JEE MAIN - Chemistry (2021 - 22th July Evening Shift - No. 18)

Assume a cell with the following reaction

$$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$$

$$E_{cell}^\Theta = 2.97$$ V

Ecell for the above reaction is ______________ V. (Nearest integer)

[Given : log 2.5 = 0.3979, T = 298 K]
Answer
3

Explanation

$$E = {E^o} - {{0.059} \over 2}\log {{[C{u^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$$

$$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$$

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