Calculate the E°cell using these electrodes whose half reactions are:
Fe(OH)2(s) + 2e– → Fe(s) + 2OH–(aq) E° = -0.88V
NiO2(s) + 2H2O(l) + 2e– → Ni(OH)2(s) + 2OH–(aq) E° = +0.49V
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Calculate the E°cell using these electrodes whose half reactions are:
Fe(OH)2(s) + 2e– → Fe(s) + 2OH–(aq) E° = -0.88V
NiO2(s) + 2H2O(l) + 2e– → Ni(OH)2(s) + 2OH–(aq) E° = +0.49V
Calculate the E°cell using these electrodes whose half reactions are:
Fe(OH)2(s) + 2e– → Fe(s) + 2OH–(aq) E° = -0.88V
NiO2(s) + 2H2O(l) + 2e– → Ni(OH)2(s) + 2OH–(aq) E° = +0.49V
E°cell = (E°cathode (SRK) – E°anode (SRP))
(SRP means Standard Reduction Potential)
= E°Ni+/Ni2+ – E°Fe2+/Fe
= +0.049V – (-0.88V)
= 1.37V