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प्रश्न
Calculate ΔrG0 and log Kc for the following cell:
\[\ce{Ni(s) + 2Ag^+(aq) -> Ni^{2+}(aq) + 2Ag(s)}\]
Given that \[\ce{E^0_{cell}}\] = 1.05 V, 1F = 96,500 C mol–1.
उत्तर
According to the equation,
\[\ce{Ni + 2Ag^+ -> Ni^{2+} + 2Ag}\]
ΔG = –nFE0
Where ΔG = Gibb's free energy
ΔG = –2 × 96500 × 1.05
N = No. of electrons gained or lost = 2
ΔG = –202.650 kJ
F = Faraday's constant = 96500
E0 = Standard emf = 1.05V
The equation shows the relationship between Gibb's free energy and Equilibrium constant.
`E_(cell)^0 = 0.0591/n log K_c`
`log K_c = - (1.05 xx 2)/0.0591`
log Kc = 35.3
Kc = 3.41 × 1035
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