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

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

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 ?