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The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is :

Cu(s) + Sn2+(aq.) $\longrightarrow$ Cu2+ (aq.) + Sn(s);

($E_{Sn^{2+}|Sn}^0$ = –0.16 V, $E_{Cu^{2+}|Cu}^0$ = 0.34 V, Take F = 96500 C mol–1)

Numerical Value Type

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Electrochemistry, Gibbs free energy

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Ans. (96500)

Sol. $E_{cell}^0=E_{Sn^{2+}/Sn}^0-E_{Cu^{2+}/Cu}^0$

= – 0.16 –0.34

= – 0.50V

$\Delta G^0 =-nF\ E_{cell}^0$

= – 2 × 96500 × (–0.5)

= 96500J

= 96.5 KJ = 96500 J

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