Resistance of 0.2 M solution of an electrolyte is $50{\kern 1pt} {\kern 1pt} \Omega $ . The specific conductance of the solution is 1.3 S ${{\rm{m}}^{ - 1}}$. If resistance of the 0.4 M solution of the same electrolyte is 260Ω , its molar conductivity is

$6.25 \times {10^{ - 4}}{\rm{S}}{\kern 1pt} {\kern 1pt} {{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$

$6250{\kern 1pt} {\kern 1pt} {\rm{S}}{\kern 1pt} {\kern 1pt} {{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$

$625 \times {10^{ - 4}}{\rm{S}}{\kern 1pt} {\kern 1pt} {{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$

$62.5{\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S}}{\kern 1pt} {\kern 1pt} {{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$

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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Hard

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91 Hard

91 of 94Question permalink

Understanding +4 | -1

Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are

View answer and explanation

Correct option: A

Fe is oxidised to

\mathrm{F}{\mathrm{e}^{2 + }}

and dissolved oxygen in water is reduced to

\mathrm{O}{\mathrm{H}^ - }

Option A (Fe is oxidised to

\mathrm{F}{\mathrm{e}^{2 + }}

and dissolved oxygen in water is reduced to

\mathrm{O}{\mathrm{H}^ - }

) is correct because iron is oxidized at the anode and oxygen is reduced at the cathode in the electrochemical process of corrosion.

Solution approach

Correct Answer + Core Reason: Fe is oxidised to $\mathrm{F}{\mathrm{e}^{2 + }}$ and dissolved oxygen in water is reduced to $\mathrm{O}{\mathrm{H}^ - }$ . This describes the electrochemical process of iron corrosion.

Step-by-Step Solution:

Step 1: Understand that corrosion of iron is an electrochemical process.

Step 2: Identify the anode and cathode reactions.
- At the anode, iron is oxidized: $Fe \to F{e^{2 + }} + 2{e^ - }$
- At the cathode, oxygen is reduced: ${O_2} + 2{H_2}O + 4{e^ - } \to 4O{H^ - }$

Step 3: Combine the half-cell reactions to describe the overall process.

Why other options are wrong:
- Option B: Iron is oxidized to $Fe^{2+}$ , not $Fe^{3+}$ , and water is not reduced to $O_2^{2-}$ .
- Option C: Water is not reduced to $O_2^-$ .
- Option D: Water is not reduced to $O_2$ .

Key Concepts:
- Corrosion
- Electrochemical cell
- Oxidation and reduction

Memory Aids:
- Remember the anode and cathode reactions in iron corrosion.

Quick Tips:
- Understand the electrochemical nature of corrosion and the role of oxidation and reduction.

92 Hard

92 of 94Question permalink

Understanding +4 | -1

Statement-I: The voltage of mercury cell remains constant for long period of time.

Statement - II: The net cell reaction does not involve active species(ions) .

View answer and explanation

Correct option: A

Both Statements-I and II are correct and Statement - II is correct explanation for Statement - I.

Option A is correct because the voltage of a mercury cell remains constant due to the net cell reaction not involving ions in solution, which means the concentration of the active species remains constant.

Solution approach

Correct Answer + Core Reason: Both Statements-I and II are correct and Statement - II is correct explanation for Statement - I. The constant voltage is due to the absence of ions in the net cell reaction.

Step-by-Step Solution:

Step 1: Analyze Statement-I: The voltage of a mercury cell remains constant for a long period of time.\nThis is a characteristic feature of mercury cells due to their unique construction and the nature of the reactions involved.\nTherefore, Statement-I is true.

Step 2: Analyze Statement-II: The net cell reaction does not involve active species(ions).\nThe net cell reaction in a mercury cell is: $Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)$ . This reaction does not involve any ions in the solution; it only involves solid and liquid phases.\nTherefore, Statement-II is true.

Step 3: Determine if Statement-II is the correct explanation for Statement-I.\nSince the net cell reaction does not involve ions, the concentration of the active species remains constant during the cell's operation. This constant concentration leads to a constant cell potential (voltage) according to the Nernst equation.\nTherefore, Statement-II is the correct explanation for Statement-I.

Why other options are wrong:
- Option B: Statement-II is the correct explanation for Statement-I.
- Option C: Statement-II is true, so this option is incorrect.
- Option D: Statement-I is true, so this option is incorrect.

Key Concepts:
- Mercury Cell
- Cell Potential
- Nernst Equation

Memory Aids:
- Remember the net cell reaction of a mercury cell to understand why its voltage remains constant.

Quick Tips:
- Understand the construction and reactions of different types of electrochemical cells to analyze their properties.

93 Hard

93 of 94Question permalink

Application +4 | -1

The resistance of a decinormal solution of a salt occupying a volume between two platinum electrodes 1.80 cm apart and

5.4\mathrm{c}{\mathrm{m}^2}

in area was formed to be 32 ohm. The specific and equivalent conductivity respectively in their proper units are :

View answer and explanation

Correct option: C

0.0104 and 104.0

Option C (0.0104 and 104.0) is correct because it calculates specific conductivity using the formula and equivalent conductivity by multiplying with 1000.

