WAEC - Chemistry (2001 - No. 25)
A feasible cell was constructed by joining the two half cells below. Cu\(^{2+}\) (aq) + 2e\(^-\) → Cu(s) E\(^0\) = +0.34 V; Zn\({^2+}\) + 2e\(^-\) → Zn\((_s)\) E\(^0\) = - 0.76V. What is the standard e.m.f. of the cell?
- 1.1V
- 0.42V
+0.42V
+1.1V
Explanation
The standard e.m.f. of a cell is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode:
E\(^0\)\(_{cell}\) = E\(^0\)\(_{cathode}\) - E\(^0\)\(_{anode}\)
Since E\(^0\)\(_{Cu^{2+}}\) > E\(^0\)\(_{Zn^{2+}}\), it can be inferred that Copper is reduced at the cathode while Zinc is oxidized at the anode.
E\(^0\)\(_{cell}\) = E\(^0\)\(_{Cu^{2+}}\) - E\(^0\)\(_{Zn^{2+}}\)
= + 0.34 - (- 0.76)
= + 0.34 + 0.76
= +1.10V
The standard e.m.f. of the cell is +1.10V
Comments (0)
