Question

The redox reaction of peroxydisulfate with iodide has been used for many years as part of the iodine clock reaction which introduces students to kinetics. If E�cell = 1.587 V and E� of the cathode half-cell is 0.536 V, what is E� of the anode half-cell?
S2O82�(aq) + 2H+ + 2I�(aq) ? 2HSO4�(aq) + I2(aq)

a. ?1.051 V
b. ?2.123 V
c. 1.051 V
d. 2.123 V
e. none of the answers is correct

Answer 1

In letters a and b, the "?" is a "-".

Answer:

a. - 1.051 V

Explanation:

In a redox reaction, one substance will oxide (lose electrons) and the other will reduce (gain electrons). The potential of reduction can be calculated by placing a redox reaction with H₂, which has a potential equal to 0.

For a spontaneous reaction, the electrons go from the anode (oxidation) to the cathode (reduction), and then the potential of the cell is the cathode potential lesse the anode potential:

E°cell = E°c - E°a

1.587 = 0.536 - E°a

E°a = 0.536 - 1.587

E°a = -1.051 V