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Nernst Equation

In the electr...

Question

In the electrochemical cell:
$Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) C u$ , the emf of this Daniell cell is $E_{1}$ . When the concentration of $Z n S O_{4}$ is changed to 1.0 M and that of $C u S O_{4}$ changed to 0.01 M, the emf changes to $E_{2}$ . From the followings, which one is the relationship between $E_{1}$ and $E_{2}$ ? (Given, RT/F = 0.059)

A

E_{1} < E_{2}

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B

E_{1} > E_{2}

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C

E_{2} = - E_{1}

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D

E_{1} = E_{2}

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Solution

The correct option is B $E_{1} > E_{2}$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$E_{1} = E^{0} - \frac{0.059}{2} l o g \frac{0.01}{0}$
$E_{1} = E^{\circ} - \frac{0.059}{2} (- 2) = E^{\circ} + 0.059$
$E_{2} = E^{\circ} - \frac{0.059}{2} l o g \frac{1}{0.01} = E^{\circ} - 0.059$
Hence, $E_{1} > E_{2}$

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