First slide

Nernst Equation

Question

In the electrochemical cell :

$Zn |{ZnSO}_{4} (0 .01 M) ∥ {CuSO}_{4} (1 .0 M)| Cu$ the emf of this Daniel cell is $E_{1}$ . When the concentration of ${ZnSO}_{4}$ is changed to 1.0 M and that of ${CuSO}_{4}$ changed to 0.01 M,the emf changes to $E_{2}$ . From the following, which one is the relationship between $E_{1} and E_{2}$
(Given, RT/F = 0.059)

Moderate

Solution

$E_{cell} = E_{cell} ° - \frac{0.059}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} E_{1} = E ° - \frac{0.059}{2} \log \frac{0.01}{1} E_{1} = E ° - \frac{0.059}{2} (- 2) = E ° + 0.059 E_{2} = E ° - \frac{0.059}{2} \log \frac{1}{0.01} = E ° - 0.059 Hence, E_{1} > E_{2}$

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