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Characteristics of Chemical Equilibrium

At 298K, th...

Question

At $298 K$ , the equilibrium constant of reaction.
${Z n}^{+ 2} + 4 {N H}_{3} ⇌ {[Z n {({N H}_{3})}_{4}]}^{+ 2}$ is $10^{9}$
If $E_{{[Z n {({N H}_{3})}_{4}]}^{+ 2} / Z n + 4 {N H}_{3}}^{o} = - 1.03 V$ . The value $E_{Z n / {Z n}^{+ 2}}$ will be:

0.7645V

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-1.1V

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+1.1V

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none of these

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Solution

The correct option is B none of these
The reaction is,
$Z n \to {Z n}^{2 +} + 2 e$
Since, given ${E^{0}}_{{[Z n {({N H}_{3})}_{4}]}^{2 +} / Z n + 4 {N H}_{3}} = - 1.03 V$
$K_{e q} = 10^{9}$
We know,
$E = E^{0} + \frac{0.059}{n} l o g K_{e q}$
$E = - 1.03 + \frac{0.059}{2} \times 9$ since, the reaction is occured transfaring two electron.
$E = - 0.7645 V$

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