Question

The standard EMF for the cell reaction,
\[\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}\], 
is 1.10 V at 25°C. The EMF of the cell reaction, when 0.1 M Cu²⁺ and 0.1 M Zn²⁺ solutions are used at 25°C is:

Options

• 1.10 V

• 0.110 V

• −1.10 V

• −0.110 V

Answer 1

1.10 V

Explanation:

Given: Standard EMF of the cell, \[\ce{E^{\circ}_{cell}}\] = 1.10 V

Temperature, T = 25°C = 298 K

[Cu²⁺] = 0.1 M 

[Zn²⁺] = 0.1 M 

Number of electrons transferred, n = 2

The given cell reaction is:

\[\ce{Zn + Cu^{2+} -> Cu + Zn^{2+}}\]

By using the Nernst equation:

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

= \[\ce{1.10 - \frac{0.0591}{2} log \frac{0.1}{0.1}}\]

= \[\ce{1.10 - \frac{0.0591}{2} log (1)}\]

= \[\ce{1.10 - \frac{0.0591}{2} \times 0}\]

= 1.10 V

∴ The EMF of the cell under the given conditions is 1.10 V.