# balanced equation

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#1
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
0
6 months ago
#2
(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 ...
0
#3
(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
0
6 months ago
#4
(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
0
#5
(Original post by charco)
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
0
6 months ago
#6
(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-
3I-(aq) + 2MnO4- + H2O ==> 3IO- + 2MnO2 + 2OH-

and that's the balanced equation.
1
#7
(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-
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 ??
0
6 months ago
#8
(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
0
#9
thank you so much, you really helped!
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