Question

Balance the following equation in acidic conditions. Phases are optional. 2 Cu + 2 NO^- + 8H^+ rightarrow 3 Cu^2+ + 2 NO_2 + 4H_2O Tip: If you need to clear your work and reset the equation, click the button that looks like two red arrows.

Answer 1

Given redox reactin:

2Cu + 2NO₃^- + 8H⁺ $\rightarrow$ 3Cu²⁺ + 2NO₂ + 4H₂O

Oxidation half reaction (OHR):

Cu $\rightarrow$ Cu²⁺

Cu $\rightarrow$ Cu²⁺ + 2e^- (balancing charge)

Reduction half reaction (RHR):

NO₃^- $\rightarrow$ NO₂

NO₃^- $\rightarrow$ NO₂ + H₂O (balancing oxygen atoms)

NO₃^- + 2H⁺ $\rightarrow$ NO₂ + H₂O (balancing hydrogen atoms)

NO₃^- + 2H⁺ + e^- $\rightarrow$ NO₂ + H₂O (balancing charge)

Multipy OHR with 1, RHR with 2 and add both of them, then we will get the following balanced equation.

Cu + 2NO₃^- + 4H⁺ $\rightarrow$ Cu²⁺ + 2NO₂ + 2H₂O

Note: Please cross check the given equation, where both Cu and N are involving in oxidation. I have mentioned the given equation in a feasible manner. Please confirm once again. If any doubts, my doors are opened!!!