Using the standard reduction potentials listed in appendix e, calculate the equilibrium constant for the reaction at 298 k: 3 ce4+ (aq) + bi (s) + h2o (l) → 3 ce3+ (aq) + bio+ (aq) + 2 h+ (aq)

2.9×10^65

Explanation:

E°cell= E°cathode- E°anode

E°cell= 1.61 V- 0.32 V= 1.29V

The anode is oxidized while the cathode is reduced hence Bi/BiO+ is the anode while Ce4+/Ce3+ is the cathode

Recall that

R=8.314JK-1

F= 96500 C

n=3 moles of electrons

T=298K

E°cell= RT/nF ln K

lnK= E°cell × nF/RT

ln K = 1.29 × 3×96500/8.314×298

ln K= 150.7

K= 2.9×10^65