Electrochemistry Chapter-Wise Test 8

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the standard emf of a cell with the reaction \( Pb(s) + Hg_2^{2+}(aq) \rightarrow Pb^{2+}(aq) + 2Hg(l) \)? (Given: \( E^\circ_{Pb^{2+}/Pb} = -0.13 \, V \), \( E^\circ_{Hg_2^{2+}/Hg} = 0.79 \, V \))

\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.79 - (-0.13) = 0.92 \, V \).

0.92 V
0.66 V
1.05 V
-0.92 V
1

How many electrons are involved in the reduction of 1 mole of \( Fe^{3+} \) to \( Fe \)?

\( Fe^{3+} + 3e^- \rightarrow Fe \). 3 electrons are required per mole.

1
3
2
4
2

What is the emf of a cell \( Cd(s) | Cd^{2+}(0.1 \, M) || Ag^+(0.001 \, M) | Ag(s) \) at 298 K? (Given: \( E^\circ_{Cd^{2+}/Cd} = -0.40 \, V \), \( E^\circ_{Ag^+/Ag} = 0.80 \, V \))

\( E^\circ_{cell} = 0.80 - (-0.40) = 1.20 \, V \).

\( E_{cell} = 1.20 - \frac{0.059}{2} \log \frac{0.1}{0.001} = 1.20 - 0.059 = 1.141 \, V \).

1.20 V
1.259 V
1.10 V
1.141 V
4

How many coulombs are required to reduce 1 mole of \( Br_2 \) to \( Br^- \)? (F = 96500 C/mol)

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

Charge = \( 2 \times 96500 = 193000 \, C \).

96500 C
48250 C
289500 C
193000 C
4

How many coulombs are required to reduce 1 mole of \( MnO_4^- \) to \( Mn^{2+} \) in acidic medium? (Faraday constant = 96500 C/mol)

\( MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \).

Charge = \( 5 \times 96500 = 482500 \, C \).

482500 C
96500 C
193000 C
289500 C
1

A cell \( Cu(s) | Cu^{2+}(0.01 \, M) || Cl_2(g)(0.5 \, atm) | Cl^-(0.05 \, M) | Pt(s) \) operates at 298 K. What is the cell potential? (Given: \( E^\circ_{Cu^{2+}/Cu} = 0.34 \, V \), \( E^\circ_{Cl_2/Cl^-} = 1.36 \, V \))

\( E^\circ_{cell} = 1.36 - 0.34 = 1.02 \, V \).

\( E_{cell} = 1.02 - \frac{0.059}{2} \log \frac{[Cu^{2+}][Cl^-]^2}{P_{Cl_2}} = 1.02 - 0.0295 \log \frac{0.01 \times 0.0025}{0.5} = 1.02 + 0.053 = 1.073 \, V \).

1.02 V
0.967 V
1.15 V
1.073 V
4

What is the potential of a hydrogen electrode in a solution with pH = 2 at 298 K? (Given: \( E^\circ_{H^+/H_2} = 0.00 \, V \))

\( E = E^\circ - \frac{0.059}{n} \log \frac{1}{[H^+]} \), \( [H^+] = 10^{-2} \), \( n = 1 \).

\( E = 0 - 0.059 \times 2 = -0.118 \, V \).

0.00 V
-0.059 V
-0.177 V
-0.118 V
4

In a lead storage battery during discharge, what is the total change in oxidation state of lead from reactants to products at both electrodes?

Anode: \( Pb (0) + SO_4^{2-} \rightarrow PbSO_4 (+2) + 2e^- \), Change = +2.

Cathode: \( PbO_2 (+4) + SO_4^{2-} + 4H^+ + 2e^- \rightarrow PbSO_4 (+2) + 2H_2O \), Change = -2.

Total change = \( |+2| + |-2| = 4 \).

4
2
6
0
1

In a lead storage battery during charging, what is the oxidation state of lead in the anode product?

Charging anode: \( PbSO_4 + 2e^- \rightarrow Pb + SO_4^{2-} \). Pb has oxidation state 0.

0
+2
+4
+1
1

What is the emf of the cell \( Pb(s) | Pb^{2+}(0.005 \, M) || Ag^+(0.05 \, M) | Ag(s) \) at 298 K? (Given: \( E^\circ_{Pb^{2+}/Pb} = -0.13 \, V \), \( E^\circ_{Ag^+/Ag} = 0.80 \, V \))

\( E^\circ_{cell} = 0.80 - (-0.13) = 0.93 \, V \).

\( E_{cell} = 0.93 - \frac{0.059}{2} \log \frac{0.005}{0.05^2} = 0.93 - 0.0295 = 0.9005 \, V \).

0.93 V
0.9005 V
0.9595 V
0.87 V
2

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