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Electrochemical Cells, Galvanic Cells

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 Daniel 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, $\frac{R T}{F} = 0.059)$ .

A

E_{2} = 0 \neq E_{1}

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B

E_{1} = E_{2}

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C

E_{1} < E_{2}

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D

E_{1} > E_{2}

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Solution

The correct option is B $E_{1} > E_{2}$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
$E_{1} = 1.1 - \frac{0.059}{2} l o g \frac{0.01}{1.0}$
$E_{1} = 1.1 - \frac{0.059}{2} \times (- 2)$
$E_{1} = 1.1 - (- 0.0592)$
$E_{1} = 1.16 V$
Again,

$E_{2} = 1.1 - \frac{0.059}{2} l o g \frac{1.0}{0.01}$
$E_{1} = 1.1 - \frac{0.059}{2} \times (2)$
$E_{1} = 1.1 - (0.059)$
$E_{2} = 1.041 V$

Hence, $E_{1} > E_{2}$

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