Chemistry Balancing Redox Reaction: FeS2+Na2O2→Na2FeO4+Na2SO4+Na2O"

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The discussion focuses on balancing the redox reaction FeS2 + Na2O2 → Na2FeO4 + Na2SO4 + Na2O. It highlights the electron transfer involved, specifically that Fe2+ is oxidized to Fe6+ and sulfur is oxidized from S-1 to S6+, totaling 18 electrons. To balance the reaction, the third half-reaction involving oxygen is multiplied by 9 to account for the electron transfer. There is a debate about whether to apply the multiplication to both sides of the semi-reaction or just the right side. Ultimately, the participants agree to multiply only the right side while balancing the left side accordingly.
DottZakapa
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Homework Statement
Balance redox reaction
Relevant Equations
FeS2+Na2O2->Na2FeO4+Na2SO4+Nao
FeS2+Na2O2->Na2FeO4+Na2SO4+Na2O

Fe2+ --->Fe+6+4e-
2S-1 --->2S+6+14e-
2O- + 2e- ---> 2O
The first two have a total of 18 electrons so
In order to balance the electrons i should multiply by 9 the third semireaction
(2O- + 2e- ---> 2O-2)x9

now this 9 is going to multiply the left hand in the semi reaction equation, isn't also multiplying the right hand side?
 
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It is, yes. But is this really the way you want to handle this ? Why not solve for $$a\text{A}+b\text{B}\rightarrow c\text{C}+d\text{D} + e\text{E}\ ?$$
 
DottZakapa said:
...now this 9 is going to multiply the left hand in the semi reaction equation, isn't also multiplying the right hand side?
Yes
 
sorted
i multiply by 9 just the right hand side
while i balance as usual the left hand side due to the multiple presence of the oxygen
 

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