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Question
Calculate e.m.f of the following cell at 298 K:
2Cr(s) + 3Fe2+ (0.1M) → 2Cr3+ (0.01M) + 3 Fe(s)
Given: E°(Cr3+ | Cr) = – 0.74 VE° (Fe2+ | Fe) = – 0.44 V
Solution
The balanced chemical reaction can be written as follows:
2 Cr(s) + 3 Fe2+(0.1 M) → 2 Cr3+(0.01 M) + 3 Fe(s)
Given:
`E_(Cr^(3+)"/Cr")^o=-0.74V`
`E_(Fe^(2+)"/Fe")^o=-0.44V`
The cell can be represented as follows:
`Cr|Cr^(3+)(0.01M)||Fe^(2+)(0.1M)|Fe(s)`
`E_(Cell)^o " can be calculated as follows:"`
`E_(Cell)^o=E_(Fe^(2+)"/Fe")^o-E_(Cr^(3+)"/Cr")^o`
`E_(Cell)^o=-0.44-(-0.74)=0.30V`
`E_(Cell)=E_(Cell)^o-(2.303RT)/(nF)log(([Cr^(3+)]^2)/([Fe^(2+)]^3))`
`E_(Cell)=0.30-0.0591/6log((0.01)^2/(0.1)^2)`
ECell = 0.30 + 0.01
ECell = 0.31 V
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