Question

Given, standard electrode potentials,
`{:(Fe^(3+)+3e^(-) rarr Fe,,,E^(@) = -0.036 " volt"),(Fe^(2+)+2e^(-)rarr Fe,,,E^(@) = - 0.440 " volt"):}`
The standard electrode potential `E^(@)` for `Fe^(3+) + e^(-) rarr Fe^(2+)` is :
A. `-0.476 ` V
B. `-0.404` V
C. `0.404` V
D. `+0.772` V

Answer 1

Correct Answer - D
`Fe^(2+) + 2e^(-) to Fe `,
`DeltaG_(1)^(@) = - n FE_(1)^(@)`
= `-2 F(-0.44) = 0.88 F " " … (i)`
`Fe^(3+) + 3e^(-) to Fe`,
`DeltaG_(2)^(@) = - nFE_(2)^(@)`
`= -3 F (-0.036) = 0.108 F " " … (ii)`
Eqn. (ii) and (i) gives
`Fe^(3+) + 3e^(-) to Fe^(2+)`
`DeltaG^(@) = DeltaG_(2)^(@) - Delta G_(1)^(@) = 0.018 F - 0.88 F = -0.772` F
`therefore Delta G^(@) = - n F E_(cell)^(@) = -0.772`
`therefore E_(cell)^(@) = (0.772)/(n) = (0.772)/(1) = 0.772` V