Consider the cell described below at 287 K:
Fe | Fe2+ (0.829 M) || Cd2+ (0.953 M) | Cd
Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.301 mol/L.
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