Question

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

Answer 1

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

Answer 2

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 :)

Answer 3

pleas help :)

Answer 4

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!

Answer 5

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.
\[MnO_4^-+8H^+ + 5e^- \rightarrow Mn^{2+}+4H_2O\]\[2Cl^- \rightarrow Cl_2 +2e^-\]

Answer 6

was the first step I did is correct?

Answer 7

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

Answer 8

:)

Answer 9

for Cl
0-(-1)=1e- loss/atom x 1= 1e- loss

for Mn
2-7=-5e- gain/atom x 1 = 5e- gain

^Yes, correct

Answer 10

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..

Answer 11

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

Answer 12

what?

Answer 13

*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!)

Answer 14

I think I got it now. Thanks :)

Answer 15

Welcome :)

Answer 16

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 ?

Answer 17

I think so...

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

It looks better now :)