chemistry - redox reaction

Question

The same coefficient (X) must be there for Cu and for Cu^2+ in the following equation.

XCu + 2NO3^- + 8H^+ → XCu^2+ + 2NO + 4H2O

A. X=1 B. X=2 C. X=3

the answer is c, X=3.

Please explain how to solve this

Answer 1

There is a short way and a long way to answer this.

Short way:

Find number of electrons transferred going from NO₃^- to NO. Oxidation # of N in NO3^- is +5. Oxidation number of N in NO is +2. So 3 electrons were transferred but there are 2 NO3^- and 2 NO, so that's a total of 6 electrons. Those 6 electrons had to come from some place, and they came from Cu. Since Cu is 2+ on the right side, you need a 3 as a coefficient to make up the 6 electrons. Thus, the coefficient for Cu is 3 on both sides of the equation.

Long way:

You could write both half reactions (oxidation and reduction) and balance each in acidic conditions. This would take a little more effort than the short method above, but you'd find the same answer.