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

Answer 1

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)`

E_Cell = 0.30 + 0.01 

E_Cell = 0.31 V