##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 18.40

Write equations for the half–reactions and the overall cell reaction, and calculate Eocell for each of the voltaic cells diagrammed below.

(a) Pt|I2(s)|I(aq)||Cl(aq)|Cl2(g)|Pt

(b) Pt|PbO2(s)|Pb2+(aq), H+(aq)||S2O82–(aq),SO42–(aq)|Pt

Split each cell into it's half–reactions, find the potential for the half–reaction from the table of standard reduction potentials, then use these to find the cell voltage.

(a) Pt|I2(s)|I(aq)||Cl(aq)|Cl2(g)|Pt

Anode (oxidation) reaction: 2 I(aq)I2(s) + 2 e

Eooxidation = –Eoreduction =–(0.535) = –0.535 V

Cathode (reduction) reaction: Cl2(g) + 2 e2 Cl(aq)

Eoreduction = +1.358 V

Overall cell reaction: Cl2(g) + 2 I(aq)I2(s) + 2 Cl(aq)

Eocell = Eooxidation + Eoreduction = –0.535 + 1.358 = +0.823 V

(b) Pt|PbO2(s)|Pb2+(aq), H+(aq)||S2O82–(aq),SO42–(aq)|Pt

Anode (oxidation) reaction: Pb2+(aq) + 2 H2O(l)PbO2(s) + 4 H+(aq) + 2 e

Eooxidation = –Eoreduction =–(1.455) = –1.455 V

Cathode (reduction) reaction: S2O82–(aq) + 2 e 2 SO42–(aq)

Eoreduction = +2.01 V

Overall cell reaction: Pb2+(aq) + 2 H2O(l) + S2O82–(aq)PbO2(s) + 4 H+(aq) + 2 SO42–(aq)

Eocell = Eooxidation + Eoreduction = –1.455 + 2.01 = +0.56 V