Calculate the potential of the half-cell containing 0.1 M
`K_(2)Cr_(2)O_(7(aq)), 0.2M Cr_((aq))^(3+)and 1.0xx10^(-4) M H_((aq))^(+).` The half-reaction
`Cr_(2) O_(7(aq))^(2-)+ 14 H_((aq))^(+)+6e^(-)to 2 Cr_((aq))^(3+)+7H_(2)O_((l))`
`(E^(0) of Cr_(2)O_(7)^(2-) //Cr^(3+)=1.33V)`

Explanation

Given half reaction
`Cr _(2)O_(7)^(2-) +14 H^(+)+6e^(-)to 2 Cr ^(+3) +7 H _(2)O`
`E^(0)or Cr_(2)O_(7)^(2-)//Cr ^(+3)=1.33V`
`E=E^(0) -(0.059)/(n)log Q`
`E=E^(0) -(0.059)/(6)log ""([Cr^(+3)][H_(3)O])/([Cr _(2)O_(7)^(-2)][H^(+)]^(14))`
`=1.33-(0.059)/(6) log ""((0.2)^(2))/((0.1)(1xx10^(-4))^(14))`
`=1.33-(0.059)/(6)log ""(0.04)/(0.1xx1xx10^(-56))`
`=1.33-0.00985xx53.6021=0.802V`