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

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moongazer Group TitleBest ResponseYou've already chosen the best response.0
KMnO4 + KCl + H2SO4 > K2SO4 + MnSO4+Cl2+H2O
 2 years 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 :)
 2 years ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
pleas help :)
 2 years 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!
 2 years 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. \[MnO_4^+8H^+ + 5e^ \rightarrow Mn^{2+}+4H_2O\]\[2Cl^ \rightarrow Cl_2 +2e^\]
 2 years ago

moongazer Group TitleBest ResponseYou've already chosen the best response.0
was the first step I did is correct?
 2 years 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.
 2 years 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
 2 years 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..
 2 years 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!
 2 years 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!)
 2 years ago

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

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Welcome :)
 2 years 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 ?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I think so...
 2 years 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!
 2 years 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
 2 years ago

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

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