Question

Calculate the standard e.m.f. of the following cell:
Zn $\mid$ Zn$^{2+}$ $\parallel$ Cu$^{2+}$ $\mid$ Cu
Given: $E^\circ_{(Zn^{2+}/Zn)} = -0.76 \, V$ and $E^\circ_{(Cu^{2+}/Cu)} = +0.34 \, V$

Hint:

Always apply the formula $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ for standard cell potential calculations.

Answer 1

Step 1: Identify electrodes.
- Oxidation occurs at the anode (Zn electrode).
- Reduction occurs at the cathode (Cu electrode).
Step 2: Formula for standard e.m.f.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Step 3: Substitute values.
\[ E^\circ_{\text{cell}} = (+0.34 \, V) - (-0.76 \, V) \] \[ E^\circ_{\text{cell}} = 0.34 + 0.76 = 1.10 \, V \] Conclusion:
The standard e.m.f. of the given cell is: \[ \boxed{1.10 \, V} \]