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)

Question

Octavio Vargas

Chemistry

26 February, 19:20

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)

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Abel · Answer 1

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