For the electrode `Cu//Cu^(2+),log[Cu^(2+)]` (along X-axis) is plotted against `E_(red)` (along Y-axis). The plot obatained is shown in figure. The electrode potential of the half cell `Cu|Cu^(2+)(0.1M)` will be

Options

A. `-0.34+(0.0591)/(2)V`
B. `0.34V`
C. `0.34+(0.0591)/(2)V`
D. `-0.34-(0.0591)/(2)V`

Answer

B

Explanation

The electrode reaction (reduction reaction) is
`Cu^(2+)+2e^(-)toCu`
`E_(cell)=E_(red)^(@)+(0.0591)/(2)"log"[Cu^(2+)]`
Thus, a plot of log`[Cu^(2+)]` vs `E_(red)` will have intercept`=E_(red)^(@)=0.34V` (Given in the plot) We have to calculate oxidation potential when
`[Cu^(2+)]=0.1M`
`E_(o x)=E_(o x)^(@)-(0.0591)/(2)log(0.1)`
`=-0.34-(0.0591)/(2)(-1)=-0.34+(0.0591)/(2)`