Standard electrode potentials are :
`Fe^(2+) "|" Fe (E^(@) = - 0.44 V), Fe^(3+) "|" Fe^(2+) (E^(@) = 0.77 V) Fe^(2+), Fe^(3+)` and Fe blocks are kept together, then

Options

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

Answer

B

Explanation

For e.m.f. to be positive, the following half cell reactions will occur :
`Fe rarr Fe^(2+) + 2e^(-) E^(@)= +0.44 V`
`2Fe^(3+)+2e^(-) rarr 2Fe^(2+) " " E^(@)= +0.77V`
Overall reaction :
`Fe +2Fe^(3+) rarr 3Fe^(2+)E^(@)= +1.21 V`
`:.Fe^(3+)` will decrease.