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Derive expression to calculate emf (reduction) of the following half cells at 25^@C Fe^(3+)|Fe^(2+) (Pt) |
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Answer» SOLUTION :`Fe^(3+)+E LEFTRIGHTARROW Fe^(2+)` (reduction) `E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^2+)^@-0.0591/1 log ""([Fe^(2+)])/([Fe^(3+)])` `E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^2+)^@ +0.0591 log""([Fe^(2+)])/([Fe^(3+)])` |
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