The standard electrode potentials `(E^(@)) " for " Ocl^(-)//Cl^(-) " and " Cl^(-)//(1)/(2) Cl_(2)` respectively are 0.94 V and `-1.36 V`. The `E^(@)` value for `Ocl^(-)// (1)/(2) Cl^(2)` will be:

Options

A. `-4.42 V`
B. `-2.20V`
C. `0.52 V`
D. `1.04V`

Answer

C

Explanation

`OCl^(-)rarrCl^(-),E_(OCl^(-)//Cl^(-))^@=0.94V`
`Cl^(-)rarr1/2Cl_2+e^(-),E_(Cl^(-)//Cl_2)^(@) = -1.36V`
`2e^(-) +overset(+1)(OCl^(-))+H_2Orarroverset(+1)(Cl^(-)+2OH^(-))` (Reduction) …(i)
`OCl^( -) and Cl^( -)` are basic in nature, so balancing the reduction half reaction in basic medium.
`overset(+1)(Cl^(-))rarr1/2overset(0)(Cl_2)+e^(-)` (Oxidation) …(ii)
Applying eq. (i) + eq. (ii)
`OCl^(-)+H_2O+e^(-)rarr1/2Cl_2+2OH^(-)" "....(iii)`
This is the required equation to get the value for `E_("OCl"^(-)//1/2Cl_2)^(@)`
If the Gibb’s free energy for eq. (i), (ii) and (iii) are respectively `G_1,G_2 and G_3` then:
`- 1 xxFxxE^@ = -2xx F xx 0.94 - 1xx Fxx (-1.36)`
`-E^@=-1.88 + 1.36 E implies E^(@) = +0.52 V`