Two students use same stock-solution of ZnSO₄ and solution of CuSO₄. The emf of one cell is 0.03 V higher than the other. The conc. of CuSO₄ in the cell with higher emf value is 0.5 M. Find out the concentration of CuSO₄ in the other cell (2.303 RT/F = 0.06).

Solution

The cell may be represented as

Zn \ {Zn}^{2 +} (C_{1}) ∥ {Cu}^{2 +} (C) \ Cu E_{cell} = E_{1} Zn \ {Zn}^{2 +} (C_{2}) ∥ {Cu}^{2 +} (C = 0.5 M) \ Cu E_{cell} = E_{2} Given E_{2} - E_{1} = 0.03 and C_{1} = C_{2} E_{Cell} = E_{Cell}^{0} - \frac{2.303}{nF} RT \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} for 1 st cell E_{1} = E_{cell}^{0} - \frac{0.06}{2} \log \frac{[C_{1}]}{[C]} for 2 nd cell E_{2} = E_{cell}^{0} - \frac{0.06}{2} \log \frac{[C_{2}]}{[0.5]} E_{2} - E_{1} = \frac{0.06}{2} \log \frac{C_{2}}{C} \times \frac{0.5}{C_{1}} Thus = 0.03 = \frac{0.06}{2} \log \frac{0.5}{C} (∵ C_{1} = C_{2}) C = 0.05 M

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