ElectrochemistryMedium
Question
The standard reduction potentials Eo, for the half reactions are as
Zn → Zn2+ + 2e-, Eo = + 0.76V
Fe → Fe2+ + 2e-, Eo = + 0.41V
The emf for the cell reaction,
Zn → Zn2+ + 2e-, Eo = + 0.76V
Fe → Fe2+ + 2e-, Eo = + 0.41V
The emf for the cell reaction,
Options
A.- 0.35 V
B.+ 0.35 V
C.+ 1.17 V
D.- 1.17 V
Solution
Fe2+ + Zn2+ + Fe
i. e, Zn acts as anode and Fe as cathode.
and Eocell = EoO.P.(Zn/Zn2+) + EoR.P>(Fe2+/Fe)
= (0.76) + (-0.41) = 0.35V
i. e, Zn acts as anode and Fe as cathode.
and Eocell = EoO.P.(Zn/Zn2+) + EoR.P>(Fe2+/Fe)
= (0.76) + (-0.41) = 0.35V
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