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Part B.
3Ce 4+ (aq)+ Bi(s)+ H2O(l) = 3Ce 3+ (aq) + BiO + (aq) + 2H + (aq)
From the table of standard reduction potential,
Ce 4+ (aq) + e =Ce 3+ 1.61
Bi(s) + H2O(l) =BiO + (aq) + 2H + (aq) -0.32
E=3*1.61 - 0.32= 4.51
ln k=nFE/RT
k=10^(2*451/59.2)=1.7*10^15
******************************************************
Part C.
N2H +5 (aq) + 4Fe(CN)6 3-(aq) = N2 (g) + 5H + (aq) + 4Fe(CN)6 4- (aq)
From the table of standard reduction potential,
N2H5 + (aq) = N2 (g) + 5H + (aq) +0.23
Fe(CN)6 3-(aq) +e = Fe(CN)6 4- (aq) 0.355
E=0.23 +(4*0.355)=1.65
ln k=nFE/RT
k=10^(2*165/59.2)=3.75*10^5
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