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

Concept: The electrode potential of an electrode under non-standard conditions is calculated using the Nernst equation. The Nernst equation establishes a relationship between the electrode potential and the concentration of ions participating in the electrode reaction. For a metal-metal ion electrode, \[ M^{n+}+ne^- \rightarrow M \] the Nernst equation at \(25^\circ C\) is \[ E=E^\circ-\frac{0.0591}{n}\log\frac{[M]}{[M^{n+}]} \] Since the activity of a pure solid metal is taken as unity, the equation becomes \[ E=E^\circ+\frac{0.0591}{n}\log[M^{n+}] \]

Step 1: Write the electrode reaction For zinc electrode, \[ Zn^{2+}+2e^- \rightarrow Zn(s) \] Here, \[ n=2 \] and \[ E^\circ=-0.76\,V \]

Step 2: Write the Nernst equation \[ E=E^\circ+\frac{0.0591}{2}\log[Zn^{2+}] \] Given concentration, \[ [Zn^{2+}]=0.01=10^{-2} \] Substituting, \[ E=-0.76+\frac{0.0591}{2}\log(10^{-2}) \]

Step 3: Simplify the logarithmic term \[ \log(10^{-2})=-2 \] Therefore, \[ E=-0.76+\frac{0.0591}{2}(-2) \] \[ E=-0.76-0.0591 \] \[ E=-0.8191\,V \]

Final Answer \[ \boxed{E=-0.819\,V} \] Thus, the electrode potential of the zinc electrode in \(0.01\,M\) \(ZnSO_4\) solution is approximately \[ \boxed{-0.82\,V} \]