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# Redox Equations watch

1. Write the equation for the redox reaction.

Iodine oxidising sodium thioslphate (Na2S2O3) to sodium tetrathionate (Na2S4O6), the iodine is reduced to sodium iodide, NaI

Well, the answer the book gives me is I2 + 2Na2S2O2 → 2NaI + Na2S4O6.

but I saw that the sodium was a spectator ion in these compounds and wrote the ionic equations, and it was

I2 + 2S2O32- → 2I- + S4O62-

Am i technically correct? or wrong?
thanks =)
2. (Original post by Hi, How are you ?)
Write the equation for the redox reaction.

Iodine oxidising sodium thioslphate (Na2S2O3) to sodium tetrathionate (Na2S4O6), the iodine is reduced to sodium iodide, NaI

Well, the answer the book gives me is I2 + 2Na2S2O2 → 2NaI + Na2S4O6.

but I saw that the sodium was a spectator ion in these compounds and wrote the ionic equations, and it was

I2 + 2S2O32- → 2I- + S4O62-

Am i technically correct? or wrong?
thanks =)
You are absolutely correct ..
3. (Original post by charco)
You are absolutely correct ..
Thanks, one more question if you don't mind,

in the compound, N2H4

the book says that the oxidation number of nitrogen is -2, but I thought that when calculating oxidation numbers, you look at the more electroneagtive element, see what group it is in, and since nitrogen is in group 5, each nitrogen atom has an oxidation number of -3, so in that compound it should be -6, and each hydrogen atom should be +1.5. can you please explain why this is not so. Thanks.
4. (Original post by Hi, How are you ?)
Thanks, one more question if you don't mind,

in the compound, N2H4

the book says that the oxidation number of nitrogen is -2, but I thought that when calculating oxidation numbers, you look at the more electroneagtive element, see what group it is in, and since nitrogen is in group 5, each nitrogen atom has an oxidation number of -3, so in that compound it should be -6, and each hydrogen atom should be +1.5. can you please explain why this is not so. Thanks.
Nitrogen has a variable oxidation state, but hydrogen can only be 1, 0, or -1.

In this case as hydrogen is more electropositive than nitrogen it takes the positive oxidation state. Four hydrogens = +4, this must be cancelled by the two nitrogen atoms, hence each nitrogen is -2.

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