`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will be

Options

A. `-0.34 + (0.0591)/(2) V`
B. `0.34 + 0.0591` V
C. 0.34 V
D. None

Answer

A

Explanation

`E_(Cu | Cu^(2+)) = E_(Cu | Cu^(2+))^(@) - (0.0591)/(2) "log" ([Cu^(2+)])/([Cu(s)])`
if `[Cu (s) ] = 1 and [Cu^(2+) ] = 0.1` M
or `OA = E_(Cu^(2+) | Cu)^(@) therefore E_(Cu| Cu^(2+))^(@) = -0.34 V`
Now `E_(Cu| Cu^(2+)) = -0.34 - (0.0591)/(2)"log" 0.1`
`= -0.34 + (0.0591)/(2)`