Question

For the redox reaction, 

\[\ce{Zn_{(s)} + Cu^2+ (0.1 M) -> Zn^2+ (1M) + Cu_{(s)}}\],

taking place in a cell, \[\ce{E{^{\circ}_{cell}}}\] is 1.10 volt. E_cell for the cell will be: \[\ce{(2.303 \frac{RT}{F} = 0.0591)}\]

विकल्प

• 2.14 V

• 1.80 V

• 1.07 V

• 0.82 V

Answer 1

1.07 V

Explanation:

Given: The given reaction is:

\[\ce{Zn_{(s)} + Cu^2+ (0.1 M) -> Zn^2+ (1M) + Cu_{(s)}}\]

\[\ce{E^{\circ}_{cell}}\] ​= 1.10 V

[Cu²⁺] = 0.1 M

[Zn²⁺] = 1.0 M

n = 2 (electrons transferred)

2.303⋅RTF = 0.0591 V at 25°C

By using the Nernst equation:

\[\ce{E_{cell} = E{^{\circ}_{cell}} - \frac{0.0591}{n} log_10 \frac{[Zn^{2+}]}{[Cu^{2+}]}}\]

∴ \[\ce{E_{cell} = 1.10 - \frac{2.303 RT}{2} log_10 \frac{1}{0.1}}\]

= \[\ce{1.10 - \frac{0.0591}{2}}\]

= 1.07 V