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^- + 2Cu^2+ + 2OH^− ⟶ NO3^- + 2Cu^+ + H2O ;)
NO2^- + 2Cu^2+ + 2OH^− ⟶ NO3^- + 2Cu^+ + H2O ;)
To balance a redox reaction in basic solution, you need to follow these steps:
Step 1: Write the unbalanced equation.
NO2– + Cu2+ ⟶ NO3– + Cu
Step 2: Separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: NO2– ⟶ NO3–
Reduction half-reaction: Cu2+ ⟶ Cu
Step 3: Balance the atoms in each half-reaction.
For the oxidation half-reaction, we have:
2NO2– ⟶ 2NO3–
For the reduction half-reaction, we have:
Cu2+ ⟶ Cu
Step 4: Balance the charges in each half-reaction.
In the oxidation half-reaction, we have an excess of negative charges on the left side. To balance the charges, add 4OH– ions to the right side. This will neutralize the excess negative charge and create water molecules.
2NO2– ⟶ 2NO3– + 4OH–
In the reduction half-reaction, the charges are already balanced.
Cu2+ + 2e– ⟶ Cu
Step 5: Balance the number of electrons transferred in both half-reactions.
To do this, we need to multiply the oxidation half-reaction by 2 so that the number of electrons transferred is equal in both half-reactions.
4NO2– ⟶ 4NO3– + 8OH–
Cu2+ + 2e– ⟶ Cu
Step 6: Combine the half-reactions.
To combine the two half-reactions and cancel out any common species, multiply each half-reaction by the appropriate factor so that the number of electrons transferred is equal.
Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 4.
8NO2– ⟶ 8NO3– + 16OH–
4Cu2+ + 8e– ⟶ 4Cu
Step 7: Combine the two half-reactions, canceling out any common species.
8NO2– + 4Cu2+ + 8e– ⟶ 8NO3– + 16OH– + 4Cu
Step 8: Simplify the equation and eliminate any unnecessary terms.
Simplify the equation by dividing through by the common coefficient of 4.
2NO2– + Cu2+ + 4e– ⟶ 2NO3– + 4OH– + Cu
Therefore, the balanced equation for the given redox reaction in basic solution is:
2NO2– + Cu2+ ⟶ 2NO3– + 4OH– + Cu
yhtr
I'll get you started.
2OH^- + NO2^- ==> NO3^- + 2e + H2O
Cu^2+ + 2e ==> Cu although I assume the Cu^2+ would be present as Cu(OH()2