Can someone help balance the following equation?
H2S + KMnO4 + H2SO4 = K2SO4 + S8 + MnSO4 + H2O
I was told that there should be a total of 64 H2O when it is balanced correctly.
Chemical Equation Balance
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The confusing part is that sulfur present as sulfuric acid and as H2S, and only H2S participates in redox reaction. Identify the oxidizer (KMnO4) and the reducer (H2S).
Write the process with electron transfer:
8 S(2-) – 16 e- = S8 (0)
Mn(7+) + 5 e- = Mn(2+)
Now to assure the reducer and oxidizer exchange with equal number of electrons we have to multiply the first equation to 5 and the second to 16, so they will exchange 80 electrons in the process:
8 S(2-) – 16 e- = S8 (0) x 5
Mn(7+) + 5 e- = Mn(2+) x 16
This will give you initial ratio between reducer and oxidizer in the equation:
40 H2S + 16 KMnO4 + * H2SO4 = * K2SO4 + 5 S8 + 16 MnSO4 + * H2O
Equalize other ingredients and finish the task:
40 H2S + 16 KMnO4 + 24 H2SO4 = 8 K2SO4 + 5 S8 + 16 MnSO4 + 64 H2O
Write the process with electron transfer:
8 S(2-) – 16 e- = S8 (0)
Mn(7+) + 5 e- = Mn(2+)
Now to assure the reducer and oxidizer exchange with equal number of electrons we have to multiply the first equation to 5 and the second to 16, so they will exchange 80 electrons in the process:
8 S(2-) – 16 e- = S8 (0) x 5
Mn(7+) + 5 e- = Mn(2+) x 16
This will give you initial ratio between reducer and oxidizer in the equation:
40 H2S + 16 KMnO4 + * H2SO4 = * K2SO4 + 5 S8 + 16 MnSO4 + * H2O
Equalize other ingredients and finish the task:
40 H2S + 16 KMnO4 + 24 H2SO4 = 8 K2SO4 + 5 S8 + 16 MnSO4 + 64 H2O
Remember safety first! Check MSDS and consult with professionals before performing risky experiments.
expert wrote:The confusing part is that sulfur present as sulfuric acid and as H2S, and only H2S participates in redox reaction. Identify the oxidizer (KMnO4) and the reducer (H2S).
Write the process with electron transfer:
8 S(2-) – 16 e- = S8 (0)
Mn(7+) + 5 e- = Mn(2+)
I would go about this in a different way entirely.
You have already identified the KMnO4 as the oxidant and you should know how to construct the half equation for the oxidation process. If you don't then this is an area to work on.
MnO4(-) + 8H+ + 5e --> Mn2+ + 4H2O
Now construct the half equation for the oxidation of H2S
8H2S --> 16H+ + S8 + 16e
Now look for the simplest common denominator of 5 & 16 = 80
So, the first half equation must be multiplied by 16 and the second by 5. This will equalise the electrons and allow the equations to be added together.
16MnO4(-) + 128H+ + 80e --> 16Mn2+ + 64H2O
40H2S --> 80H+ + 5S8 + 80e
------------------------------------------add together and cancel electrons
16MnO4(-) + 128H+ + 40H2S --> 16Mn2+ + 64H2O + 80H+ + 5S8
now cancel out hydrogen terms
16MnO4(-) + 48H+ + 40H2S --> 16Mn2+ + 64H2O + 5S8
Now put in the balancing potassium and sulphate ions (H+ forms part of H2SO4)
16MnO4(-) + 112H+ + 40H2S --> 16Mn2+ + 64H2O + 5S8
--------------------------------------------
16KMnO4 + 24H2SO4 + 40H2S --> 16MnSO4 + 64H2O + 5S8 + 8K2SO4