Question

Calculate the standard cell potential for each of the electrochemical cells. Part A 2Ag+(aq)+Pb(s)?2Ag(s)+Pb2+(aq) Express your answer using two significant figures. Part B 2ClO2(g)+2I?(aq)?2ClO?2(aq)+I2(s) Express your answer using two significant figures. Part C O2(g)+4H+(aq)+2Zn(s)?2H2O(l)+2Zn2+(aq)

Answer 1

Part a

The two half cell reactions are

Oxidation reaction

Pb(s) = Pb2+(aq) + 2e-

Eox = 0.13 V

Reduction reaction

2Ag+(aq) + 2e- = 2Ag(s)

Ered = 0.80

Overall cell reaction

2Ag+(aq)+Pb(s) = 2Ag(s)+Pb2+(aq)

standard cell potential

E°cell = Eox + Ered

= 0.13 + 0.80

= 0.93 V

Part b

The two half cell reactions are

Oxidation reaction

2I-(aq) = I2(s) + 2e-

Eox = - 0.54V

Reduction reaction

2ClO2(g) + 2e- = 2ClO-2(aq)

Ered = 0.95 V

Overall cell reaction

2ClO2(g)+2I-(aq)?2ClO-2(aq)+I2(s)

standard cell potential

E°cell = Eox + Ered

= - 0.54 + 0.95

= 0.41 V

Part c

The two half cell reactions are

Oxidation reaction

Zn(s) = Zn2+(aq) + 2e-

Multiply by 2

2Zn(s) = 2Zn2+(aq) + 4e-

Eox = 0.76 V

Reduction reaction

O2(g) + 4H+(aq) + 4e- = 2H2O(l)

Ered = 1.23 V

Overall cell reaction

O2(g)+4H+(aq)+2Zn(s) = 2H2O(l)+2Zn2+(aq)

standard cell potential

E°cell = Eox + Ered

= 0.76 + 1.23

= 1.98 V