Electrochemistry Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

How many Faradays are required to reduce 1 mole of \( Cu^{2+} \) to \( Cu^+ \)?

\( Cu^{2+} + e^- \rightarrow Cu^+ \). 1 mole requires 1F.

2 F
3 F
1 F
0.5 F
3

A hydrogen electrode in a solution with \( P_{H_2} = 2 \, atm \) and \( [H^+] = 0.002 \, M \) operates at 298 K. What is its potential? (Given: \( E^\circ_{H^+/H_2} = 0.00 \, V \))

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

\( E = 0 - 0.0295 \log \frac{2}{0.000004} = 0 - 0.0295 \times 5.699 = -0.1681 \, V \).

0.00 V
-0.118 V
-0.236 V
-0.1681 V
4

A cell \( Fe(s) | Fe^{2+}(0.02 \, M) || I_2(s) | I^-(0.1 \, M) | Pt(s) \) operates at 298 K. What is the cell potential? (Given: \( E^\circ_{Fe^{2+}/Fe} = -0.44 \, V \), \( E^\circ_{I_2/I^-} = 0.54 \, V \))

\( E^\circ_{cell} = 0.54 - (-0.44) = 0.98 \, V \).

\( E_{cell} = 0.98 - \frac{0.059}{2} \log \frac{[Fe^{2+}][I^-]^2}{1} = 0.98 - 0.0295 \log (0.02 \times 0.01) = 0.98 + 0.068 = 1.048 \, V \).

0.98 V
0.912 V
1.10 V
1.048 V
4

In a dry cell, what is the approximate cell potential?

The dry cell (Leclanché cell) has a potential of approximately 1.5 V.

1.35 V
2.0 V
1.1 V
1.5 V
4

What is the potential of a hydrogen electrode in a solution with \( [H^+] = 10^{-5} \, M \) at 298 K? (Given: \( E^\circ_{H^+/H_2} = 0.00 \, V \))

\( E = E^\circ - \frac{0.059}{n} \log \frac{1}{[H^+]} \), \( [H^+] = 10^{-5} \), \( n = 1 \).

\( E = 0 - 0.059 \times 5 = -0.295 \, V \).

0.00 V
-0.118 V
-0.177 V
-0.295 V
4

In a dry cell, the cathode reaction produces MnO(OH). If 0.87 g of MnO\(_2\) (atomic mass Mn = 55, O = 16) is reduced, how many coulombs are consumed? (F = 96500 C/mol)

\( MnO_2 + H^+ + e^- \rightarrow MnO(OH) \). Molar mass = \( 55 + 32 = 87 \, g/mol \).

Moles = \( \frac{0.87}{87} = 0.01 \, mol \), Charge = \( 0.01 \times 96500 = 965 \, C \).

1930 C
4825 C
965 C
2895 C
2

How many coulombs are required to deposit 0.965 g of barium from a BaCl\(_2\) solution? (Atomic mass of Ba = 137 g/mol, F = 96500 C/mol)

\( Ba^{2+} + 2e^- \rightarrow Ba \). 1 mol Ba (137 g) requires 2F.

Moles = \( \frac{0.965}{137} = 0.00704 \, mol \), Charge = \( 0.00704 \times 2 \times 96500 = 1358.24 \, C \).

1358.24 C
9650 C
679.12 C
19300 C
1

What is the emf of the cell \( Sn(s) | Sn^{2+}(0.002 \, M) || Pb^{2+}(0.02 \, M) | Pb(s) \) at 298 K? (Given: \( E^\circ_{Sn^{2+}/Sn} = -0.14 \, V \), \( E^\circ_{Pb^{2+}/Pb} = -0.13 \, V \))

\( E^\circ_{cell} = -0.13 - (-0.14) = 0.01 \, V \).

\( E_{cell} = 0.01 - \frac{0.059}{2} \log \frac{0.002}{0.02} = 0.01 + 0.0295 = 0.0395 \, V \).

0.01 V
0.0395 V
-0.0195 V
0.05 V
2

How many electrons are transferred in the reduction of 1 mole of \( NO_3^- \) to \( NH_4^+ \) in acidic medium?

\( NO_3^- + 10H^+ + 8e^- \rightarrow NH_4^+ + 3H_2O \). 8 electrons per mole.

6
8
4
10
2

What is the emf of a cell \( Fe(s) | Fe^{2+}(0.1 \, M) || H^+(0.001 \, M) | H_2(g)(1 \, bar) | Pt(s) \) at 298 K? (Given: \( E^\circ_{Fe^{2+}/Fe} = -0.44 \, V \), \( E^\circ_{H^+/H_2} = 0.00 \, V \))

\( E^\circ_{cell} = 0.00 - (-0.44) = 0.44 \, V \).

\( E_{cell} = 0.44 - \frac{0.059}{2} \log \frac{0.1}{0.001^2} = 0.44 - 0.0885 = 0.3515 \, V \).

0.44 V
0.5285 V
0.40 V
0.3515 V
4

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