For the cell Zn(s)| ZnSO _ 4 (aq)|| CuSO _ 4 (aq)|Cu(s) , when the...

For the cell $Zn\left(s\right)\left|{ZnSO}_{4}\left(aq\right)\right|\left|{CuSO}_{4}\left(aq\right)\right|Cu\left(s\right)$ , when the concentration of ${Zn}^{2 +}$ is $10$ times the concentration of ${Cu}^{2 +}$ , the expression for $\Delta G\left(inJ{mol}^{- 1}\right)$ is

$[ F =$ Faraday's constant, $R =$ universal gas constant, $T =$ temperature, ${E}^{o}cell = 1.1V ]$

Options

A $2 .303 \times RT + 1 .1 F$
B $2 .303 \times RT - 2 .2 F$
C $1 .1 F$
D $- 2 .2 F$

Correct answer

B. $2 .303 \times RT - 2 .2 F$

Step-by-step solution

$\Delta G = \Delta G ° + 2 .303 \times RT \times logQ$ ,

$\Delta G = - nFE ° + 2 .303 \times RT \times logQ$ .

Given, $E ° = 1 .1 V$ and $n = 2$ ,

$\Delta G = \left(- 2 \times 1 .1 \times F\right) + 2 .303 \times RT \times log\dfrac{\left[{Zn}^{+ 2}\right]}{\left[{Cu}^{+ 2}\right]}$ ,

\Delta G = - 2 .2 F + 2 .303 \times RT

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