Question

For a particular redox reaction NO2– is oxidized to NO3– and Cu2 is reduced to Cu . Complete and balance the equation for this reaction in basic solution. Phases are optional. NO2- + Cu2+ ----> NO3- + Cu+

Answer 1

Given, the reaction,

NO₂^-(aq) + Cu²⁺(aq) NO₃^-(aq) + Cu⁺(aq)

Step1) Separating the half reactions,

NO₂^-(aq) NO₃^-(aq) ----- Oxidation half-reaction

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

Step 2) Balance the oxygens by adding H₂O(l) and balance the hydrogens by adding H⁺(aq) to both half reactions,

H₂O(l) + NO₂^-(aq) NO₃^-(aq) + 2H⁺(aq)

Cu²⁺(aq) Cu⁺(aq)

Step 3) Balance the charge by adding electrons to both the half-reactions,

H₂O(l) + NO₂^-(aq) NO₃^-(aq) + 2H⁺(aq) + 2e^-

1e^- + Cu²⁺(aq) Cu⁺(aq)

Step 4) Balance the electrons by multiplying to the each half-reactions by a simple whole number,

Multiplying by 1 to the first half-reaction and multiplying by 2 to the second half reaction,

H₂O(l) + NO₂^-(aq) NO₃^-(aq) + 2H⁺(aq) + 2e^- x 1

1e^- + Cu²⁺(aq) Cu⁺(aq) x 2

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

H₂O(l) + NO₂^-(aq) NO₃^-(aq) + 2H⁺(aq) + 2e^-

2e^- + 2Cu²⁺(aq) 2Cu⁺(aq)

Step 5) Add the both half reactions,

H₂O(l) + NO₂^-(aq) + 2Cu²⁺(aq) NO₃^-(aq) + 2H⁺(aq) + 2Cu⁺(aq)

Step 6) Now, add 2OH^-(aq) on both the sides, since we need to balance the reaction in basic condition,

H₂O(l) + NO₂^-(aq) + 2Cu²⁺(aq) + 2OH^-(aq) NO₃^-(aq) + 2H⁺(aq) + 2Cu⁺(aq) + 2OH^-(aq)

Thus,

H₂O(l) + NO₂^-(aq) + 2Cu²⁺(aq) + 2OH^-(aq) NO₃^-(aq) + 2H₂O(l) + 2Cu⁺(aq)

NO₂^-(aq) + 2Cu²⁺(aq) + 2OH^-(aq) NO₃^-(aq) + H₂O(l) + 2Cu⁺(aq) ---------(Answer)