 94Question permalink

Understanding +4 | -1

In an electrolytic cell of

\mathrm{Ag}\left| {\mathrm{AgN}{\mathrm{O}_3}} \right|\mathrm{Ag}

, when current is passed, the concentration of

\mathrm{AgN}{\mathrm{O}_3}

:

View answer and explanation

Correct option: C

Remains same

Option C (Remains same) is correct because in an electrolytic cell with silver electrodes and silver nitrate electrolyte, silver ions are deposited at the cathode and silver from the anode dissolves into the solution, maintaining a constant concentration.

Solution approach

Correct Answer + Core Reason: Remains same. The concentration of $\mathrm{AgN}{\mathrm{O}_3}$ remains the same because the silver ions deposited at the cathode are replenished by the silver ions released from the anode.

Step-by-Step Solution:

Step 1: Understand the process of electrolysis.

Step 2: At the anode (Ag electrode), silver is oxidized: $\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^+(aq) + {e^ - }$

Step 3: At the cathode (Ag electrode), silver ions are reduced: $\mathrm{Ag}^+(aq) + {e^ - } \longrightarrow \mathrm{Ag}(s)$

Step 4: The rate of oxidation at the anode is equal to the rate of reduction at the cathode.

Step 5: Therefore, the concentration of $\mathrm{AgN}{\mathrm{O}_3}$ remains constant.

Why other options are wrong:
- Option A: The concentration does not increase because the silver ions are being deposited at the cathode.
- Option B: The concentration does not decrease because the silver ions are being replenished at the anode.
- Option D: The concentration remains the same, so none of these is incorrect.

Key Concepts:
- Electrolysis
- Oxidation and reduction at electrodes
- Electrolytic cell

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

Quick Tips:
- In electrolytic cells with the same metal as the electrode and electrolyte, the concentration of the electrolyte remains constant.

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