Cu ==> Cu^2+ + 2e is correct.
Stepwise: acid medium.
NO3^- ==> NO2
N changes from +5 on the left to +4 on the right so it gained 1 e.
NO3^- + e ==> NO2
Count the charge. -2 on the left; zero on the right. Add H^+ to balance the charge.
NO3^- + e + 2H^+ ==> NO2
Add water to balance the H^+
NO3^- + e + 2H^+ ==> NO2 + H2O
It should be balanced but I like to check them to see. 1N on both sides. 3O on both sides. 2 H on both sides. zero charge on both sides. N changed from +5 + e to +4. Everything ok.
Balance the following redox reaction occurring in acidic solution
Cu(s) + NO3-(aq)--> Cu2+(aq) +NO2(g)
This is what I have done so far
Cu-->Cu2+ + 2e-
2H + NO-3 --> NO2 + H2O
Need help please
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