Question

Calculate the electrode potential of a half-cell for zinc electrode dipping in \(0.01\,M\) \(ZnSO_4\) solution at \(25^\circ C\). \[ \text{Given : } E^\circ_{Zn^{2+}/Zn}=-0.76\,V \] \[ \log 10 = 1 \]

Hint:

For metal-ion electrodes, decreasing the concentration of metal ions makes the electrode potential more negative than its standard electrode potential.

Answer 1

Step 1: Nernst equation for the zinc half-cell.
For $Zn^{2+} + 2e^- \rightarrow Zn(s)$ with $n = 2$: \[ E = E^\circ + \frac{0.0591}{2}\log[Zn^{2+}] \]
Step 2: Substitute the given values.
$E^\circ = -0.76\,V$, $[Zn^{2+}] = 0.01 = 10^{-2}$, so $\log(10^{-2}) = -2$: \[ E = -0.76 + \frac{0.0591}{2} \times (-2) = -0.76 - 0.0591 \]
Step 3: Final electrode potential.
$E = -0.8191\,V$
\[ \boxed{E \approx -0.82\,V} \]