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Calculate e.m.f of the following cell at...

Calculate e.m.f of the following cell at 298 K :
`2Cr(s) + 3Fe^(2+) (0.1M) rarr 2Cr^(3+)(0.01M) + 3Fe(s)`
Given : `E^(@) (Cr^(3+)|Cr) = - 0.74 V E^(@) (Fe^(2+) | Fe) = - 0.44V`

Text Solution

Verified by Experts

The half-cell reactions can be written as-
`Fe^(2+) +2e^(-) to Fe`
`E_(el) = -0.44 +((0.0591)/(2)) (log [Fe^(2+)])`
Substituting the concentration value and calculating -
`E_(el) = -0.44 +((0.0591)/(2)) (log 0.01)`
`= -0.44 -0.0591`
`= -0.4991 V " " ...(i)`
And
`Cr^(3+) +3e^(-) to Cr`
`E_(el) = -0.74 +(0.0591)/(3) (log[Cr^(3+)])`
Substituting the concentration value and calculating -
`E_(el) = -0.74 + (0.0591)/(3) (log 0.1)`
`= -0.74 -0.0197`
`= -0.7597V" " ...(ii)`
As Fe comes after Cr in tendency for reduction to occur in the electrochemical activity series. So, `Fe^(2+)` will reduce and Cr will oxidise.
`3Fe^(2+)(aq) +2Cr(s) to 2Cr^(3+) (aq) +3Fe(s)`
So, `E_("cell") ` will be -
`E_("cell") = -0.4991 V -(-0.7597)V`
`=0.2606V`
`DeltaG` of the reaction is given by -
`DeltaG_(r ) = -nFE_("cell")`
Here the total number of electrons involved in balanced chemical reaction are 6.
Hence `DeltaG_(r )= -nFE_("cell")`
`= -6 xx 96500 xx (0.2606V)`
`= -150887.4 "J mol"^(-1)`
`= -1508"kJ mol"^(-1)`
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