The cell reaction for the given cell is $P t (H_{2} (P_{1})) | p H = 2 | | p H = 3 | P t (H_{2} (P_{2}))$
where , $P_{1} = 1 a t m$ and $P_{2} = 1 a t m$ .

spontaneous

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non-spontaneous

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in equiliibrium

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either of these

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Solution

The correct option is B non-spontaneous
According to Nernst equations, $E = E^{o} - \frac{0.0591}{n} l o g Q$
$E_{c e l l} = E^{o} + \frac{0.059}{2} l o g \frac{[H^{+}]_{R H E}^{2} P_{1}}{[H^{+}]_{L H E}^{2} P_{2}}$
$= \frac{0.059}{2} l o g \frac{10^{- 3} \times 10^{- 3}}{10^{- 2} \times 10^{- 2}}$
$= - 0.059 V$
As emf value is negative so reaction is non-spontaneous in forward direction.

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