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The E⁰ at 25⁰C for the following reaction at the indicated concentration is 1.50 V .calculate the $∆$ G in kJ at 25⁰C ; Cr(s)+3Ag⁺(aq,0.1M) $\to$ Ag(s)+Cr³⁺⁺(aq,0.1M)

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-140.94

-212

-295

-422.83kJ

answer is D.

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Detailed Solution

$△$ G= $△$ G⁰+ RT lnQ
- nFE⁰+ 2.303 RT log $\frac{[C r^{+ 3}]}{{[A g^{+}]}^{3}}$

$\begin{array}{l} △ G=-3 x 96,500 x1.5 + 8.314 x 298 x 2.303 log \frac{0.1}{{(0.1)}^{3}} \end{array}$
=422838.3J
=-422.83kJ

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