Redox reactions involve simultaneous oxidation and reduction through electron transfer or oxidation number change. Important Questions Class 11 Chemistry Chapter 7 use this idea to test oxidants, reductants, equations, and electrode processes.
Redox is one of the most scoring concepts in Class 11 Chemistry because every answer follows a rule. Important Questions Class 11 Chemistry Chapter 7 help students practise oxidation number, electron transfer, reaction classification, redox titration, and balancing equations. CBSE 2026 questions from this chapter often test definitions, oxidation states, oxidising agents, reducing agents, acidic medium balancing, basic medium balancing, and electrode processes.
Key Takeaways
- Oxidation Number: It helps identify oxidation and reduction in covalent and ionic reactions.
- Electron Transfer: Oxidation means electron loss, while reduction means electron gain.
- Reaction Types: Redox reactions include combination, decomposition, displacement, and disproportionation reactions.
- Balancing Methods: Oxidation number method and half-reaction method balance redox equations.
Important Questions Class 11 Chemistry Chapter 7 Structure 2026
| Principle |
Application |
Unit |
| Oxidation and Reduction |
Identify oxidised and reduced species |
Redox Reactions |
| Oxidation Number |
Find oxidant and reductant |
Redox Reactions |
| Half-Reaction Method |
Balance ionic redox equations |
Redox Reactions |
Class 11 Chemistry Chapter 7 Important Questions Overview
Class 11 Chemistry Chapter 7 Important Questions focus on the NCERT 2026 idea that oxidation and reduction occur together. Students should practise definitions, electron transfer, oxidation number, redox balancing, and electrode processes.
Q1. What is a redox reaction?
A redox reaction is a reaction in which oxidation and reduction occur simultaneously.
Example:
Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s)
- Zinc loses electrons and forms Zn^2+.
- Cu^2+ gains electrons and forms Cu.
- Therefore, zinc gets oxidised and Cu^2+ gets reduced.
Final Result: Zn is oxidised,
and Cu^2+ is reduced.
Q2. Why are redox reactions important in Chemistry?
Redox reactions explain energy production, corrosion, batteries, metallurgy, and biological processes.
Examples include fuel burning, caustic soda manufacture, metal extraction, dry cell operation, and corrosion. These reactions also explain hydrogen economy and ozone-related environmental chemistry.
Q3. What does Important Questions Class 11 Chemistry Chapter 7 mainly test?
Important Questions Class 11 Chemistry Chapter 7 mainly test oxidation number, oxidising agents, reducing agents, and equation balancing.
Students must identify oxidation number changes before writing the final answer. Most redox answers depend on correct signs and charge balance.
Redox Reactions Class 11 Important Questions on Oxidation and Reduction
Redox Reactions Class 11 Important Questions often begin with classical definitions. Students must connect oxygen transfer, hydrogen transfer, electron transfer, and oxidation number change.
Q4. What is oxidation according to the classical concept?
Oxidation means addition of oxygen or an electronegative element.
It can also mean removal of hydrogen or an electropositive element.
Examples:
2Mg(s) + O2(g) → 2MgO(s)
S(s) + O2(g) → SO2(g)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Final Result: Magnesium, sulphur, and methane get oxidised.
Q5. What is reduction according to the classical concept?
Reduction means removal of oxygen or an electronegative element.
It can also mean addition of hydrogen or an electropositive element.
Examples:
2HgO(s) → 2Hg(l) + O2(g)
CH2=CH2(g) + H2(g) → CH3-CH3(g)
2FeCl3(aq) + H2(g) → 2FeCl2(aq) + 2HCl(aq)
Final Result: HgO, ethene, and FeCl3 undergo reduction.
Q6. What is oxidation and reduction Class 11 in electron transfer terms?
Oxidation means loss of electrons, while reduction means gain of electrons.
Example:
Na(s) → Na^+(g) + e^-
Cl2(g) + 2e^- → 2Cl^-(g)
- Sodium loses one electron.
- Chlorine gains electrons.
- Sodium gets oxidised, and chlorine gets reduced.
Final Result: Oxidation is electron loss, and reduction is electron gain.
Q7. Identify oxidation and reduction in H2S + Cl2 → 2HCl + S.
H2S gets oxidised, while Cl2 gets reduced.
