Redox reactions (1 Viewer)

Fortify

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Throw us a question, best off starting with questions.
 

spammy679

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lol the teacher threw us an unexpected test yesterday and everyone failed :)

what about these questions:

a)for each write reduction, oxidation half equations and overall redox reaction

1) Tin(II) ions reduce dichromate (Cr2O7 2-) ions in acidic (H+) solution to form chromium(III) ions and water. The tin(II) ions are oxidised by the dichromate ions to form(IV) ions.​

2) Copper metal reduces an acidified solution of nitrate ions (nitric acid) to nitrogen(IV) oxide and water. The copper metal is oxidised by the nitrate ions to copper(II) ions​

b) determine which species is the reductant and oxidant:

1) MnO2 + 4HCl <--> MnCl2 + 2H2O + Cl2

2) MnO4(-) + 8H(+) + 5Fe(2+) <--> Mn(2+) + 4H2O + 5fe(3+)​

3) SOCl2 + 2H(+) + 4Li <--> S + 2Cl(-) + H2O + 4Li(+)​

4)3SO2 + 2H(+) + Cr2O7 (2-) <--> 3SO4 (2-) + 2Cr(2+) + H2O​
 

dc1337

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a)
1)O: Sn2+ > Sn4+ + 2e-
R: Cr2O72- + 14H+ +6e- > 2Cr3+ + 7H2O
N: 3Sn2+ + Cr2O72- + 14H+ > 3Sn4+ + 2Cr3+ + 7H2O
2) O: Cu > Cu2+ + 2e-
R: 2H+ + NO3 + e- > NO2 + H2O
N: 4H+ + 2NO3- > NO2 + H2O + Cu2+

b)
1) Reductant: Mn Oxidant: Cl
2) Red: Mn Oxi: Fe
3) Red: S Oxi: Li
4) Red: Cr Oxi: S

all are ions so just add the charge
 
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kbsb

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Heres an easy way to remember it...(but u have to get the initial equation rite)
For example: Magnesium and hydrochloric acid:

  • Neutral Species Equation-(you must get this rite to procceed)<O:p</O:p
Mg <SUB>(s)</SUB> + 2HCl <SUB>(aq)</SUB> -> MgCl<SUB>2(aq)</SUB> + H<SUB>2(g)</SUB><O:p</O:p<O:p</O:p
  • Complete Ionic Equation (you are simply breaking the ionic compounds into ions)
Mg <SUB>(s)</SUB> + 2H<SUP>+</SUP><SUB>(aq)</SUB> <SUP></SUP>+ 2Cl<SUP>-</SUP> <SUB>(aq)</SUB> -> Mg<SUP>2+</SUP><SUB>(aq)</SUB> <SUP></SUP>+ 2Cl<SUP>-</SUP><SUB> (aq)</SUB> + H<SUB>2(g)</SUB><O:p</O:p<O:p</O:p
  • Net Ionic Reaction (you can eliminate the 2Cl<SUP>-</SUP> <SUB>(aq)</SUB>because as it is on both sides it is called the spectator ion)
Mg <SUB>(s)</SUB> + 2H<SUP>+</SUP><SUB>(aq)</SUB> -> Mg<SUP>2+</SUP><SUB>(aq)</SUB> + H<SUB>2(g)</SUB><O:p</O:p

<O:p</O:p
  • Redox Half Reactions (You re simply breaking the above up into 2 half equations)
Mg <SUB>(s)</SUB> -> Mg<SUP>2+</SUP><SUB>(aq)</SUB> + 2e<SUP>-</SUP><SUB> …………………………</SUB>Oxidation Half Reaction<O:p></O:p>
2H<SUP>+</SUP><SUB>(aq)</SUB> + 2e<SUP>-</SUP><SUB> </SUB>-> + H<SUB>2(g) …………………………</SUB>Reduction Half Reaction<O:p></O:p>


Hope it helped :)
 

Beth2010

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Mg <sub>(s)</sub> -> Mg<sup>2+</sup><sub>(aq)</sub> + 2e<sup>-</sup><sub> …………………………</sub>Oxidation Half Reaction<o>:p></o>:p>
2H<sup>+</sup><sub>(aq)</sub> + 2e<sup>-</sup><sub> </sub>-> + H<sub>2(g) …………………………</sub>Reduction Half Reaction<o>:p></o>:p>
one way to remember these if its a reduction or oxidisation is to think of the length of each side of the half equation. if it gets shorter (aka reduces) from whats on the left to whats on the right then its a reduction reaction so if it gets longer its an oxidisation.
just a handy little trick that might come in useful :)
 

Pwnage101

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There is a RED CAT and AN OX on an OIL RIG


(REDuction occurs at the CAThode)

(OXidation occurs at the ANode)

(Oxidation Is the Loss and Reduction Is the Gaining of electrons)
 

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