Calculate the potential for a half cell reaction containing `0.1 M K_(2)Cr_(2)O_(7), 0.20 M Cr^(3+)(aq)` and `1.0xx10^(-4) M H^(+)(aq)`. The half cell
Calculate the potential for a half cell reaction containing `0.1 M K_(2)Cr_(2)O_(7), 0.20 M Cr^(3+)(aq)` and `1.0xx10^(-4) M H^(+)(aq)`. The half cell reaction is :
`Cr_(2)O_(7)^(2-)(aq)+14 H^(+)(aq)+6e^(-)to2 Cr^(3+)(aq)+7H_(2)O (l)`
The standard electrode potential `(E^(@))=1.33 V`.
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According to Nernst equation
`E=E^(@)-((0.0591 V))/(n)"log"([Cr^(3+)(aq)]^(2))/([Cr_(2)O_(7)^(2-)(aq)][H^(+)(aq)]^(14)`
Substituting the data in the above relationship, we get
`E=(1.33V)-((0.0591 V))/(6)"log"(0.20)^(2)/((0.1)xx(1xx10^(-4))^(14))=(1.33 V)-((0.0591))/(6)"log"
(4xx10^(55))`
`E=(1.33V)-((0.0591 V))/(6)"log"(0.20)^(2)/((0.1)xx(1xx10^(-4))^(14))=(1.33 V)-((0.0591))/(6)"log"
(4xx10^(55))`
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