Balance the following redox reaction
MnO4- +H2S> Mn+2 +S
I've assigned the oxidation numbers for everything, i just cant write out the half reactions
I got this, but I don't know if its right. Also how can i balance the H and O?
Oxidation: MnO4- + 8H+ > Mn+2 4H2O
Reduction: H2S > S
Can you tell me if what i did is right? and if its not can you tell me what i should do?
MnO4- +H2S> Mn+2 +S
I've assigned the oxidation numbers for everything, i just cant write out the half reactions
I got this, but I don't know if its right. Also how can i balance the H and O?
Oxidation: MnO4- + 8H+ > Mn+2 4H2O
Reduction: H2S > S
Can you tell me if what i did is right? and if its not can you tell me what i should do?
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You are half way there on the MnO4^- half equation, you just need to do the electrons.
(The method below is for reactions under acidic conditions. Basic conditions are different, some of my recent answers show balancing of basic conditions if you want some examples.)
MnO4^- ----> Mn^2+
balance O by adding H2O to the other side of the arrow
MnO4^- ----> Mn^2+ + 4H2O
Now, balance the H by adding H+ to the other side of the arrow (for acidic conditions)
MnO4^- + 8H+ ----> Mn^2+ + 4H2O
all the atoms now balance, so work out the electron transfer.
To do this add up the charges on each side of the arrow. Then add the right amount of electrons to the MOST POSITIVE side that will result in an equal charge on both sides
LHS = 1 x MnO4^- + 8 x H+ = 7+
RHS = 1 x Mn^2+ = 2+
add 5 e to the LHS, this makes the charge on both sides 2+
MnO4^- + 8H+ + 5e -----> Mn^2+ + 4H2O
this is the balanced reduction (gain of electrons) half equation
Oxidation half equation
H2S ----> S
balance H
H2S ----> S + 2H+
balance charge with electrons
LHS = neutral
RHS = 2 x H+ = 2+
add 2 e to the RHS
H2S -----> S + 2H+ + 2e
this is the balanced oxidation half equation
Now, the oxidation half equation transfers 2 electrons and the reduction half equation transfers 5.
The electrons need to balance as well, so multiply each half equation by the number of electrons transferred in the other.
2 x (MnO4^- + 8H+ + 5e -----> Mn^2+ + 4H2O)
gives us 2MnO4^- + 16H+ + 10e -----> 2Mn^2+ + 8H2O
5 x (H2S -----> S + 2H+ + 2e)
gives us 5H2S -----> 5S + 10H+ + 10e
(note there are now an equal number of electrons (10) in each)
then add them together
2MnO4^- + 16H+ + 10e -----> 2Mn^2+ + 8H2O
5H2S -----> 5S + 10H+ + 10e
--------------------------------------…
2MnO4^- + 5H2S + 16H+ + 10e --> 2Mn2+ + 8H2O + 5S + 10H+ + 10e
cancel out the stuff that occurs on both sides
2MnO4^- + 5H2S + 6H+ --> 2Mn2+ + 8H2O + 5S
(The method below is for reactions under acidic conditions. Basic conditions are different, some of my recent answers show balancing of basic conditions if you want some examples.)
MnO4^- ----> Mn^2+
balance O by adding H2O to the other side of the arrow
MnO4^- ----> Mn^2+ + 4H2O
Now, balance the H by adding H+ to the other side of the arrow (for acidic conditions)
MnO4^- + 8H+ ----> Mn^2+ + 4H2O
all the atoms now balance, so work out the electron transfer.
To do this add up the charges on each side of the arrow. Then add the right amount of electrons to the MOST POSITIVE side that will result in an equal charge on both sides
LHS = 1 x MnO4^- + 8 x H+ = 7+
RHS = 1 x Mn^2+ = 2+
add 5 e to the LHS, this makes the charge on both sides 2+
MnO4^- + 8H+ + 5e -----> Mn^2+ + 4H2O
this is the balanced reduction (gain of electrons) half equation
Oxidation half equation
H2S ----> S
balance H
H2S ----> S + 2H+
balance charge with electrons
LHS = neutral
RHS = 2 x H+ = 2+
add 2 e to the RHS
H2S -----> S + 2H+ + 2e
this is the balanced oxidation half equation
Now, the oxidation half equation transfers 2 electrons and the reduction half equation transfers 5.
The electrons need to balance as well, so multiply each half equation by the number of electrons transferred in the other.
2 x (MnO4^- + 8H+ + 5e -----> Mn^2+ + 4H2O)
gives us 2MnO4^- + 16H+ + 10e -----> 2Mn^2+ + 8H2O
5 x (H2S -----> S + 2H+ + 2e)
gives us 5H2S -----> 5S + 10H+ + 10e
(note there are now an equal number of electrons (10) in each)
then add them together
2MnO4^- + 16H+ + 10e -----> 2Mn^2+ + 8H2O
5H2S -----> 5S + 10H+ + 10e
--------------------------------------…
2MnO4^- + 5H2S + 16H+ + 10e --> 2Mn2+ + 8H2O + 5S + 10H+ + 10e
cancel out the stuff that occurs on both sides
2MnO4^- + 5H2S + 6H+ --> 2Mn2+ + 8H2O + 5S