For `Cr_2O_7^(2-) + 14 H^(+) + 6e^(-) to 2Cr^(+3) + 7H_2O, E^(@) = 1.33 V " At" [ Cr_2O_7^(2-)]=4.5` millimoles , `[Cr^(+3)]=15` millimole , E is 1.067V. The pH of the solution is nearly equal to

Options

A. 2
B. 3
C. 4
D. 5

Answer

A

Explanation

`E=E^(@)-(0.0591)/(n) log. ([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`1.067=1.33- (0.0591)/(6) log. ((15xx10^(-3))^(2))/((4.5xx10^(-3))[H^(+)]^(14))`
`1.067=1.33-0.0098 log. ((50xx10^(-3)))/([H^(+)]^(14))`
`0.0098 log. ((50xx10^(-3)))/([H^(+)]^(14))=1.33-1.067`
`0.263 V`
or `log. ((50xx10^(-3)))/([H^(+)]^(14))=(0.263)/(0.0098)=26.83`
`log 50 xx 10^(-3) - log [H^(+)]^(14)=26.83`
`-1.3-14log[H^(+)]=26.83`
or `-14log[H^(+)]=26.83+1.3=28.13`
or `-log[H^(+)]=(28.13)/(14)=2.01`
`:.pH=2`.