Question

Calculate the electrode potential of a half-cell for a zinc electrode dipping in 0·01 M $\mathrm{ZnSO_4}$ solution at 25°C. [Given : $\mathrm{E^\circ_{Zn^{2+}/Zn} = -0.76\,V}$, log 10 = 1]

Hint:

Nernst: E = E° − (0.059/2) log(1/[Zn²⁺]).

Answer 1

Concept: When the ion concentration is not the standard 1 M, we cannot use $\mathrm{E^\circ}$ directly. We correct it using the Nernst equation.

Step 1: Write the Nernst equation
For the half-reaction $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$, two electrons are involved, so $n = 2$:\[ E = E^\circ - \frac{0.059}{2}\log\frac{1}{[Zn^{2+}]} \]

Step 2: Put in the numbers
Here $\mathrm{E^\circ = -0.76\,V}$ and $\mathrm{[Zn^{2+}] = 0.01\,M}$, so $\dfrac{1}{0.01} = 100$ and $\log 100 = 2$:\[ E = -0.76 - \frac{0.059}{2}\times 2 \]

Step 3: Work it out
\[ E = -0.76 - 0.0295\times 2 = -0.76 - 0.059 = -0.819\ \text{V} \]

Answer: The electrode potential is $\mathrm{E = -0.819\ V}$.