At `25^(@)C`, the reduction potential of `Cu^(2+)|Cu` half - cell is 0.28V. If `E^(@)` of `Cu^(2)|Cu` half-cell is `+0.34V`, find the molar concentration of `Cu^(2+)`ion in the half - cell.

Explanation

The reduction occurring in the half - cell :
`Cu^(2+)(aq)+2e rarr Cu(s)`
The Nernst equation for this reaction is
`E_(Cu^(2+)|Cu)=E_(Cu^(2+)|Cu)^(@)-(0.059)/(n)log.([Cu])/([Cu^(2+)])`
Since 2 mol of electrons are involved in the half - cell reaction,
n = 2. For pure `Cu,[Cu]=1`
`therefore" "E_(Cu^(2+)|Cu)=E_(Cu^(2+)|Cu)^(@)+(0.059)/(2)log[Cu^(2+)]`
`"or, "0.28=0.34+(0.059)/(2)log[Cu^(2+)]`
`"or, "log.[Cu^(2+)]=-2.033" "therefore" "[Cu^(2+)]=9.26xx10^(-3)(M)`