`E^(@)` in the given diagram is

Options

A. `0.51 V`
B. `0.61 V`
C. `0.71 V`
D. `0.81 V`

Answer

B

Explanation

`ClO_(3)^(-)+2H_(2)O+4e^(-)rarr ClO^(-)+4OH^(-) Delta G_(1)^(@)`
`ClO^(-)+H_(2)O+e^(-)rarr(1)/(2)Cl_(2)+2OH^(-) Delta G_(2)^(@)`
`(1)/(2)Cl_(2)+e^(-)rarr Cl^(-) Delta G_(3)^(@)`
`ClO_(3)^(-)+3H_(2)O+6e^(-)rarr Cl+6OH^(-)`
`Delta G^(@)=Delta G_(1)^(@)+Delta G_(2)^(@)+Delta G_(3)^(@) , Delta G = -nF E^(@)`
`-6FE^(@)=-4Fxx0.54-1Fxx0.45-1Fxx1.07`
`E^(@)=+(3.68)/(6)=0.61`