Given `E_(Mn^(7+)//Mn^(2+))^(@)` and `E_(Mn^(4+)//Mn^(2+))^(@)` are `1.51 V` and `1.23 V`. Calculate `E_(Mn^(7+)//Mn^(4+))^(@)`.

Options

A. `0.3` V
B. `0.1` V
C. `1.7` V
D. `2.1` V

Answer

C

Explanation

`Mn^(7+) + 5e^- to Mn^(2+) , E^(@) = 1.5 V`
`Delta G_(1)^(@) =- n FE^(@) = - 5 xx F xx 1.5 = -7.5 F " " …. (i)`
`Mn^(4+) + 2e^(-) to Mn^(2+) , E^(@) = 1.2 V `
`Delta G_(2)^(@) = - n FE^(@) = - 2 xx F xx 1.2 = -2.4 F " " … (ii)`
Substracting eq. (ii) from eq. (i) , we get
`Mn^(7+) + 3e^(-) to Mn^(4+) , E^(@) = ?`
`Delta G_(3)^@ = -n FE^(@) = - 3 FE^(@)`
`Delta G_(3)^@ = Delta G_(1)^(@) - Delta G_(2)^(@)`
- `3 FE^(@) = -7.5 F - (-2.4F) = 5.1 F`
`E^(@) = (5.1)/(3) = 1.7 V `