For this problem, we have to calculate for the cell potential of the reaction:
Fe3+(aq) + 3e- → Fe(s).
This particular reaction appears to be the overall reaction of the two reduction reactions:
(1) Fe3+ (aq) + e- → Fe2+ (aq) E° = +0.77 V
(2) Fe2+ (aq) + 2e- → Fe(s) E° = -0.45 V
Calculate the half-cell potential for the reaction: Fe3+(aq) + 3e- → Fe(s). Show all work and circle your final answer.
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