Given that `E_(Fe^(2+)//Fe)^(.)=-0.44V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V` if `Fe^(2+),Fe^(3+)` and `Fe` solid are kept together then

Options

A. `Fe^(3+)` increases
B. `Fe^(3+)` decreases
C. `Fe^(2+)//Fe^(3+)` remains unchanged
D. `Fe^(2+)` decreases

Answer

B

Explanation

For the emf to be +ve, the following half-cell reactions will occur.
`FetoFe^(2+)+2e^(-),E^(@)=+0.44V`
`underline(2Fe^(3+)+2e^(-)2Fe^(2+)," "E^(@)=+0.77V" ")`
Overall reaction:
`Fe+2Fe^(3+)to3Fe^(2+)," "E_(cell)^(@)=0.121V`
Thus, `Fe^(3+)` will decrease.