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))`.

Explanation

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`