Question

Calculate E at 25° C for a galvanic cell based on the reaction:

Zn(s) + Cu²⁺(aq)→Zn ²⁺(aq) + Cu (s)    in which [Zn²⁺]=0.55M and [Cu²⁺]=1.02M

Answer 1

Zn(s) + Cu²⁺(aq) → Zn ²⁺(aq) + Cu (s)

[Zn²⁺] = 0.55 M

[Cu²⁺] = 1.02 M

The standard potentials of the half cell reactions are given as:

At cathode:   Cu²⁺(aq) + 2e- →   Cu (s)   .........Eo = 0.34 V

At anode:     Zn(s) →    Zn ²⁺(aq) + 2e-   ..........Eo = - 0.76 V

Eo cell = Eo cathode - Eo anode

= 0.34 - (-0.76)

= - 1.10 V

Now, using nernst equation:

E cell = Eo cell - RT/nF ln [Reduction] / [Oxidation]

= - 1.10 - (8.314 * 298) / (2 * 96485) ln (1.02 / 0.55)

= - 1.10 - 0.0128 * ln 0.561

= - 1.10 - 0.0128 * (-0.578)

= - 1.10 + 0.0074

= - 1.09 V