A galvanic cell is made up to S.H.E and another hydrogen electrode. The other hydrogen should be immersed in which of the following solutions to get maximum value of e.m.f :

0.1 M H C l

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0.1 M 3 C H_{3} C O O H

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0.1 M H_{2} S O_{4}

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0.1 M H_{3} P O_{4}

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Solution

The correct option is B $0.1 M 3 C H_{3} C O O H$
The Nernst equation for $E_{H_{2}}$ for reduction is $E_{H_{2}} = E_{0}^{H_{2}} - \frac{0.0592}{n} l o g \frac{P_{H_{2}}}{{⌊ H^{+} ⌋}^{2}}$ , but $E_{0}^{H 2} = 0.$
When concentration is less, the value of e.m.f is high. Hence for acetic acid (a weak electrolyte with low degree of dissociation) the $H^{+}$ ion concentration is low and the e.m.f of cell is maximum.

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