# Adjusting redox reactions

$$\ce{KCl + KMnO4 + H2SO4 -> K2SO4 + MnSO4 + H2O + Cl2}$$

Attempt. I want to adjust this redox reaction, I've figured out that Chlorine and Manganese are the 2 elements that change thier oxidation number. But I've encountered the problem that if I do:

$$\ce{KCl -> 1/2 Cl2}$$

I don't have any ideas on how to adjust the K atoms, similar stuff with the K and S on the manganese half reaction. Is this impossible to do? Beforehand, I've been doing redox reactions where none of this issue ocurred and I could adjust O and H by adding $$\ce{H2O}$$ or $$\ce{H+}$$ but what do I do here?

• Be careful in confusing 0 and O in formulas. Most of chemists still think 0 means zero. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use MathJax with mhchem – Poutnik Apr 11 at 11:18
• If each child wants just 1/2 of an apple, how many children you can serve with 1 apple ? Basically, all the enumeration of chemical equations can be reduced to counting until 20 and other basic stuff of elementary school arithmetics. – Poutnik Apr 11 at 11:24
• ..what are you talking about? that doesn't help me at all – Xetrez Apr 11 at 11:32
• Try to apply it. It is very easy, if you forget it is chemistry. If apples fit in 3 rows with 5 columns, how many rows are needed if there are just 3 columns ? – Poutnik Apr 11 at 12:05
• Cl2 had 2 chlorine atoms. KCl has 1. How many KCl you need for 1 Cl2 ? Do you see ? Just children, apples, rows and columns. Many people lose elementary school skills when it comes to chemistry. – Poutnik Apr 11 at 12:12

## 1 Answer

Apparently the discussion does not go much ahead. So I will try to speed it up and show how it goes. First try to establish the half-reaction with ions, not with molecules and formula units. We will go back to neutral compounds later on and get rid of the positive and negative charges at the end.

Here, in the first half-reaction, the ion $$\ce{MnO4-}$$ is reduced to $$\ce{Mn^{2+}}$$, and the $$4$$ lost oxygen atoms are transformed into $$\ce{H2O}$$ molecules, here $$\ce{4H2O}$$ to be added on the other side of the arrow. These $$4$$ water molecules introduce unwanted $$\ce{H}$$ atoms, that must appear on the other side of the arrow as $$\ce{8 H+}$$ ions In the second half-reaction, the ion $$\ce{Cl-}$$ is oxidized to $$\ce{Cl2}$$. Hopefully you know how to build these half-equations. If you don't, tell us, and we will explain it in more details. Here are these half-equations : $$\ce{MnO4- + 8 H+ + 5 e- -> Mn^{2+} + 4 H2O} \\ \ce{2 Cl- -> Cl2 + 2 e-}$$ Now you have to add these two half-equations after multiplication of both in order to make the electrons disappear. This can be done by multiplying the first half-equation by $$2$$ and the second by $$5$$. Look. $$\ce{2 MnO4- + 16 H+ + 10 e- -> 2 Mn^{2+} + 8 H2O} \\ \ce{10 Cl- -> 5 Cl2 + 10 e-}$$ This produces $$10$$ electrons on both sides, so that the electrons can be skipped after summing. $$\ce{2 MnO4- + 10 Cl- + 16 H+ -> 2 Mn^{2+} + 5 Cl2 + 8 H2O}$$ The final equation is obtained. But now you would like to have the ions disappear, so that only neutral species appear in the equation. So you must add $$2+10 = 12$$ ions $$\ce{K+}$$ ions on the left hand side to compensate the anions and form neutral potassium salts. You also must add $$8$$ ions $$\ce{SO4^{2-}}$$ to compensate the $$\ce{H+}$$ ions. Afterwards you must take care of these new ions and make some neutral salts on the right hand side. So the two $$\ce{Mn^{2+}}$$ ions recombine with $$2$$ of the $$\ce{8 SO4^{2-}}$$ ions, forming $$\ce{8 MnSO4}$$. The remaining ions produce $$\ce{6 K2SO4}$$ . This gives : $$\ce{2 KMnO4 + 10 KCl + 8 H2SO4 -> 2 MnSO4 + 5 Cl2 + 6 K2SO4 + 8 H2O}$$ Is it OK ?