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Equivalent, Molar Conductivity and Cell Constant

Calculate the...

Question

Calculate the cell potential of the following cell at $25^{o} C$ .
$A g (s) | A g^{+} (a q, 0.001 M) | | A g^{+} (a q, 0.5 M) | A g (s)$

0.52 V

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0.03 V

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1.09 V

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0.159 V

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Solution

The correct option is D $0.159 V$
From cell representation,
$[A g^{+}] = 0.001 M \Rightarrow$ Anode
$[A g^{+}] = 0.5 M \Rightarrow$ Cathode

Cell reaction will be,
$A g^{+} (a q, 0.5 M) \to A g^{+} (a q, 0.2 M)$

By nernst equation,

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{1} l o g \frac{[0.001]}{[0.5]}$

$E_{c e l l} = E_{c e l l}^{0} + \frac{0.0591}{1} l o g \frac{[0.5]}{[0.001]}$

For concentration cells, cathode and anode consist of same metal and its solution.
Thus,
$E^{0} = 0$

$E_{c e l l} = \frac{0.0591}{1} l o g \frac{[0.5]}{[0.001]}$

$E_{c e l l} = \frac{0.0591}{1} l o g 5 \times 10^{2}$

$E_{c e l l} = \frac{0.0591}{1} \times 0.69 + \frac{0.0591}{1} \times 2$

$E_{c e l l} \approx 0.159 V$

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