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प्रश्न
Calculate standard Gibbs energy change at 25°C for the cell reaction.
\[\ce{Cd_{(s)} + Sn^{2+}_{ (aq)} -> Cd^{2+}_{ (aq)} + Sn_{(s)}}\]
\[\ce{E^0_{cd}}\] = –0.403 V, \[\ce{E^0_{Sn}}\] = –0.136 V
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उत्तर
Given: \[\ce{E^0_{cd}}\] = –0.403 V, \[\ce{E^0_{Sn}}\] = –0.136 V, ΔG0 = ?
As \[\ce{E^0_{Sn}}\] is greater than \[\ce{E^0_{cd}}\]
∴ \[\ce{E^0_{cell}}\] = \[\ce{E^0_{Sn}}\] − \[\ce{E^0_{cd}}\]
= –0.136 – (–0.403)
= 0.267 V
n = 2 mol eΘ
ΔG0 = \[\ce{-{n}FE^0_{cell}}\]
= –2 × 96500 × 0.267
= –51.53 kJ
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