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EMF

Two students ...

Question

Two students use same stock solution of $Z n S O_{4}$ and a solution of $C u S O_{4}$ . The emf of one cell is $0.03 V$ higher than that of other. The concentration of $C u S O_{4}$ in the cell with higher emf value is $0.5 M$ . Find out the concentration of $C u S O_{4}$ in the other cell. $(\frac{2.303 R T}{F} = 0.06)$ .

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Solution

Student I.
$Z n (s) | Z n S O_{4} (C_{1} | | 0.5 M C u S O_{4} | C u (s)$

$E_{1} = E^{\circ} - \frac{0.06}{2} log \frac{C_{1}}{0.5} . . . . (i)$

Student II. $Z n (s) | Z n S O_{4} (C_{1}) | | C u S O_{4} (C_{2}) | C u (s)$

$E_{2} = E^{\circ} - \frac{0.06}{2} log \frac{C_{1}}{C_{2}} . . . . (i i)$

$E_{1} - E_{2} = \frac{0.06}{2} [log \frac{C_{1}}{C_{2}} - log \frac{C_{1}}{0.5}]$

$0.03 = \frac{0.06}{2} [log (\frac{0.5}{C_{2}})]$

$1 = log \frac{0.5}{C_{2}}$

$C_{2} = 0.05 M$ .

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