This page contains the AQA AS Chemistry C7 Oxidation ,Reduction and redox reaction Questions and their Kerboodle answers for revision and understanding Oxidation ,Reduction and redox reaction .This page also contains the link to the notes and video for the revision of this topic.
Banner 17.1 Oxidation and reduction AQA AS Chemistry C7 Oxidation,Reduction And Redox Reaction Kerboodle Answers: Page No.124
1.Ans-
- Bromine
- Calcium
- Calcium
- Bromine
- CaàCa2++2e– Br2 + 2e–à2Br –
- Bromine
- Calcium
7.2-Oxidation states AQA AS Chemistry C7 Oxidation,Reduction And Redox Reaction Kerboodle Answers: Page No.127
1. Ans-
- Pb-> +4 Cl-> -1
- C=+4 Cl=-1
- Na=+1 N=+5 O=-2
2.Ans- O=-2 before and after O=-2
3.Ans- Before O=0 and after O=-2
4. Ans- Before Fe=+2 and Fe=+3
5. Ans-
- P=+5
- N=+5
- N=-3
7.3-Redox equations AQA AS Chemistry C7 Oxidation,Reduction And Redox Reaction Kerboodle Answers: Page No.130
1. Ans-
- Before Fe=+2 and after +3 Before Cl=0 and after -1
- Iron has been oxidized as it loses electrons
- Chlorine has been reduced as it gains electrons.
- Fe2+àFe3+ + e– Cl2 + 2e- à2Cl–
2.Ans-
- i)3 Cl2 + 6 NaOH = NaClO3 + 5 NaCl + 3 H2O
ii)Sn + 4 HNO3 = SnO2 + 4 NO2 + 2 H2O
- i) Cl + e- àCl– and Clà Cl5+ + 5e–
ii)SnàSn4++4e- N5++ e- àN4+
Practice questions– Page no.134
1.Ans-
- The reducing agent reduces other substaces in a reaction by providing electrons and oxidising itself.
- 0
- i) +3 ii)-3 iii)-1
- i) PbO2 + 4H+ + 2e− → Pb2+ + 2H2O
ii) 2Cl– -2e- àCl2
iii) PbO2 + 4 HCl → PbCl2 + Cl2 + 2 H2O
2.Ans-
- An oxidising agent is the one that oxidises other substances by accepting the electrons and thereby reducing itself.
- i) Sulfur in SO2=+4 and in SO4-2 = +6
ii)Br2+ 2e- à 2Br –
iii) SO2 + 2H2O -> SO42- + 4H+ + 2e–
iv) Br2 + SO2 + 2H20 -> 2HBr + H2SO4
3.Ans-
- An oxidising agent is a substance that gains electron to oxidise the other substances.
- Oxidation state refers to the number of electrons lost or gained to form that compound.
- i) +4 ii)+5 iii)-3
- i)CuàCu2++2e–
ii) NO3– + 4H+ + 3e– ——> NO + 2H2O
iii)2NO3– + 8H+ + 3Cu ——> 2NO + 4H2O + 3Cu2+
4.Ans-
- NO=+2 NO3– = +5
- +5 +2
NO3– + 4H+ +3e– —–> NO + 2H2O
- 0 +1
Ag —–> Ag+ + e–
- NO3– + 4H+ +3Ag —–> NO + 2H2O + 3Ag+
5.Ans-
- i) I2= 0 HIO3=+5
ii) I2 + 10 HNO3 = 2 HIO3 + 10 NO2 + 4 H2O
- 5NaI + NaIO3 + 3H2SO4 → 3I2 + 3Na2SO4 + 3H2O
Iodine is the oxidising agent.
- i)Iodine
ii) SO42- + 8H+ + 6e- à S(s) + 4H2O
- H2S
- i) Ag+ +I–àAgI
ii)The precipitate will not dissolve.
iii)To remove any other ions like carbonates etc that could interfere with the test.
- i) The reducing agent is a substance that provides electrons to the other substance and reduces them.
ii) Cl2 +2e- à2Cl–
iii)Iodide ions react differently from chloride ions as it has greater reducing power than chlorine and therefore in reactions it further reduces the products formed.
Banner 2