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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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  1. moongazer Group Title
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    KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O

    • 2 years ago
  2. moongazer Group Title
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    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 :)

    • 2 years ago
  3. moongazer Group Title
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    pleas help :)

    • 2 years ago
  4. Callisto Group Title
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    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
  5. Callisto Group Title
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    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
  6. moongazer Group Title
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    was the first step I did is correct?

    • 2 years ago
  7. moongazer Group Title
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    @Callisto I think I slightly understood it. I'll try to answer by myself.

    • 2 years ago
  8. moongazer Group Title
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    :)

    • 2 years ago
  9. Callisto Group Title
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    for Cl 0-(-1)=1e- loss/atom x 1= 1e- loss for Mn 2-7=-5e- gain/atom x 1 = 5e- gain ^Yes, correct

    • 2 years ago
  10. Callisto Group Title
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    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..

    • 2 years ago
  11. Callisto Group Title
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    Btw, ''for Cl'' you're talking one Cl only, but there are two actually!

    • 2 years ago
  12. moongazer Group Title
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    what?

    • 2 years ago
  13. Callisto Group Title
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    *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
  14. moongazer Group Title
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    I think I got it now. Thanks :)

    • 2 years ago
  15. Callisto Group Title
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    Welcome :)

    • 2 years ago
  16. moongazer Group Title
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    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
  17. Callisto Group Title
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    I think so...

    • 2 years ago
  18. Callisto Group Title
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    When I learnt this method, the tutor said it's a fast method, but very easy to make mistakes!

    • 2 years ago
  19. moongazer Group Title
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    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

    • 2 years ago
  20. moongazer Group Title
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    because the atom of Cl on the left side is now 2

    • 2 years ago
  21. Callisto Group Title
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    It looks better now :)

    • 2 years ago
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