`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 of these

Answer

A

Explanation

`E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@) -(0.059)/(1)log[Cu^(2+)]`
If `log [Cu^(2+)] = 0`, i.e., `[Cu^(2+)] = 1`,
Then `E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@)`
or Intercept `= E_(Cu^(2+)//Cu) = 0.34`
Now `E_(Cu//Cu^(2+))^(@) = -0.34 - (0.059)/(2) log 0.1`
`= -0.34 + (0.059)/(2) V`