Answer to Question #154596 in General Chemistry for Shana

Question

How to balance this redox reaction using half-reaction method: Cu(s)+NO3-(aq)->Cu+2(aq)+NO2(g) in acidic medium

Answer 1

Cu(s) + NO$_3^-$ (aq) $\to$ Cu⁺²(aq) + NO₂(g)

Oxidation no. Cu =0 , N=+5 ,O= -2 $\to$ Cu=+2, N=+4 ,O= -2

reduction half equation NO$_3^-$ $\to$ NO₂ ( Oxidation no. Of N decreses +5 to +4)

Oxidation half equation Cu $\to$ Cu⁺² ( Oxidation no. Of Cu increases 0 to +2)

Balance oxidation number by adding electrons. 

Cu $\to$ Cu⁺² + 2e^-

Balance oxidation number by adding electrons. 

NO$_3^-$ + e^- $\to$ NO₂

Balance charge by adding H⁺ ions in acidic solution. 

NO$_3^-$ + 2H⁺ + e^- $\to$ NO₂

Balance hydrogen by adding H₂O molecules. 

NO$_3^-$ + 2H⁺ + e^- $\to$ NO₂ + H₂O

Combine the two half-equations in such a way as to eliminate electrons

We generally do this by multiplying each half-equation by the number of electrons in the other 

half-equation:

Cu(s) + 2 NO$_3^-$ (aq) + 4H⁺(aq) $\to$ Cu⁺²(aq) + 2NO₂(g) + 2H₂O

This equation is now balanced: one Cu atom is on both sides of the equation, two N atoms are 

on both sides, six O atoms are on both sides, and four H atoms are on both sides. The total 

charge is +2 on both sides of the equation.