# Is this balanced redox equation correct? [closed]

I'm trying to balance an equation using oxidation numbers

$$\ce{MnO4^- + H+ + Cl- -> Mn^2+ + Cl2 + H2O}$$

my suggestion:

$$\ce{MnO4- + 8H+ + 5Cl- → Mn^2+ + 2.5Cl2 + 4H2O}$$

the 2.5 $$\ce{Cl2}$$ is throwing me off, would like to make sure my methodology is correct.

• Could you edit the question and tell us what your methodology is (or how you check whether the equation makes sense)? How many electrons are being transferred, and from where to where? If the 2.5 is bothering you, just multiply all coefficients (including the explicit ones in front of the manganese species) by two. – Karsten Theis Jun 11 '19 at 1:03

The balanced equation seems correct. Multiplying by $$2$$ we get the following:

$$\ce{2 MnO4- + 16 H+ + 10 Cl- → 2 Mn^2+ + 5 Cl2 + 8 H2O}$$

which is also correct.

Actually, if you balance given redox reaction strictly following the half-reactions method where you have to equate the number of transferred electrons, you won't end up with fractional coefficients:

\begin{align} \ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- &→ \overset{+2}{Mn}^2+ + 4 H2O} &|\cdot 2 \tag{red}\\ \ce{2 \overset{-1}{Cl}^- &→ \overset{0}{Cl}_2 + 2 e-} &|\cdot 5 \tag{ox}\\ \hline \ce{2 MnO4- + 16 H+ + 10 Cl- &→ 2 Mn^2+ + 8 H2O + 5 Cl2} \tag{redox} \end{align}

As with any equation, you sure can divide both parts by any real number which is not zero. Dividing the balanced redox equation above by two yields in your solution.