1.

The standard electrode potential for the following reaction is + 1.33V.What is potential at pH=2.0? Cr_2O_7^(2-) (aq. 1M) + 14H^(+)(aq) + 6e^(-) to 2Cr^(3+) (aq. 1M0 + 7H_2O(l)

Answer»

`+1.820V `
` + 1.990 V `
`+1.608 V `
`+1.0542V`

Solution :`E^(-) = E^(0)(-0.0941)/(n) log Q , E = 1.33 - (0.0591)/(6) log ""([Cr^(+3)]^(2))/([Cr_2O_7^(-2)][H^(oplus)]^(14))`
`= 1.33 - (0.0591)/(6) log""((1)^(2))/((1) XX (10^(-2))^(14))= 1.33 + (0.0591)/(6) log 10^(-28) = 1.33 + (0.0591)/(6) xx (-28)`
` = 1.33 to 0.2788 = 1.05212 V `


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