Balance the equation as a redox reaction in acidic solution.
Cu(s)+NO3^-1(aq)-->Cu^2+(aq)+NO(g)
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]
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 ?????
use the rexod spontaneity rule to predict whether yhe follow mixtures will show evidence of a reaction.Zinc metal in a solution aluminum ion.
To balance the given equation, you need to follow these steps:
Step 1: Identify the oxidation states of each element in the equation.
In this case, copper (Cu) has an oxidation state of 0 in its solid form (Cu(s)), and a charge of +2 in its aqueous form (Cu^2+(aq)). Nitrogen (N) has an oxidation state of +5 in the nitrate ion (NO3^-1(aq)), and a charge of +2 in nitrogen dioxide (NO(g)). Oxygen (O) has an oxidation state of -2 in most compounds.
Step 2: Separate the reaction into half-reactions for oxidation and reduction.
In the given equation, copper undergoes a reduction from an oxidation state of 0 to +2, while nitrogen undergoes an oxidation from +5 to +2.
The reduction half-reaction can be written as: Cu^2+(aq) + 2e^- → Cu(s)
The oxidation half-reaction can be written as: 2NO3^-1(aq) → 2NO(g) + 3O2 + 2e^-
Step 3: Balance the atoms in each half-reaction.
For the reduction half-reaction, there is already an equal number of copper atoms on both sides of the equation.
For the oxidation half-reaction, there are two nitrosyl (NO) molecules on the product side, so you need to put a coefficient of 2 in front of NO3^-1(aq).
Step 4: Balance the charges in each half-reaction by adding electrons (e^-).
In the reduction half-reaction, there are two excess electrons (2e^-) on the left side, while there are no electrons on the right side.
In the oxidation half-reaction, there is a deficit of two electrons (2e^-) on the product side, so you need to add two electrons to the left side.
Step 5: Multiply each half-reaction by the necessary factor to make the number of electrons equal in both half-reactions.
In this case, you need to multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 to make the number of electrons equal in both.
The balanced equation for the redox reaction in acidic solution is:
2Cu(s) + 4NO3^-1(aq) + 8H+(aq) → 2Cu^2+(aq) + 4NO(g) + 6H2O(l)