Chemistry · General chemistry II · Worked example

Use the Nernst equation

Find the potential of the zinc–copper cell at 298 K when [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M.

Write Q

For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), the solids are left out.

Q = \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} = \frac{1.0}{0.010} = 100

n = 2 and log 100 = 2.

E = 1.10 - \frac{0.0592}{2} log 100 = 1.04 V

\begin{array}{c} E = 1.10 - \frac{0.0592}{2} log 100 \\ = 1.04 V \end{array}

Scarce Cu²⁺ lowers the potential. As the cell discharges, Q keeps rising and E keeps falling, until E = 0 when Q reaches K.

E = 1.04 V.

Find E when [Zn²⁺] = 0.10 M and [Cu²⁺] = 1.0 M.

Show the answer and explanation

Q = 0.10/1.0 = 0.10 and log 0.10 = −1, so E = 1.10 V + 0.0296 V = 1.13 V.

1.10 - \frac{0.0592}{2} log (0.10) \approx 1.13

In Electrochemistry & charge, set Q = 0.010, with the concentrations reversed. The potential rises above the standard value, to 1.16 V.

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