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For the following reaction, calculate the equilibrium constant at $25^{\circ}C$

$Cu\left ( s \right )+2Ag^{+}\left ( aq \right )\rightarrow Cu^{2+}\left ( aq \right )+2Ag\left ( s \right )$

Given: $E^{\circ}=0.5V,R=8.314\; J/mol.K,\; F=96500\; C$
Option: 1
$8.2\times10^{18}$

Option: 2
$8.2\times10^{14}$

Option: 3
$8.2\times10^{16}$

Option: 4
$4.1\times10^{16}$

Answers (1)

best_answer

As we have learnt,

Relationship between $\Delta \mathrm{G}$ or $\Delta \mathrm{G}^{\circ}$ with $\mathrm{E}$ or $\mathrm{E}^{\circ}$

Free energy change $(\Delta \mathrm{G})$ in an electrochemical cell can be related to electrical work done (E) in cell as follows

$\Delta \mathrm{G}=-\mathrm{nFE}$

when we use standard conditions than

$\Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^0$

Here $(\mathrm{E^o})$ = standard E.M.F of the cell

n = No. of moles of e- transferred

F = Faraday's constant

So,

Here n is 2 because 2 electrons have been exchanged.

$\Delta G^{\circ}=-2\times 96500\times 0.5$

$\Delta G^{\circ}=-96500\; J/mol$

Also $\mathrm{\Delta G^{\circ}=-RT\; ln\left ( Kc \right )}$

$\mathrm{\Rightarrow K_c=antilog\left [ \frac{-\Delta G^{\circ}}{RT} \right ]}$

Putting values we get,

$\mathrm{K_c=8.2\times10^{16}}$

Hence, Option number (3) is correct.

Posted by

sudhir.kumar

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