Can someone please help with this
The cathode and anode in the given cell representation are,
Cu2+ (aq) + 2 e- -----> Cu (s) , Cathode ( reduction) Eocathode = 0.34 v
Fe (s) ---> Fe3+ (aq) + 3 e- , Anode ( oxidation) Eoanode = -0.036 v
Balancing the half equations for same number of electron transfer we have,
3 Cu2+ (aq) + 6 e- -----> 3 Cu (s)
2 Fe (s) ---> 2 Fe3+ (aq) + 6 e-
Thus, number of electron transferred is 6.
We know that, Gibbs energy change is given by,
ΔGo = - nFEocell
Where, ΔGo
n is the number of electrons transferred= 6
F is Faradays constant = 96458 C
Eocell is standard cell potential = Eocathode - Eoanode
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