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Nernst Equation

Consider the ...

Question

Consider the reaction at $T = 298 K$ , ${C l}_{2} (g) + {2 B r}^{-} (a q) ⟶ {2 C l}^{-} (a q) + {B r}_{2} (a q)$ . The emf of the cell, when $[{C l}^{-}] = [{B r}^{-}] = 0.01 M$ and ${C l}_{2}$ gas is 1 atm pressure, will be ( $E_{c e l l}^{o}$ for the above reaction is $= 0.29 V$ ) :-

0.54

volt

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0.35

volt

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0.24

volt

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0.29

volt

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Solution

The correct option is C $0.29$ volt
$E_{c e l l} = 0.29 - \frac{0.0591}{2} log \frac{[{C l}^{-}]^{2} [{B r}_{2}]}{[{B r}^{-}]^{2} [{C l}_{2}]}$
$E_{c e l l} = 0.29 - \frac{0.0591}{2} log \frac{[10^{- 2}]^{2} [1]}{[10^{- 2}]^{2} [1]}$

$E_{c e l l} = 0.29 + \frac{0.0591}{2} \times 0 = 0.29 V$

Option D is correct.

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298 K

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