Electrochemistry Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A cell \( Co(s) | Co^{2+}(0.01 \, M) || Br_2(l) | Br^-(0.05 \, M) | Pt(s) \) operates at 298 K. What is the cell potential? (Given: \( E^\circ_{Co^{2+}/Co} = -0.28 \, V \), \( E^\circ_{Br_2/Br^-} = 1.07 \, V \))

\( E^\circ_{cell} = 1.07 - (-0.28) = 1.35 \, V \).

\( E_{cell} = 1.35 - \frac{0.059}{2} \log \frac{[Co^{2+}][Br^-]^2}{1} = 1.35 - 0.0295 \log (0.01 \times 0.0025) = 1.35 + 0.073 = 1.423 \, V \).

1.35 V
1.277 V
1.423 V
1.50 V
3

During electrolysis of aqueous NaBr with inert electrodes, 0.112 L of Br\(_2\) gas (STP) is produced at the anode. How many coulombs were passed? (F = 96500 C/mol)

Anode: \( 2Br^- \rightarrow Br_2 + 2e^- \). 1 mol Br\(_2\) (22.4 L) requires 2F.

Moles = \( \frac{0.112}{22.4} = 0.005 \, mol \), Charge = \( 0.005 \times 2 \times 96500 = 965 \, C \).

1930 C
4825 C
965 C
2895 C
3

What is the time (in seconds) required to deposit 0.355 g of cobalt from a CoSO\(_4\) solution using a current of 0.2 A? (Atomic mass of Co = 59 g/mol, F = 96500 C/mol)

\( Co^{2+} + 2e^- \rightarrow Co \). 1 mol Co (59 g) requires 2F.

Moles = \( \frac{0.355}{59} = 0.006017 \, mol \), Charge = \( 0.006017 \times 2 \times 96500 = 1161.24 \, C \).

\( t = \frac{Q}{I} = \frac{1161.24}{0.2} = 5806.2 \, s \).

5806.2 s
2903.1 s
9650 s
1161.24 s
1

How many coulombs are required to deposit 6.75 g of potassium from a KCl solution? (Atomic mass of K = 39 g/mol, F = 96500 C/mol)

\( K^+ + e^- \rightarrow K \). 1 mol K (39 g) requires 1F.

Moles = \( \frac{6.75}{39} = 0.1731 \, mol \), Charge = \( 0.1731 \times 96500 = 16700 \, C \).

16700 C
96500 C
33400 C
8350 C
1

A cell reaction has \( E^\circ_{cell} = 1.36 \, V \) and an equilibrium constant of \( 10^{46} \) at 298 K. How many electrons are transferred?

\( E^\circ_{cell} = \frac{0.059}{n} \log K_c \), \( 1.36 = \frac{0.059}{n} \times 46 \), \( n = \frac{2.714}{1.36} \approx 2 \).

1
3
2
4
3

In electrolysis, 0.635 g of Cu (atomic mass 63.5 g/mol) is deposited from CuSO\(_4\) using 0.5 A for 3860 s. What is the percentage efficiency of the process? (F = 96500 C/mol)

Charge = \( 0.5 \times 3860 = 1930 \, C \).

Theoretical mass = \( \frac{1930}{96500} \times \frac{63.5}{2} = 0.02 \times 31.75 = 0.635 \, g \).

Efficiency = \( \frac{\text{Actual}}{\text{Theoretical}} \times 100 = \frac{0.635}{0.635} \times 100 = 100\% \).

90%
100%
95%
85%
2

What is the emf of the cell \( Cr(s) | Cr^{3+}(0.002 \, M) || Ag^+(0.05 \, M) | Ag(s) \) at 298 K? (Given: \( E^\circ_{Cr^{3+}/Cr} = -0.74 \, V \), \( E^\circ_{Ag^+/Ag} = 0.80 \, V \))

\( E^\circ_{cell} = 0.80 - (-0.74) = 1.54 \, V \).

\( E_{cell} = 1.54 - \frac{0.059}{3} \log \frac{[Cr^{3+}]}{[Ag^+]^3} = 1.54 - 0.01967 \log \frac{0.002}{0.000125} = 1.54 - 0.025 = 1.515 \, V \).

1.54 V
1.515 V
1.565 V
1.49 V
2

A cell reaction has \( \Delta_r G^\circ = -285900 \, J/mol \) and involves 3 electrons at 298 K. What is the standard emf? (F = 96500 C/mol)

\( \Delta_r G^\circ = -n F E^\circ_{cell} \), \( -285900 = -3 \times 96500 \times E^\circ_{cell} \).

\( E^\circ_{cell} = \frac{285900}{3 \times 96500} = 0.987 \, V \approx 0.99 \, V \).

0.95 V
1.03 V
0.99 V
1.10 V
3

During the electrolysis of aqueous CuSO\(_4\) using copper electrodes, what mass of copper is deposited at the cathode if a current of 1 A flows for 965 seconds? (Faraday constant = 96500 C/mol, Atomic mass of Cu = 63.5 g/mol)

Charge \( Q = I \times t = 1 \times 965 = 965 \, C \).

For \( Cu^{2+} + 2e^- \rightarrow Cu \), 2F (193000 C) deposits 63.5 g Cu.

Mass = \( \frac{63.5 \times 965}{2 \times 96500} = 0.3175 \, g \).

0.3175 g
0.635 g
1.27 g
0.159 g
1

The degree of dissociation of a weak electrolyte is 0.25, and its molar conductivity is 95 S cm\(^2\) mol\(^{-1}\). What is its \( \Lambda_m^\circ \)?

\( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \), \( 0.25 = \frac{95}{\Lambda_m^\circ} \), \( \Lambda_m^\circ = 380 \, S \, cm^2 \, mol^{-1} \).

190 S cm\(^2\) mol\(^{-1}\)
47.5 S cm\(^2\) mol\(^{-1}\)
95 S cm\(^2\) mol\(^{-1}\)
380 S cm\(^2\) mol\(^{-1}\)
4

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