A solution containing 4.5mM of `Cr_(2)O_(7)^(2-)` and 15mM of `Cr^(3+)` shows a pH of 2.0. Calculate the potential of half reaction. (Standardd potential of the reaction `Cr_(2)O_(7)^(2-)rarrCr^(2+)` is 1.33 V)

Explanation

We know,
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr 2Cr^(3+)+7H_(2)O`
`E=E_(Cr_(2)^(6+)//Cr^(3+))^(@)+(0.059)/(6)"log"([Cr_(2)O_(7)^(2-)][H^(+)]^(14))/([Cr^(3+)]^(2))`
`E=1.33+(0.059)/(6)"log"([(4.5)/(1000)]xx[10^(-2)]^(14))/([(15)/(1000)]^(2))`
`=1.33+(0.059)/(6)log 20xx10^(-28)`
`=1.33+(0.059)/(6)[log 20-28log 10]`
`=1.33+(0.059)/(6)[1.3010-28]`
`=1.33-0.26=1.07V`