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# Redox equation help !! watch

1. hi, for part b I don't get how it is worked out. I managed to work out the oxidation number but I am not sure what to do after this step. Do I balance the oxidation number first or balance the number of atoms ? Attachment 613366613368thanks
Attached Images

2. Hi there,

based on changes in oxidation number,

1) you need to multiply the "+1" by 2 to balance out the decrease by 2 hence you would have 2 x Mn(+7) species

2) then seeing the equation, you would have total 3 x Mn on RHS, so you need 3 x Mn on LHS

3) then seeing how the rest of the charge species must be untouched since you just balanced the changes in oxidation number, work out the net charge on LHS and net charge on RHS (without the hydroxide), then decide how many hydroxide you need to balance overall charge

4) then balance equation off by counting atoms, and you should get 2 x H2O
(you should get 4 hydroxide on the RHS also from step 3)
3. (Original post by coconut64)
hi, for part b I don't get how it is worked out. I managed to work out the oxidation number but I am not sure what to do after this step. Do I balance the oxidation number first or balance the number of atoms ? Attachment 613366613368thanks

Disproportionation means simultaneous oxidation and reduction of the same species.

The simplest and best way to approach this is to write one half-equation for the oxidation of the manganate species and a second half-equation for the reduction of the manganate species.

Then you deal with these two half-equations as with all combination of half-equations by multiplying through by a suitable integer in each equation to equalise the electrons and then add them together.
4. (Original post by shengoc)
Hi there,

based on changes in oxidation number,

1) you need to multiply the "+1" by 2 to balance out the decrease by 2 hence you would have 2 x Mn(+7) species

2) then seeing the equation, you would have total 3 x Mn on RHS, so you need 3 x Mn on LHS

3) then seeing how the rest of the charge species must be untouched since you just balanced the changes in oxidation number, work out the net charge on LHS and net charge on RHS (without the hydroxide), then decide how many hydroxide you need to balance overall charge

4) then balance equation off by counting atoms, and you should get 2 x H2O
(you should get 4 hydroxide on the RHS also from step 3)
I don't follow step 2. how will get obtain +1 by multiplying MnO4 by 2? Attachment 613746613748
Attached Images

5. (Original post by coconut64)
I don't follow step 2. how will get obtain +1 by multiplying MnO4 by 2? Attachment 613746613748
I suspect your written formula doesnt match what the question said.

it mentioned Manganate (VI) to Manganate(VII)

MnO4- is Mn (+7)

MnO4 2- is Mn(+6)

+6 to +7 is increase by 1, not balanced by decrease of 2 with the other species.

in any case, as i first advised, and also as advised by Charco, half equations are what I would usually do, rather than using changes in oxidation numbers, but both methods are similar.

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