How do we count charges in this example of a Redox reaction equation? My brother(younger) had a doubt regarding this chemistry book example so I let him here :

Question :

Balance the equation : $\ce{H+ + MnO4- + Fe^2+ -&gt; Fe^3+ + Mn^2+}$

Solution steps(as given by the textbook):

Step 1 :  $$\ce{MnO4-  -&gt;  Mn^2+}$$

Step 2 : $$\ce{MnO4-  -&gt; Mn^2+ + 4 H2O}$$

Step 3 :  $$\ce{MnO4- + 8 H+  -&gt; Mn^2+ + 4 H2O}$$

Step 4 : $$\ce{MnO4- + 8 H+ + 5 e- -&gt; Mn^2+ + 4 H2O}\tag{1}$$

Step 5 : $$\ce{Fe^2+ -&gt; Fe^3+ + e-}\tag{2}$$

Step 6 : $5\times(2) + (1)$ and thus we have the balanced equation:

$$\ce{MnO4- + 8 H+ + 5 Fe^2+ -&gt; 5 Fe^3+ + Mn^2+ + 4 H2O}$$

Explanation given in the textbook for step 4:  we equalise the charge by adding 5 electrons on left.

My doubt regarding step 4: how do I know that in step 4 that 5 electrons do actually equalize the charge in the given equation? More accurately how do I know that there is any surplus or deficit of charge in any given equation atleast in this case?
If you look at the charges in step 4, the left side adds up to 7 and the right side adds up 2. The MnO4 on the left has a -1 charge, and the 8 hydrogens add a +8 charge, adding up to a +7 charge on the left overall. The Mn on the right has a +2 charge so that is +2.

To balance out this charge, they must be equal. How can +7 be made equal to +2? By subtracting 5. In this case, adding 5 electrons creates a -5 charge that means both sides of the equation have a +2 charge.
You just need to add up the charges like this for a given equation to balance it.