Efficiency of the following cell is $84$ %.
$A (s) + B^{2 +} (a q .) ⇌ A^{2 +} (a q .) + B (s); △ H = - 285 k J$
Then the standard electrode potential of the cell will be___________.

1.20 V

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2.40 V

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1.10 V

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1.74 V

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Solution

The correct option is D $1.74 V$
Efficiency of cell = 84%
so, the total energy that has to be produced = $\frac{285}{84} * 100$
so, $\frac{28500}{84} = n F E_{c e l l}^{0}$
$\frac{28500}{84} = 2 * 96500 E_{c e l l}^{0}$
$E_{c e l l}^{0} = 1.74 V$

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