`4.5 mM Cr_(2)O_(7)^(2-)` and 15mM `Cr^(3+)` is mixed to produce a solution of pH = 2.0. What will be the reduction potential of the half - cell? Given : For the change `Cr_(2)O_(7)^(2-) rarr Cr^(3+)`, the standard reduction potential is 1.33V`.

Explanation

`Cr_(2)O_(7)^(2-)+14H^(+)+6e rarr 2Cr^(3+)+7H_(2)O`
`E=E^(@)-(0.059)/(6)log.([Cr^(3+)])/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`=1.33-(0.059)/(6)log.([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)])-(14xx0.059)/(6)pH`
`=1.33-(0.059)/(6)log.((15)^(2))/(4.5)-(14xx0.059xx2)/(6)=1.038V`