In: Chemistry
Calculate the equilibrium constant for each of the reactions at 25 ∘C.
Part B
O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)
Express your answer using two significant figures.
2Cu(s) ------------------------------> 2Cu^2+ (aq) + 4e^- E0 = -0.34v
O2(g) + 2H2O(l) + 4e^- ----------------> 4OH^- (aq) E0 = 0.4v
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O2(g) + 2H2O(l) + 2Cu(s) --------------------> 2Cu^2+ (aq) + 4OH^- (aq) E0cell = 0.06v
n = 4
G0 = -nE0cell *F
= -4*0.06*96500 = -23160J
G0 = -RTlnK
-23160 = -8.314*298*2.303logK
-23160 = -5705.8483logk
logK = -23160/-5705.8483
logK = 4.06
K = 10^4.06 = 1.15*10^4 >>>>>answer