What is the potential of an electrode which originally contained `0.1 M NO_(3)^(-)` and `0.4 M H^(+)` and which has been treated by `8%` of the cadmium necessary to reduce all the `NO_(3)^(-)` to `NO(g)` at 1 bar ? Given : `NO_(3)^(-)+4H^(+)+3e^(-)rarr NO+2H_(2)O`, `E^(@)=0.96 , log 2 = 0.3`A. `0.84 V`B. `1.08 V`C. `1.23 V`D. `1.36 V`

What is the potential of an electrode which originally contained `0.1

1.	What is the potential of an electrode which originally contained `0.1 M NO_(3)^(-)` and `0.4 M H^(+)` and which has been treated by `8%` of the cadmium necessary to reduce all the `NO_(3)^(-)` to `NO(g)` at 1 bar ? Given : `NO_(3)^(-)+4H^(+)+3e^(-)rarr NO+2H_(2)O`, `E^(@)=0.96 , log 2 = 0.3`A. `0.84 V`B. `1.08 V`C. `1.23 V`D. `1.36 V`
Answer» Correct Answer - A After addition of Cd and it s oxidation into `Cd^(2+)` `{:(NO_(3)^(-)(aq)+4H^(+)(aq)+3e^(-)rarr NO(g)+2H_(2)O(l)),(0.1-x " " 0.4-4x " , where" x=0.08):}` `[NO_(3)^(-)]` remaining `=0.02 M, [H^(+)]` remaining `=0.08 M` `E_(NO_(3)^(-)NO)=E_(NO_(3)^(-)NO)^(@)-(0.0591)/(3)"log"(1)/([NO_(3)^(-)][H^(+)]^(4))` `=0.96-(0.0591)/(3)"log"(1)/((0.02)(0.08)^(4))`

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