Answer:
b) 1.40 V
Explanation:
Oxidation half equation;
2Fe(s) → 2Fe3+(aq) + 6 e-
Reduction half equation;
3Cl2(g) + 6 e- → 6 Cl-(aq)
E°cathode= 1.36 V
E°anode= -0.036 V
E°cell= E°cathode - E°anode
E°cell= 1.36 -(-0.036)
E°cell= 1.36 + 0.036
E°cell= 1.396 V
E°cell= 1.40 V