Electrochemistry Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A current of 1 A deposits 0.635 g of Cu from CuSO\(_4\) in 1930 s. What is the time required to deposit 0.54 g of Al from Al\(_2\)(SO\(_4\))\(_3\) with the same current? (Atomic masses: Cu = 63.5 g/mol, Al = 27 g/mol, F = 96500 C/mol)

Cu: \( Cu^{2+} + 2e^- \rightarrow Cu \), Moles = \( \frac{0.635}{63.5} = 0.01 \, mol \), Charge = \( 0.01 \times 2 \times 96500 = 1930 \, C \), matches \( 1 \times 1930 \).

Al: \( Al^{3+} + 3e^- \rightarrow Al \), Moles = \( \frac{0.54}{27} = 0.02 \, mol \), Charge = \( 0.02 \times 3 \times 96500 = 5790 \, C \).

\( t = \frac{5790}{1} = 5790 \, s \).

5790 s
1930 s
3860 s
9650 s
1

The limiting molar conductivity of CaCl\(_2\) is 258 S cm\(^2\) mol\(^{-1}\). If \( \lambda^\circ_{Ca^{2+}} = 119 \, S \, cm^2 \, mol^{-1} \), what is \( \lambda^\circ_{Cl^-} \)?

\( \Lambda_m^\circ = \lambda^\circ_{Ca^{2+}} + 2\lambda^\circ_{Cl^-} \).

\( 258 = 119 + 2\lambda^\circ_{Cl^-} \), \( 2\lambda^\circ_{Cl^-} = 139 \), \( \lambda^\circ_{Cl^-} = 69.5 \, S \, cm^2 \, mol^{-1} \).

69.5 S cm\(^2\) mol\(^{-1}\)
119 S cm\(^2\) mol\(^{-1}\)
139 S cm\(^2\) mol\(^{-1}\)
258 S cm\(^2\) mol\(^{-1}\)
1

The conductivity of a 0.025 M NaNO\(_3\) solution is 0.003 S cm\(^{-1}\), and its molar conductivity is 120 S cm\(^2\) mol\(^{-1}\). What is the degree of dissociation if \( \Lambda_m^\circ = 126.5 \, S \, cm^2 \, mol^{-1} \)?

\( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{120}{126.5} = 0.9486 \approx 0.95 \).

0.90
0.95
1.00
0.85
2

What is the time (in seconds) required to deposit 0.405 g of calcium from molten CaCl\(_2\) using a current of 1.5 A? (Atomic mass of Ca = 40.5 g/mol, F = 96500 C/mol)

\( Ca^{2+} + 2e^- \rightarrow Ca \). 1 mol Ca (40.5 g) requires 2F.

Moles = \( \frac{0.405}{40.5} = 0.01 \, mol \), Charge = \( 0.01 \times 2 \times 96500 = 1930 \, C \).

\( t = \frac{Q}{I} = \frac{1930}{1.5} = 1286.67 \, s \).

1286.67 s
965 s
1930 s
2573.33 s
1

What is the emf of a cell with the reaction \( Mg(s) + 2Ag^+(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s) \) at standard conditions, given \( E^\circ_{Mg^{2+}/Mg} = -2.36 \, V \) and \( E^\circ_{Ag^+/Ag} = 0.80 \, V \)?

\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-2.36) = 3.16 \, V \).

3.16 V
1.56 V
2.36 V
-3.16 V
1

A current of 0.5 A is passed through a CuSO\(_4\) solution for 9650 s, and then the same current is passed through an AlCl\(_3\) solution for 6433.33 s. What is the total mass of metal deposited? (Atomic masses: Cu = 63.5 g/mol, Al = 27 g/mol, F = 96500 C/mol)

Charge = \( 0.5 \times 9650 = 4825 \, C \).

Cu: \( Cu^{2+} + 2e^- \rightarrow Cu \), Mass = \( \frac{4825}{96500} \times \frac{63.5}{2} = 0.05 \times 31.75 = 1.5875 \, g \).

Charge = \( 0.5 \times 6433.33 = 3216.665 \, C \).

Al: \( Al^{3+} + 3e^- \rightarrow Al \), Mass = \( \frac{3216.665}{96500} \times \frac{27}{3} = 0.03333 \times 9 = 0.3 \, g \).

Total mass = \( 1.5875 + 0.3 = 1.8875 \, g \).

1.8875 g
1.5875 g
2.1875 g
1.2875 g
1

The conductivity of a 0.01 M solution is 0.00141 S cm\(^{-1}\). What is its molar conductivity?

\( \Lambda_m = \frac{\kappa \times 1000}{c} = \frac{0.00141 \times 1000}{0.01} = 141 \, S \, cm^2 \, mol^{-1} \).

14.1 S cm\(^2\) mol\(^{-1}\)
141 S cm\(^2\) mol\(^{-1}\)
1.41 S cm\(^2\) mol\(^{-1}\)
1410 S cm\(^2\) mol\(^{-1}\)
2

The emf of a cell \( Ni(s) | Ni^{2+}(0.001 \, M) || Ag^+(0.002 \, M) | Ag(s) \) at 298 K is (Given: \( E^\circ_{Ni^{2+}/Ni} = -0.25 \, V \), \( E^\circ_{Ag^+/Ag} = 0.80 \, V \))?

\( E^\circ_{cell} = 0.80 - (-0.25) = 1.05 \, V \).

\( E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Ni^{2+}]}{[Ag^+]^2} = 1.05 - \frac{0.059}{2} \log \frac{0.001}{(0.002)^2} = 0.989 \, V \).

1.05 V
0.95 V
0.989 V
1.10 V
3

A cell \( Zn(s) | Zn^{2+}(0.001 \, M) || Br_2(l) | Br^-(0.002 \, M) | Pt(s) \) operates at 298 K. What is the cell potential? (Given: \( E^\circ_{Zn^{2+}/Zn} = -0.76 \, V \), \( E^\circ_{Br_2/Br^-} = 1.07 \, V \))

\( E^\circ_{cell} = 1.07 - (-0.76) = 1.83 \, V \).

\( E_{cell} = 1.83 - \frac{0.059}{2} \log \frac{[Zn^{2+}][Br^-]^2}{1} = 1.83 - 0.0295 \log (0.001 \times 0.000004) = 1.83 + 0.148 = 1.978 \, V \).

1.83 V
1.682 V
2.10 V
1.978 V
4

The molar conductivity of LiCl at infinite dilution is 115.0 S cm\(^2\) mol\(^{-1}\). If \( \lambda^\circ_{Li^+} = 38.7 \, S \, cm^2 \, mol^{-1} \), what is \( \lambda^\circ_{Cl^-} \)?

\( \Lambda_m^\circ = \lambda^\circ_{Li^+} + \lambda^\circ_{Cl^-} \).

\( 115.0 = 38.7 + \lambda^\circ_{Cl^-} \), \( \lambda^\circ_{Cl^-} = 76.3 \, S \, cm^2 \, mol^{-1} \).

76.3 S cm\(^2\) mol\(^{-1}\)
38.7 S cm\(^2\) mol\(^{-1}\)
115.0 S cm\(^2\) mol\(^{-1}\)
19.3 S cm\(^2\) mol\(^{-1}\)
1

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