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For the half - cell Fe^(3+), Fe^(2+)|Pt,...

For the half - cell `Fe^(3+), Fe^(2+)|Pt`, the concentration of `Fe^(2+) and Fe^(3+)` ions are 0.1 and 0.01 (M) respectively. Determine the reduction potential of the half - cell at `25^(@)C`. Given : `E_(Fe^(3+), Fe^(2+)|Pt)^(@)=+0.77V`

Text Solution

Verified by Experts

The reduction reaction occurring in the half - cell :
`Fe^(3+)(aq)+e rarr Fe^(2+)(aq)`
The Nernst equation for this reaction is,
`E_(Fe^(3+)|Fe^(2+))=E_(Fe^(3+)|Fe^(2+))^(@)-(0.059)/(1)log.([Fe^(2+)])/([Fe^(3+)])`
`=0.77-(0.059)/(1)log.(0.1)/(0.01)=0.711V`
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