Half-Reactions

Like mentioned earlier, the reaction can be split into two separate reactions: the one being reduced and the one being oxidized.

Step 1

Well we first gotta figure out which compounds are being reduced and which ones are being oxidized and then split it into the half reactions that we will use for our balancing. In this case example, copper (Cu) is being oxidized going from 0 to +2 oxidation state,

While N is reduced from a +5 oxidation state to a +2 oxidation state

Image 1

Step 2

Next thing to do is start balancing each half reaction separately. The general rule is that you must use what is already in the solution to help balance the equation which includes H2O (because it is an aqueous solution) and H+ions (for acidic solutions). Since this is acidic, we will be using H+ ions to help us balance the reactions.

Using the explicitly chosen neutralHNO₃ bookkeeping reagent, add2H₂O to the product side ofHNO₃→NO to balance oxygen: HNO₃→NO+2H₂O. ThreeH⁺ are then needed on the reactant side to balance hydrogen, and three electrons balance charge. The completed reduction half-reaction isHNO₃+3H⁺+3e⁻→NO+2H₂O. The original3H⁺ coefficient is valid for this neutral-form convention; usingNO₃⁻ instead would change the hydrogen/charge bookkeeping.

Oxygen-balanced skeleton:HNO₃→NO+2H₂O. Hydrogen and charge are balanced in the next steps.

The oxygen-balanced skeleton has oneH on the left and four on the right. Add3H⁺ to the reactant side, using the chosen neutralHNO₃ convention.

HNO₃+3H⁺→NO+2H₂O. This is hydrogen-balanced but still needs electrons to balance charge.

The reactant side has charge+3 and the products are neutral. Add3e⁻ to the reactants. The original3H⁺ is correct forHNO₃; it must not be mixed with aNO₃⁻ starting formula.

HNO₃+3H⁺+3e⁻→NO+2H₂O. Both sides haveN1,O3,H4and net charge0.

Step 3

Now, lets look at the oxidized half reaction. Now there is no balancing of oxygen and hydrogen that needs to be done. The only balancing requirements for Cu is the charge, therefore we must add electrons to balance the charges on both sides,

Cu→Cu²⁺+2e⁻. Copper loses two electrons and is oxidized.

Step 4

We will want to be able to cancel the electrons so we must equalize the electrons for both reactions. Both half reactions have a common number of 6 electrons so we must multiply the reduced reaction by 2 and the oxidized

Scaled oxidation:3Cu→3Cu²⁺+6e⁻.

reaction by 3 to give the following half reactions:

Oxidized:

3Cu→3Cu²⁺+6e⁻.

Reduced:

2HNO₃+6H⁺+6e⁻→2NO+4H₂O.

Step 5

Add both reactions to get the final balanced reaction

Sum after cancelling electrons:3Cu+2HNO₃+6H⁺→3Cu²⁺+2NO+4H₂O. Both sides containCu3,N2,O6,H8and charge+6. Equivalent fully molecular nitric-acid bookkeeping is3Cu+8HNO₃→3Cu(NO₃)₂+2NO+4H₂O.

Notice how the electrons are not part of the final reaction as they were canceled out.

Step 1

We’re going to have the same approach as before. Write down the two half reactions that are oxidized and reduced.

Reduced:

Manganese is reduced going from +7 oxidation state to +4

Oxidized:

Carbon is oxidized because it goes from a +2 oxidation state to +4

Step 2

We will still balance the O’s by adding H2O to the equations. Let’s look at the reduced half reaction first.

And we will still add H+ ions to balance the hydrogens,

Step 3

What we’re going to do differently here is add OH- ions to balance the H+ ions, thus it’ll look like,

Step 5

Now to simplify this equation, we can combine the OH- and H+ ions to make H2O atoms, which in our reduced half reaction, we take the 4 OH- and 4H+ions to make 4H2O atoms,

So we can replace this reduced half reaction further to

Step 6

We can simplify things even further by eliminating the excess H2O atoms, thus giving us,

In addition, the balancing will continue by ensuring that the charges will remain balanced, and we will do this by adding electrons to equalize the charges. We will need to add 3 electrons to counter the -4 charge on the right side

Step 7

Use the same approach for the oxidized reaction:

After balancing with H2O, H+, OH - and electrons, we get the final oxidized half reaction of:

Step 8

Next we must balance the amount of electrons like we did in the previous example so that the electrons will cancel, so we multiply the reduced half reaction by 2 and the oxidized half reaction by 3,

giving us,

(reduced)

(oxidized)

Step 9

Combine the equations, while making sure to cancel the extra OH- ions, the extra H2O molecules and eliminate the electrons to get the final balanced reaction in a basic solution which is our answer,

Notice that the only difference between this solution and the previous examples solution is that this has OH- ions in its final balanced reaction rather than H+ ions.