1.

A solution containing 4.5 mmol of Cr_2O_7^(2-) and 15 mmol of Cr^(3+) shows a pH of 2.0. calculate the potential of the half reaction Cr_2O_7^(2-) to Cr^(3+),E^@=1.33V

Answer»

Solution :The complete half cell REACTION is
`Cr_2O_7^(2-) +14H^+ +6e to 2Cr^(3+) +7H_2O, E^@=1.33V`
Given: `[Cr_2O_7^(2-)]=4.5/1000M, [Cr^(3+)]=15/1000M and [H^+]=10^-2 M`
`E=E^@-0.0591/6 log""[Cr^(3+)]^2/([Cr_2O_7^(2-)][H^+]^(14))`
`E=1.33-0.0591/6 log""(0.015)^2/((0.0045)(10^-2)^(14))`
`E=1.067V`


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