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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 |
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Answer» <P>`10^(1.32) M ` ` 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)` |
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