Copper reduces NO 3 - into NO and NO 2 depending upon the concentration of HNO 3 in solution. (Assuming fixed Cu 2 + and P NO = P NO 2 ), the HNO 3 concentration at which the thermodynamic tendency for reduction of NO 3 - into NO and NO 2 by copper is same is 10 x M . The value of 2 x is ___. (Rounded-off to the nearest integer) [Given, E Cu 2 + / Cu o = 0 . 34 V , E NO 3 - / NO o = 0 . 96 V , E NO 3 - / NO 2 o = 0 .

Step-by-step solution

If the partial pressure of NO and NO 2 gas is taken as 1 bar, then Answer is 4 , else the question is bonus. NO 3 - + 4 H + + 3 e - ⟶ NO + 2 H 2 O E NO 3 - / NO o = 0 . 96   V NO 3 - + 2 H + + e - ⟶ NO 2 + H 2 O E NO 3 - / NO 2 o = 0 . 79 Let HNO 3 = y ⇒ H + = y and NO 3 - = y for same thermodynamic tendency E NO 3 - / NO = E NO 3 - / NO 2 or, E NO 3 - / NO o - 0 . 059 3 log P NO y × y 4 = E NO 3 - / NO 2 o - 0 . 059 1 log P NO 2 y × y 2 or, 0 . 96 - 0 . 059 3 log P N 0 y 5 = 0 .