Copper reduces `NO_3^(-)` into NO and `NO_2` depending upon concentration of `HNO_3` in solution . Assuming `[Cu^(2+)] = 0.1 M ` , and `P_(NO) = P_(NO_2) = 10^(-3)` bar . At which concentration of `HNO_3 `. Thermodynamic tendency for reduction of `NO_3^(-) ` into` NO and NO_2` by copper is same ?
Given : `E_(Cu^(2+)|Cu)^(@) = + 0.34 V , E_(NO_3^(-)|NO)^(@) = + 0.96 V , E_(NO_3^(-)|NO_2)^(@) = + 0.79 V `

Options

A. `10^(1.32) M `
B. `10^(0.56) M`
C. `10^(0.66) M`
D. None

Answer

D

Explanation

`3Cu^(0) + 2NH_3^(-) + 8H^(+) to 3Cu^(+2) + 2NO + 4H_2O`
` E^(0) =E_("cathode") -E_("anode") = E_(NO_3^(-) // NO ) - E_(Cu^(+2)// NO_2) -E_(Cu^(+2)//Cu) = 0.96 - 0.34 = 0.02 `
` E_("cell") = E^(0) -(0.059)/(n) log""([Cu^(+2)]^(3) P_(NO)^(2))/([H^(+)]^(8)) = 0.62 - (0.059)/(6) log_(10)""((0.1)^(3)(10^(-3)^(2)))/([H^(+)]^(8))`
`Cu + 2NO_3^(-) + 4H^(+) to Cu^(+2) + 2NO_2 + 2H_2O , E^(0) =E_(NO_3^(-) // NO_2)^(0)-E_(Cu^(+2)//Cu) = 0.79 - 0.34 = 0.45 `
`E_("cell")^(11) = E^(0) -(0.059)/(2) log ""([Cu^(+2)]P_(NO_2)^(2))/([H^(+)]^(4)) = 0.45 - (0.059)/(2) log""((0.1)(10^(-3))^(2))/([H^(+2)]^4), E_("cell")=E_("cell")`
Then `[H^(+)] = 10^(1.32)`