Solution approach

Correct Answer + Core Reason: 0.0104 and 104.0. Specific conductivity is calculated using cell constant and resistance, and equivalent conductivity is specific conductivity multiplied by 1000.

Step-by-Step Solution:

Step 1: Define the given values: Resistance (R) = 32 ohm, distance between electrodes (l) = 1.80 cm, area of electrodes (A) = 5.4 cm², concentration = decinormal = 0.1 N

Step 2: Calculate the cell constant (G*): $G^* = \frac{l}{A} = \frac{1.80}{5.4} = \frac{1}{3} cm^{-1}$

Step 3: Calculate specific conductivity (κ): $κ = \frac{G^*}{R} = \frac{1/3}{32} = \frac{1}{96} S cm^{-1} \approx 0.0104 S cm^{-1}$

Step 4: Calculate equivalent conductivity ( $\Lambda_{eq}$ ): $\Lambda_{eq} = κ \times 1000/N = 0.0104 \times 1000/0.1 = 104.0 S cm^2 eq^{-1}$

Step 5: Final Answer: Specific conductivity = 0.0104 S cm⁻¹, Equivalent conductivity = 104.0 S cm² eq⁻¹

Why other options are wrong:
- Option A (104.1 and 0.0104): This reverses the values of specific and equivalent conductivity.
- Option B (208.2 and 0.0208): This doubles the correct values, likely due to a miscalculation.
- Option D (None of these): The correct answer is present in option C.

Key Concepts:
- Specific conductivity
- Equivalent conductivity
- Cell constant

Memory Aids:
- Cell constant: G* = l/A
- Specific conductivity: κ = 1/R * (l/A)
- Equivalent conductivity: Λeq = (κ * 1000) / N

Quick Tips:
- Always ensure correct units are used in calculations.
- Remember the formula for cell constant and its relation to specific conductivity.

94 Hard

94 of 94Question permalink

Application +4 | -1

Resistance of 0.2 M solution of an electrolyte is

50{\kern 1pt} {\kern 1pt} \Omega

. The specific conductance of the solution is 1.3 S

{\mathrm{m}^{ - 1}}

.
If resistance of the 0.4 M solution of the same electrolyte is 260Ω , its molar conductivity is

View answer and explanation

Correct option: B

6.25 \times {10^{ - 4}}\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}

Option B (

6.25 \times {10^{ - 4}}\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}

) is correct because it correctly calculates the molar conductivity using the given resistance and specific conductance.

Solution approach

Correct Answer + Core Reason: $6.25 \times {10^{ - 4}}\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}$ . Molar conductivity is calculated using specific conductance and concentration.

Step-by-Step Solution:

Step 1: Calculate the cell constant (G*) using the 0.2 M solution data: We know that specific conductance (κ) = $\frac{G^*}{R}$ , so $G^* = κ \times R = 1.3 S m^{-1} \times 50 \Omega = 65 m^{-1}$ .

Step 2: Calculate the specific conductance (κ) of the 0.4 M solution: $κ = \frac{G^*}{R} = \frac{65 m^{-1}}{260 \Omega} = 0.25 S m^{-1}$ .

Step 3: Calculate the molar conductivity ( $\Lambda_m$ ) of the 0.4 M solution: $\Lambda_m = \frac{κ}{C}$ , where C is the concentration in mol/m³. Since the concentration is given in mol/L (M), we need to convert it to mol/m³: 0.4 M = 0.4 mol/L = 0.4 mol/(10⁻³ m³) = 400 mol/m³.

Step 4: Substitute the values: $\Lambda_m = \frac{0.25 S m^{-1}}{400 mol/m^3} = 6.25 \times 10^{-4} S m^2 mol^{-1}$ .

Step 5: Final Answer: The molar conductivity of the 0.4 M solution is $6.25 \times {10^{ - 4}}\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}$ .

Why other options are wrong:
- **Option A ( $6250{\kern 1pt} {\kern 1pt} \mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}$ ): This is incorrect due to a miscalculation or incorrect unit conversion.
- Option C ( $625 \times {10^{ - 4}}\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}$ ): This is incorrect due to a miscalculation or incorrect unit conversion.
- Option D ( $62.5{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}\mathrm{mo}{\mathrm{l}^{ - 1}}$ ): This is incorrect due to a miscalculation or incorrect unit conversion.

Key Concepts:
- Specific conductance
- Molar conductivity
- Cell constant

Memory Aids:**
- Molar conductivity: $\Lambda_m = \frac{κ}{C}$
- Specific conductance: κ = G*/R

Quick Tips:
- Ensure correct units are used in calculations, especially when converting between L and m³.
- Remember the relationship between specific conductance, molar conductivity, and concentration.

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