Half Reaction Problem: Get Help Here

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To solve the half-reaction problem, equal amounts of oxygen and hydrogen must be present on both sides of the equation. Adding H2O to the left side helps achieve this balance. The oxidation state of MnO4 is 7, and it reduces to MnO2 with an oxidation state of 4, requiring three electrons for the reduction. It's important to balance the electrons in both half-reactions before addressing hydrogen and oxygen. Proper balancing is crucial for accurately representing the chemical reaction.
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Thanks for the help guys.
 
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You need equal amounts of the elements O and H on both sides of the equation. Try adding H2O on the left side and balance the elements, then balance the charges.
 
Thanks for the help guys.
 
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fasterthanwoo said:
I am still having some trouble with this.

The MnO4 is being reduced.

The oxidation state of MnO4 is 7. The oxidation state of MnO2 is 4. Therefore I need three e- to reduce?

Is it needed to have H2O on the left side of the reaction?

Thank you.

Yes. Three electrons to reduce Mn. Now balance the electrons for both half-reactions before worrying about the H's and the O's and the H2O's.
 
Thanks for the help guys.
 
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Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

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