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Question
Calculate the ΔrG0 and log Kc, for the given reaction at 298 K:
\[\ce{Ni_{(s)} + 2Ag^+_{( aq)} <=> Ni^{2+}_{( aq)} + 2Ag_{(s)}}\]
Given: `"E"_("Ni"^(2+)//"Ni")^0` = −0.25 V, `"E"_("Ag"^+//"Ag")^0` = +0.80 V, 1F = 96500 C mol−1.
Solution
\[\ce{E^0_{cell} = E^0_{C} - E^0_{A}}\]
= 0.80 − (−0.25)
= 1.05 V
ΔG0 = `-"nFE"_"cell"^0`
= −2 × 96500 × 1.05
= −2 × 105 J
ΔG0 = 2.303 RT log Kc
log Kc = `(2 xx 10^5)/(2.303 xx 8.314 xx 298)`
log Kc = 35.52
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