Question

For a particular redox reaction NO is oxidized to NO3– and Cu2 is reduced to Cu . Complete and balance the equation for this reaction in basic solution. Phases are optional.

NO + Cu²⁺ ----> NO₃^- + Cu⁺

I know how to do it for an acidic solution, please show work for a basic solution. Thanks.

Answer 1

Given, the redox reaction,

NO(g) + Cu²⁺(aq) NO₃^-(aq) + Cu⁺(aq)

Step 1) Separating the reaction into half reactions,

NO(g) NO₃^-(aq) --------- Oxidation half-reaction

Cu²⁺(aq) Cu⁺(aq) -------- Reduction half reaction

Step 2) Balancing the oxygens by adding H₂O(l) and balancing the hydrogens by adding H⁺(aq).

2H₂O(l) + NO(g) NO₃^-(aq) + 4H⁺(aq)

Cu²⁺(aq) Cu⁺(aq)

Step 3) Balancing the charges by adding electrons to both the half-reactions,

2H₂O(l) + NO(g) NO₃^-(aq) + 4H⁺(aq) + 3e^-

1e^- + Cu²⁺(aq) Cu⁺(aq)

Step 4) Balance the number of electrons by multiplying to each half reaction by the simple whole number,

Multiplying first half-reaction by 1 and multiplying second half-reaction by 3,

2H₂O(l) + NO(g) NO₃^-(aq) + 4H⁺(aq) + 3e^-x 1

1e^- + Cu²⁺(aq) Cu⁺(aq) x 3

------------------------------------------------------------------------------------

2H₂O(l) + NO(g) NO₃^-(aq) + 4H⁺(aq) + 3e^-

3e^- + 3Cu²⁺(aq) 3Cu⁺(aq)

Step 5) Adding both the half-reactions,

2H₂O(l) + NO(g) + 3Cu²⁺(aq)   NO₃^-(aq) + 4H⁺(aq) + 3Cu⁺(aq)

Step 6) Adding 4OH^-(aq) on both the sides,

4OH^-(aq) + 2H₂O(l) + NO(g) + 3Cu²⁺(aq)   NO₃^-(aq) + 4H⁺(aq) + 3Cu⁺(aq) + 4OH^-(aq)

Thus,

4OH^-(aq) + 2H₂O(l) + NO(g) + 3Cu²⁺(aq)   NO₃^-(aq) + 4H₂O(l) + 3Cu⁺(aq)