Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: G° for this reaction would be than zero. Submit AnswerRetry Entire Group

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)>2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

b is the correct