Balance the equation as a redox reaction in acidic solution.
Cu(s)+NO3^-1(aq)-->Cu^2+(aq)+NO(g)
5 answers
Tell us what method you are to use and what you don't understand about it.
Cu(s)+NO3^-1(aq)-->Cu^2+(aq)+NO(g)
Using the ion-electron method,
1) The incomplete half reactions are:
Cu(s) —> Cu^2+(aq)
NO3^-(aq) —> NO(g)
2) We use electrons, e^-, hydrogen ions, H+(aq), H2O, and OH^-(aq) to balance for electrical charge and for number of atoms.
Cu(s) —> Cu^2+(aq) + 2e^-
NO3^-(aq) + 4H^+(aq) + 3e^- —> NO(g) + 2H2O(L)
3) We reconcile the number of electrons in the two balanced half reactions (Multiply the first one by 3, and the second one by 2):
3Cu(s) —> 3Cu^2+(aq) + 6e^-
2NO3^-(aq) + 8H^+(aq) + 6e^- —> 2NO(g) + 4H2O(L)
4) We ADD the two half reactions:
3Cu(s) + 2NO3^-(aq) + 8H^+(aq) + 6e^- —> 3Cu^2+(aq) + 6e^- + 2NO(g) + 4H2O(L)
5) We simplify by doing the appropriate cancellations:
[I leave that final step for you you to do]
Using the ion-electron method,
1) The incomplete half reactions are:
Cu(s) —> Cu^2+(aq)
NO3^-(aq) —> NO(g)
2) We use electrons, e^-, hydrogen ions, H+(aq), H2O, and OH^-(aq) to balance for electrical charge and for number of atoms.
Cu(s) —> Cu^2+(aq) + 2e^-
NO3^-(aq) + 4H^+(aq) + 3e^- —> NO(g) + 2H2O(L)
3) We reconcile the number of electrons in the two balanced half reactions (Multiply the first one by 3, and the second one by 2):
3Cu(s) —> 3Cu^2+(aq) + 6e^-
2NO3^-(aq) + 8H^+(aq) + 6e^- —> 2NO(g) + 4H2O(L)
4) We ADD the two half reactions:
3Cu(s) + 2NO3^-(aq) + 8H^+(aq) + 6e^- —> 3Cu^2+(aq) + 6e^- + 2NO(g) + 4H2O(L)
5) We simplify by doing the appropriate cancellations:
[I leave that final step for you you to do]
3Cu(s) + 2NO3^-(aq) + 8H^+(aq)—> 3Cu^2+(aq)+ 2NO(g) + 4H2O(L)
POR QUE EN EL PASO 2 A LA SEGUNDA REACCION SE LE ADICIONA 4 H ?????
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