Balancing question

[jza70]

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Balanced in a basic solution
MnO^4- + AsO3^3- -> MnO2 + AsO4^3-

I can't seem to balance this correctly.
Thanks

 

[herkulease]

Yeah I am not able to balance it either.

You can't chance out the charges. when you combine the half reactions.

 

[jza70]

Thanks for trying. The question was how many moles of AsO4^3- will be on the product side after balancing the question.

The answer was 3.

 

[Baylor2011]

Are you sure you wrote that down correctly?

 

[UCB05]

Assuming the first term in your equation is a typo for permanganate (MnO4-)
Your half reactions are:

MnO4- -> MnO2
AsO3^-3 -> AsO4^-3

Add water to balance out O's, then H+ to balance out H's, then electrons to balance out total charges.

MnO4- + 4H+ + 3e- -> MnO2 + 2H2O
AsO3^-3 + H2O -> AsO4-3 + 2H+ + 2e-

2(MnO4- + 4H+ + 3e- -> MnO2 + 2H2O)
3(AsO3^-3 + H2O -> AsO4-3 + 2H+ + 2e-)

When you balance out the electrons by multiplying coefficients, you can see that AsO4^-3 will have a coefficient of 3 for the remainder of the balancing steps, and that will be your answer.

2MnO4- + 8H+ + 6e- -> 2MnO2 + 4H2O
3AsO3^-3 + 3H2O -> 3AsO4-3 + 6H+ + 6e-

2MnO4 + 3AsO3^-3 + 2H+ -> 2MnO2 + 3AsO4^-3 + H2O

Since it's in basic conditions we have to add -OH to eliminate the H+, forming water

2MnO4 + 3AsO3^-3 + 2H+ + 2-OH -> 2MnO2 + 3AsO4^-3 + H2O + 2-OH

2MnO4 + 3AsO3^-3 + 2H2O -> 2MnO2 + 3AsO4^-3 + H2O + 2-OH

2MnO4 + 3AsO3^-3 + H2O -> 2MnO2 + 3AsO4^-3 + 2-OH




If MnO^4- was in fact not a typo, then 😕

 

[jza70]

Here is the actual question

That is a 4- behind the Mn0? Thats what it seems to look like.
It looks like UCB figured it out. Thanks!

 

[UCB05]

That's gotta be a typo. I've never heard of MnO^4-, Mn would have to have a charge of -2, and it would mean both species gets oxidized.

 

[sfoksn]

I suggest you not use that exam source 😛

What is that? Ace the DAT?

 

[jza70]

Kaplan ineractive online.

 

[Baylor2011]

That's really odd..