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A half - cell is constructed by dipping ...

A half - cell is constructed by dipping a Zn- rod into a solution of `ZnSO_(4)` at `25^(@)C`. If the concentration of `Zn^(2+)` ion in the solution be 0.01 (M), calculate the oxidation potential of the half- cell, Given : `E_(Zn^(2+)|Zn)^(@)=-0.76V`.

Text Solution

Verified by Experts

The reduction occurring in the half - cell :
`Zn^(2+)(aq)+2erarrZn(s)`
The Nernst equation for this reaction is
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(@)-(0.059)/(n)log.([Zn])/([Zn^(+)])`
Since 2 mol of electrons are involved in the half - half reaction,
n = 2. For pure `Zn, [Zn]=1.` Thus,
`therefore" "E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(@)+(0.059)/(2)log[Zn^(2+)]`
`=-0.76+(0.059)/(2)log0.01=-0.819V`
Therefore, the oxidation potential of the half - cell `=-(-0.819V)`
`=+0.819V`
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