Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of h2c2o4 and h2o in the balanced reaction?

See explanation below

Explanation:

The overall reaction, is:

MnO4⁻(aq) + H₂C₂O₄(aq) > Mn²⁺(aq) + CO₂(g)

Balancing this redox reaction means that one compound is reducting while the other is oxidizing. So, we need to separate both compounds into 2 semi equations and balance both of them, per separate and then, we can join them.

As we want to balance in acid medium, means that we need to add water and H⁺ in both reactions. Doing that we have the following:

MnO₄⁻   > Mn²⁺

In this reaction, we can clearly see that it's not balanced. To balance this semi equation, let's see the elements. In the reactans we have Mn and O, but in the products we only have Mn, the atom of oxygen where could it be? As we are doing acid medium, if in the reactants we have oxygen, this oxygen can be as products in the form of water, so we add water there.

MnO₄⁻   > Mn²⁺ + H₂O

Now, the water has hydrogen atoms, and if we are in acid medium, the hydrogen can only come from the acid medium, and in this case H⁺ so:

H⁺ + MnO₄⁻ > Mn²⁺ + H₂O

Now, it's time to balance the charges. First Mn²⁺ is the lowest oxidation state of the manganese, this means that in the reactants Mn is passing from a higher state to a lower state, therefore, this compouns is reducting. How many electrons? well, in this case, we know that oxygen usually have the oxidation state -2, so the manganese would be:

(-2 * 4) + x = -1

-8 + x = -1 > x = +7

Therefore, manganese passes from 7+ to 2+, it's gaining 5 electrons so:

H⁺ + MnO₄⁻ + 5e⁻ > Mn²⁺ + H₂O

Finally, we just balance the masses and charges:

8H⁺ + MnO₄⁻ + 5e⁻> Mn²⁺ + 4H₂O

Now, we just do the same thing with the other semi equation which is oxydizing. The explanation of that, is similar to this, so I'm gonna do it directly:

C₂O₄²⁻ > CO₂

In this case, we can easily see that carbon is losing 2 electrons, so, let's put the 2 electrons on the product to balance the charges, and then, the masses:

C₂O₄²⁻ > CO₂ + 2e⁻

C₂O₄²⁻ > 2CO₂ + 2e⁻

Let's join both equations and do the sum of them:

8H⁺ + MnO₄⁻ + 5e⁻> Mn²⁺ + 4H₂O

C₂O₄²⁻ > 2CO₂ + 2e⁻

As we can see, we do not have the same electrons on both equations, we need to equal those values so:

2 * (8H⁺ + MnO₄⁻ + 5e⁻> Mn²⁺ + 4H₂O)

5 * (C₂O₄²⁻ > 2CO₂ + 2e⁻)

16H⁺ + 2MnO₄⁻ + 10e⁻> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ > 10CO₂ + 10e⁻

Now, let's sum both equations:

16H⁺ + 2MnO₄⁻ + 10e⁻> 2Mn²⁺ + 8H₂O

5C₂O₄²⁻ > 10CO₂ + 10e⁻

16H⁺ + 2MnO₄⁻ + 5C₂O₄²⁻ > 2Mn²⁺ + 10CO₂ + 8H₂O

This would be the balanced reaction, however, let's put it as it was originally with the H2 in the C2O4 and balance it: