-1
$\begingroup$

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.

$\endgroup$

closed as off-topic by Todd Minehardt, user55119, Karsten Theis, airhuff, Mathew Mahindaratne Jun 11 at 20:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 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. $\endgroup$ – Karsten Theis Jun 11 at 1:03
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.