Given below are half-cell reaction:
`Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.18 V`
`2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V`
The `E^(@)` for `3Mn^(2+) rarr Mn+2Mn^(3+)` will be:

Options

A. `-0.33 V`, the reaction will not occur
B. `-0.33 V`, the reaction will occur
C. `-2.69 V`, the reaction will not occur
D. `-2.69 V`, the reaction will occur

Answer

C

Explanation

To obtain the given reaction, we reverse the second reaction and add it to the first one: `{:(Mn^(2+)+2e^(-) rarr Mn,),(2Mn^(2+)rarr2Mn^(3+)+2e^(-),):}/(3Mn^(2+)rarrMn+2Mn+2Mn^(3+))" "{:(E^(@) = -1.81 V),(E^(@) = -1.51 V):}`
Thus, `E^("cell")^(@) = E_(SOP)^(@)+E_(SRP)^(@)`
`= (-1.18 V)+(-1.51 V)`
`= -2.69 V`
Since
`DeltaG_("cell")^(@) = -nFE_("cell")^(@)`
negative `E_("cell")^(@)` indicates positive `DeltaG_("cell")^(@)` i.e., a non-spontaneous cell reaction.