## moongazer 3 years ago How to balance this by oxidation state change method? KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O

1. moongazer

KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O

2. moongazer

oxidation numbers are enclosed with parenthesis K(+1)Mn(+7)O4(-2) + K(+1)Cl(-1) + H2(+1)S(+6)O4(-2) --> K2(+1)S(+6)O4(-2) + Cl2(0) + H2(+1)O(-2) Then I identify which elements change in oxidation numbers for Cl 0-(-1)=1e- loss/atom x 1= 1e- loss for Mn 2-7=-5e- gain/atom x 1 = 5e- gain then I looked for a multiplier for Cl 0-(-1)=1e- loss/atom x 1= 1e- loss x 5 = 5 for Mn 2-7=-5e- gain/atom x 1 = 5e- gain x 1 = 5 the multiplier are the coefficients but when I checked it, it is not balanced what should I do ? please help :) I already spent plenty of time thinking on what should I do :)

3. moongazer

pleas help :)

4. Callisto

KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O For this reaction, there is no change for oxidation number (O.N.) of H2SO4, so neglect it for a while. Like you've mentioned, Cl and Mn involve O.N. change, so write the equation for it first. $MnO_4^- + Cl^- \rightarrow Mn^{2+} + Cl_2$Balance the number of Cl first. $MnO_4^- + 2Cl^- \rightarrow Mn^{2+} + Cl_2$ ON change for Mn = (+2) - (+7) = -5 ON change for Cl = 0 - (-1) = +1 Note that $$Cl^-$$ on the left has been multiplied by 2 to , so, instead of +1, it's +2 So, multiply Cl by 5 and multiply Mn by 2 $2MnO_4^- + 10Cl^- \rightarrow 2Mn^{2+} + 5Cl_2$Balance the number of oxygen by adding $$H_2O$$ $2MnO_4^- + 10Cl^- \rightarrow 2Mn^{2+} + 5Cl_2+8H_2O$Balance the number of H by adding $$H^+$$ $2MnO_4^- + 10Cl^- +16H^+\rightarrow 2Mn^{2+} + 5Cl_2+8H_2O$ Now, we know that there are 2 $$H^+$$ in $$H_2SO_4$$, so, we can get $2MnO_4^- + 10Cl^- +8H_2SO_4\rightarrow 2Mn^{2+} + 5Cl_2+8H_2O$ Put $$SO_4^{2-}$$ with $$Mn^{2+}$$ $2MnO_4^- + 10Cl^- +8H_2SO_4\rightarrow 2MnSO_4+ 5Cl_2+8H_2O$ Add K in front of $$MnO_4^-$$ and $$Cl^-$$ and balance it on the right $2KMnO_4 + 10KCl +8H_2SO_4\rightarrow 12K^++2MnSO_4+ 5Cl_2+8H_2O$ For K and SO4, we need 2 K for 1 SO4 (Since it's K2SO4), so, reduce the coefficient of K by half on the right and then write K2SO4 instead of $$K^+$$ $2KMnO_4 + 10KCl +8H_2SO_4\rightarrow 6K_2SO_4+2MnSO_4+ 5Cl_2+8H_2O$ Hmm... Sorry if I confuse you!

5. Callisto

Note: I made some mistake in balancing the change of ON when I was typing that. Then I double check it with the safer method - writing the equations separately. $MnO_4^-+8H^+ + 5e^- \rightarrow Mn^{2+}+4H_2O$$2Cl^- \rightarrow Cl_2 +2e^-$

6. moongazer

was the first step I did is correct?

7. moongazer

@Callisto I think I slightly understood it. I'll try to answer by myself.

8. moongazer

:)

9. Callisto

for Cl 0-(-1)=1e- loss/atom x 1= 1e- loss for Mn 2-7=-5e- gain/atom x 1 = 5e- gain ^Yes, correct

10. Callisto

then I looked for a multiplier for Cl 0-(-1)=1e- loss/atom x 1= 1e- loss x 5 = 5 for Mn 2-7=-5e- gain/atom x 1 = 5e- gain x 1 = 5 ^No, some problems here..

11. Callisto

Btw, ''for Cl'' you're talking one Cl only, but there are two actually!

12. moongazer

what?

13. Callisto

*talking about I meant 2Cl^- -> Cl2 You need to balance it first 2Cl^- -> Cl2 Then, you'll know that there are actually 2 electrons lost (sorry I made a serious mistake in the deleted post!)

14. moongazer

I think I got it now. Thanks :)

15. Callisto

Welcome :)

16. moongazer

Wait, did you mean for any reaction, I need first to balance the elements that changed in oxidation number before looking for a multiplier that I will make as a coefficient ?

17. Callisto

I think so...

18. Callisto

When I learnt this method, the tutor said it's a fast method, but very easy to make mistakes!

19. moongazer

so for this one I need first to do it like this: KMnO4 + 2KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O then follow the steps oxidation numbers are enclosed with parenthesis K(+1)Mn(+7)O4(-2) + K(+1)Cl(-1) + H2(+1)S(+6)O4(-2) --> K2(+1)S(+6)O4(-2) + Cl2(0) + H2(+1)O(-2) Then I identify which elements change in oxidation numbers for Cl 0-(-1)=1e- loss/atom x 2 = 2e- loss for Mn 2-7=-5e- gain/atom x 1 = 5e- gain then I looked for a multiplier for Cl 0-(-1)=1e- loss/atom x 2= 2e- loss x 5 = 10 for Mn 2-7=-5e- gain/atom x 1 = 5e- gain x 2 = 10 I made it x 2

20. moongazer

because the atom of Cl on the left side is now 2

21. Callisto

It looks better now :)