# 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 : $\ce{H+ + MnO4- + Fe^2+ -> Fe^3+ + Mn^2+}$

Solution steps(as given by the textbook):

Step 1 : $$\ce{MnO4- -> Mn^2+}$$

Step 2 : $$\ce{MnO4- -> Mn^2+ + 4 H2O}$$

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

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

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

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

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

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?

• @Nij Please check out formatting hints for chemical equations. Using plain MathJax for chemical equations is wrong since the elements should be typeset upright, not in italics. – Jan Aug 16 '16 at 19:53