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balanced equation

I - + MnO4- + h2O -> IO- + MnO2 + OH-

what is the ratio of Mno2 to OH- in the balanced equation

i got 1:2 for this question but not sure if i was doing it right, can someone check if ive done it correctly
Original post by terr123_78787
I - + MnO4- + h2O -> IO- + MnO2 + OH-

what is the ratio of Mno2 to OH- in the balanced equation

i got 1:2 for this question but not sure if i was doing it right, can someone check if ive done it correctly

show your working and we'll check it ...
Original post by charco
show your working and we'll check it ...

when balancing the equation i got I- + MnO4- + H2O ----> IO- + MnO2 + 2OH-

so thats why i thought it was 1: 2 as MnO2 : 2OH- is 1:2
Original post by terr123_78787
when balancing the equation i got I- + MnO4- + H2O ----> IO- + MnO2 + 2OH-

so thats why i thought it was 1: 2 as MnO2 : 2OH- is 1:2

I said, "show your working". This is not your working, it's your answer!
Original post by charco
I said, "show your working". This is not your working, it's your answer

I- + MnO4- + H2O ----> IO- + MnO2 + 2OH-

I- - balanced on each side
MnO4- Mn is balanced but O4 isnt

H2O- one H is balanced, one isnt so two OH are needed, O is balanced

IO is balanced

MnO2 is balanced

therefore only a 2 is added in front of OH
Original post by terr123_78787
I- + MnO4- + H2O ----> IO- + MnO2 + 2OH-

I- - balanced on each side
MnO4- Mn is balanced but O4 isnt

H2O- one H is balanced, one isnt so two OH are needed, O is balanced

IO is balanced

MnO2 is balanced

therefore only a 2 is added in front of OH

To balance redox equations you need to follow a process:

1. Generate the reduction and oxidation half-equations.
2. Then equalise the electrons
3. Then add together the half-equations
4. Finally gather terms.

In this case:
iodide turns to iodate(I) in basic solution (this means that you can only use water and hydroxide ions to balance)

Oxidation: I-(aq) ==> IO-(aq)

I-(aq) + 2OH- ==> IO- + H2O + 2e-

Reduction: MnO4- + 2H2O ==> MnO2 + 4OH-

MnO4- + 2H2O + 3e- ==> MnO2 + 4OH-

Now equalise the electrons:

3I-(aq) + 6OH- ==> 3IO- + 3H2O + 6e-
2MnO4- + 4H2O + 6e- ==> 2MnO2 + 8OH-
----------------------------------------------------------- add together
3I-(aq) + 2MnO4- + H2O ==> 3IO- + 2MnO2 + 2OH-

and that's the balanced equation.
Original post by charco
To balance redox equations you need to follow a process:

1. Generate the reduction and oxidation half-equations.
2. Then equalise the electrons
3. Then add together the half-equations
4. Finally gather terms.

In this case:
iodide turns to iodate(I) in basic solution (this means that you can only use water and hydroxide ions to balance)

Oxidation: I-(aq) ==> IO-(aq)

I-(aq) + 2OH- ==> IO- + H2O + 2e-

Reduction: MnO4- + 2H2O ==> MnO2 + 4OH-

MnO4- + 2H2O + 3e- ==> MnO2 + 4OH-

Now equalise the electrons:

3I-(aq) + 6OH- ==> 3IO- + 3H2O + 6e-
2MnO4- + 4H2O + 6e- ==> 2MnO2 + 8OH-
----------------------------------------------------------- add together
3I-(aq) + 2MnO4- + H2O ==> 3IO- + 2MnO2 + 2OH-

and that's the balanced equation.

that would mean that the ratio of MnO2 to OH- would be 2:2 which cancels down to 1:1 ??
Original post by terr123_78787
that would mean that the ratio of MnO2 to OH- would be 2:2 which cancels down to 1:1 ??

yes
thank you so much, you really helped!
I’d usually use H2O first to balance oxygen and the H+ to balance the hydrogen. I see in this question they’re using H2O and OH- how do you know if you need to use H2O or OH for balancing oxygen or for hydrogen?


Original post by charco
To balance redox equations you need to follow a process:

1. Generate the reduction and oxidation half-equations.
2. Then equalise the electrons
3. Then add together the half-equations
4. Finally gather terms.

In this case:
iodide turns to iodate(I) in basic solution (this means that you can only use water and hydroxide ions to balance)

Oxidation: I-(aq) ==> IO-(aq)

I-(aq) + 2OH- ==> IO- + H2O + 2e-

Reduction: MnO4- + 2H2O ==> MnO2 + 4OH-

MnO4- + 2H2O + 3e- ==> MnO2 + 4OH-

Now equalise the electrons:

3I-(aq) + 6OH- ==> 3IO- + 3H2O + 6e-
2MnO4- + 4H2O + 6e- ==> 2MnO2 + 8OH-
----------------------------------------------------------- add together
3I-(aq) + 2MnO4- + H2O ==> 3IO- + 2MnO2 + 2OH-

and that's the balanced equation.

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