Electrochemistry Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A lead storage battery delivers 1.5 A for 10 hours during discharge. How many grams of PbSO\(_4\) (molar mass 303 g/mol) are formed at the anode? (F = 96500 C/mol)

Charge = \( 1.5 \times 10 \times 3600 = 54000 \, C \).

Anode: \( Pb + SO_4^{2-} \rightarrow PbSO_4 + 2e^- \). 1 mol PbSO\(_4\) requires 2F.

Faradays = \( \frac{54000}{96500} = 0.5596 \, F \), Moles = \( \frac{0.5596}{2} = 0.2798 \, mol \), Mass = \( 0.2798 \times 303 = 84.78 \, g \).

42.39 g
169.56 g
30.3 g
84.78 g
4

During electrolysis of aqueous NaI with inert electrodes, 0.224 L of H\(_2\) (STP) is produced at the cathode. How many coulombs were passed? (F = 96500 C/mol)

Cathode: \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \).

Moles = \( \frac{0.224}{22.4} = 0.01 \, mol \), Charge = \( 0.01 \times 2 \times 96500 = 1930 \, C \).

965 C
3860 C
1930 C
4825 C
3

A weak acid has \( \Lambda_m^\circ = 350 \, S \, cm^2 \, mol^{-1} \) and \( \Lambda_m = 17.5 \, S \, cm^2 \, mol^{-1} \) at 0.05 M. What is the dissociation constant \( K_a \)?

\( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{17.5}{350} = 0.05 \).

\( K_a = \frac{\alpha^2 c}{1 - \alpha} = \frac{0.05^2 \times 0.05}{1 - 0.05} = \frac{0.0025 \times 0.05}{0.95} = 1.316 \times 10^{-4} \).

1.25 \times 10^{-3}
2.5 \times 10^{-4}
5.0 \times 10^{-5}
1.316 \times 10^{-4}
4

A lead storage battery delivers 1 A for 4825 s. What is the mass of PbSO\(_4\) (molar mass 303 g/mol) formed at the cathode? (F = 96500 C/mol)

Charge = \( 1 \times 4825 = 4825 \, C \).

Cathode: \( PbO_2 + SO_4^{2-} + 4H^+ + 2e^- \rightarrow PbSO_4 + 2H_2O \), 1 mol PbSO\(_4\) requires 2F.

Faradays = \( \frac{4825}{96500} = 0.05 \, F \), Moles = \( \frac{0.05}{2} = 0.025 \, mol \), Mass = \( 0.025 \times 303 = 7.575 \, g \).

7.575 g
15.15 g
3.7875 g
30.3 g
1

The resistance of a conductivity cell with 0.001 M KCl solution is 1500 Ω, and its conductivity is 0.146 × 10\(^{-3}\) S cm\(^{-1}\). What is the cell constant?

\( \kappa = \frac{\text{cell constant}}{R} \), \( 0.146 \times 10^{-3} = \frac{\text{cell constant}}{1500} \), cell constant = 0.219 cm\(^{-1}\).

0.146 cm\(^{-1}\)
1.500 cm\(^{-1}\)
0.219 cm\(^{-1}\)
0.073 cm\(^{-1}\)
3

How many electrons are involved in the reduction of 1 mole of \( MnO_4^- \) to \( MnO_2 \) in basic medium?

\( MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \). 3 electrons per mole.

5
3
2
4
2

A hydrogen electrode operates at 298 K with \( P_{H_2} = 0.1 \, atm \) and pH = 4. What is its potential? (Given: \( E^\circ_{H^+/H_2} = 0.00 \, V \))

\( E = E^\circ - \frac{0.059}{2} \log \frac{P_{H_2}}{[H^+]^2} \), \( [H^+] = 10^{-4} \).

\( E = 0 - 0.0295 \log \frac{0.1}{10^{-8}} = 0 - 0.0295 \times 7 = -0.2065 \, V \).

0.00 V
-0.118 V
-0.177 V
-0.2065 V
4

In a lead storage battery, what is the electrolyte?

The electrolyte in a lead storage battery is dilute sulfuric acid (H\(_2\)SO\(_4\)).

PbSO\(_4\)
H\(_2\)O
PbO\(_2\)
H\(_2\)SO\(_4\)
4

What is the emf of the cell \( Mn(s) | Mn^{2+}(0.05 \, M) || H^+(0.005 \, M) | H_2(g)(1 \, bar) | Pt(s) \) at 298 K? (Given: \( E^\circ_{Mn^{2+}/Mn} = -1.18 \, V \), \( E^\circ_{H^+/H_2} = 0.00 \, V \))

\( E^\circ_{cell} = 0.00 - (-1.18) = 1.18 \, V \).

\( E_{cell} = 1.18 - \frac{0.059}{2} \log \frac{0.05}{0.005^2} = 1.18 - 0.07375 = 1.10625 \, V \).

1.18 V
1.25375 V
1.10625 V
1.05 V
3

A cell reaction has \( \Delta_r G^\circ = -212300 \, J/mol \) and involves 2 electrons at 298 K. What is the equilibrium constant?

\( \Delta_r G^\circ = -n F E^\circ_{cell} \), \( E^\circ_{cell} = \frac{212300}{2 \times 96500} = 1.10 \, V \).

\( E^\circ_{cell} = \frac{0.059}{n} \log K_c \), \( 1.10 = \frac{0.059}{2} \log K_c \), \( \log K_c = 37.29, K_c \approx 10^{37} \).

10\(^{35}\)
10\(^{36}\)
10\(^{37}\)
10\(^{38}\)
3

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