ElectrochemistryHard
Question
Given EoCr3+/Cr = - 0.72V, EoFe2+/Fe = - 0.42 V. The potential for the cell
Cr | Cr3+ (0.1M)||Fe2+(0.01M)|Fe is
Cr | Cr3+ (0.1M)||Fe2+(0.01M)|Fe is
Options
A.0.26 V
B.0.399 V
C.- 0.339 V
D.- 0.26 V
Solution
As EoCr/Cr3+ = - 0.72 V and EoFe2+/Fe = - 0.42 V
2Cr + 3Fe2+ → 3Fe + 2Cr3+
Ecell = Eocell-
=(-0.42 + 0.72)-
= 0.30 -
Ecell = 0.2606 V
2Cr + 3Fe2+ → 3Fe + 2Cr3+
Ecell = Eocell-
=(-0.42 + 0.72)-
= 0.30 -
Ecell = 0.2606 V
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