1.

If E_(1)=0.5V corresponds to Cr^(3+)+3e^(-) rarr Cr_((s)) " and " E_(2)=0.41V corresponds to Cr^(3+)+e^(-) rarr Cr^(2+) reactions, calculate the emf (E_(3)) of the reaction Cr^(2+)+2e^(-) rarr Cr_((s)).

Answer»

Solution :Given : `E_(1)=0.5V`
`Cr^(3+)+3E^(-) RARR Cr_((s)) "...(1)"`
`E_(2)=0.41V`
`Cr^(3+)+e^(-) rarr Cr^(2+) "...(2)"`
The REQUIRED reaction is,
`Cr^(2+)+2E^(-) rarr Cr_((s))`
Then,
Formula :
`E_(3)=(3E_(1)+E_(2))/2`
Solution :
`""=(3(0.5)+(0.41))/2=(1.5+0.41)/2`
`""=0.955 V`
`E_(3)=0.955V`


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