Year 12s: what will be the first way you research your future options? >>

REDOX half equations help..

hey there, I was doing redox half equations in school last week and I am really lost with them. I would really appreciate some help as of the steps I should take in these questions :frown:

:

1a) write ionic half equations for: A) Cr2+ to Cr3+ B) HI to I2 C) Sn4+ to Sn2+ D) Cl- to Cl2+

2) write ionic half equations for the following, which take place in acidic solution a)Iodate(v) ions, I03- , to iodine b_ Manganese dioxide, MnO2 , to Mn2+ ions c) VO+ ions to V02+ ions

3) Write the overall equations for the following redox reactions:

a) Manganate (vii) ions in acid solution reacting with hydrogen iodide

b) Fe3+ ions being reduced to Fe2+ ions by Sn2+ ions, which are oxidized to Sn4+ ions

c) Dichromate (vi) ions in acid solution oxidizing nitrogen dioxide , NO2 , to nitrate ions, NO3-

d) Iodine oxidising sodium thiosulfate , Na2S2O3 , to sodium tetrathionate Na2S4O6 - the iodine is reduced to sodium iodide , NaI.

Steps for answering any of these questions would be really really appreciated as I am lost with this redox topic :frown:

Thanks

Reply 1

NoManIsAnIsland

Original post by madfish

hey there, I was doing redox half equations in school last week and I am really lost with them. I would really appreciate some help as of the steps I should take in these questions :frown:

Thanks

You need to work out oxidation states for each side of the equation, and then use this to work out how many electrons you have gained or lost. For 1a: Cr2+ to Cr3+ as increased by +1 so it has lost one electron (oxidation) so the equation would be Cr2+ --> Cr3+ + e-. If the half equation is a reduction the electrons will go before the arrow.

Now if you have oxygen involved it will be balanced with water. so 2b: MnO2 to Mn2+. Firstly work out the oxidation state of Mn originally, well O = 2-, there are two O so these equals 4-... the overall charge is zero so Mn must be 4+. It has decreased by -2, so this is reduction. SO the first part of the answer would look like this: MnO2 + 2e- --> Mn2+, however we need to balance out the oxygens... two oxygens = two H2O molecules = 4H+ needed... put all this together: MnO2 + 4H+ + 2e- --> Mn2+ + 2H2O.

For the final part you need to work out the two half equations (using what I have said before) then balance out the electrons e.g. if there are two e- on the left side there must be 2 e- on the right side... so you may need to multiply the half equations to get them to balance, then you simply add the two half equations together and remove the electrons...

Anything that doesn't make sense just ask :smile: