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moongazer
Group Title
How to balance this by oxidation state change method?
KMnO4 + KCl + H2SO4 > K2SO4 + MnSO4+Cl2+H2O
 one year ago
 one year ago
moongazer Group Title
How to balance this by oxidation state change method? KMnO4 + KCl + H2SO4 > K2SO4 + MnSO4+Cl2+H2O
 one year ago
 one year ago

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moongazer Group TitleBest ResponseYou've already chosen the best response.0
KMnO4 + KCl + H2SO4 > K2SO4 + MnSO4+Cl2+H2O
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
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 27=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 27=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 :)
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
pleas help :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
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. \[MnO_4^+8H^+ + 5e^ \rightarrow Mn^{2+}+4H_2O\]\[2Cl^ \rightarrow Cl_2 +2e^\]
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
was the first step I did is correct?
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
@Callisto I think I slightly understood it. I'll try to answer by myself.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
for Cl 0(1)=1e loss/atom x 1= 1e loss for Mn 27=5e gain/atom x 1 = 5e gain ^Yes, correct
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
then I looked for a multiplier for Cl 0(1)=1e loss/atom x 1= 1e loss x 5 = 5 for Mn 27=5e gain/atom x 1 = 5e gain x 1 = 5 ^No, some problems here..
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Btw, ''for Cl'' you're talking one Cl only, but there are two actually!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
*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!)
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
I think I got it now. Thanks :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Welcome :)
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I think so...
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
When I learnt this method, the tutor said it's a fast method, but very easy to make mistakes!
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
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 27=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 27=5e gain/atom x 1 = 5e gain x 2 = 10 I made it x 2
 one year ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
because the atom of Cl on the left side is now 2
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
It looks better now :)
 one year ago
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