Given reaction:
H2S(g) + Cl2(g) → 2HCl(g) + S(s)
- H2S loses hydrogen and forms sulphur.
- Cl2 gains hydrogen and forms HCl.
- Removal of hydrogen from H2S shows oxidation.
Final Result: H2S is oxidised, and Cl2 is reduced.
Q8. Identify oxidation and reduction in 3Fe3O4 + 8Al → 9Fe + 4Al2O3.
Aluminium gets oxidised, while Fe3O4 gets reduced.
Given reaction:
3Fe3O4(s) + 8Al(s) → 9Fe(s) + 4Al2O3(s)
- Aluminium gains oxygen and forms Al2O3.
- Fe3O4 loses oxygen and forms Fe.
- Oxygen transfer confirms a redox reaction.
Final Result: Al is oxidised, and Fe3O4 is reduced.
Q9. Identify oxidation and reduction in 2Na + H2 → 2NaH.
Sodium gets oxidised, while hydrogen gets reduced.
Given reaction:
2Na(s) + H2(g) → 2NaH(s)
NaH can be written as Na^+H^-.
- Oxidation half-reaction: 2Na(s) → 2Na^+(g) + 2e^-
- Reduction half-reaction: H2(g) + 2e^- → 2H^-(g)
- Sodium donates electrons to hydrogen.
Final Result: Na is oxidised, and H2 is reduced.
Class 11 Chemistry Redox Reactions and Electron Transfer
Class 11 Chemistry Redox Reactions become easier when students write half-reactions. Each half-reaction shows either electron loss or electron gain.
Q10. What is an oxidising agent?
An oxidising agent accepts electrons and gets reduced.
Example:
Cl2(g) + 2e^- → 2Cl^-(g)
- Chlorine accepts electrons.
- Its oxidation number decreases from 0 to -1.
- Therefore, chlorine acts as an oxidising agent.
Final Result: An oxidising agent gains electrons.
Q11. What is a reducing agent?
A reducing agent donates electrons and gets oxidised.
Example:
Zn(s) → Zn^2+(aq) + 2e^-
- Zinc loses electrons.
- Its oxidation number increases from 0 to +2.
- Therefore, zinc acts as a reducing agent.
Final Result: A reducing agent loses electrons.
Q12. What is the difference between oxidising agent and reducing agent?
An oxidising agent accepts electrons, while a reducing agent donates electrons.
Example:
Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s)
- Zn donates electrons and acts as reducing agent.
- Cu^2+ accepts electrons and acts as oxidising agent.
- Both processes occur in the same reaction.
Final Result: Zn is the reducing agent, and Cu^2+ is the oxidising agent.
Q13. Explain the redox reaction between zinc and copper ion.
Zinc gets oxidised to Zn^2+, while Cu^2+ gets reduced to copper metal.
Reaction:
Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s)
- Oxidation: Zn(s) → Zn^2+(aq) + 2e^-
- Reduction: Cu^2+(aq) + 2e^- → Cu(s)
- Zinc has a higher electron-releasing tendency than copper.
Final Result: Zinc displaces copper from copper ion solution.
Q14. Why does zinc displace copper from copper sulphate solution?
Zinc displaces copper because zinc loses electrons more easily than copper.
Reaction:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Ionic form:
Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s)
- Zinc reduces Cu^2+ to Cu.
- Zinc itself changes into Zn^2+.
- The blue colour of Cu^2+ solution fades.
Final Result: Zinc has stronger reducing power than copper.
Q15. Arrange Zn, Cu, and Ag in decreasing electron-releasing tendency.
The decreasing electron-releasing tendency is Zn > Cu > Ag.
Reason:
- Zinc gives electrons to Cu^2+.
- Copper gives electrons to Ag^+.
- Silver has the least tendency to lose electrons.
Final Result: Zn > Cu > Ag.
Oxidation Number Class 11 Chemistry Important Questions
Oxidation number Class 11 Chemistry questions are rule-based. Students should apply every rule before deciding oxidant, reductant, oxidation, and reduction.
Q16. What is oxidation number?
Oxidation number is the oxidation state assigned to an element in a compound or ion.
It assumes complete electron transfer to the more electronegative atom. It helps identify oxidation and reduction in ionic and covalent reactions.
Q17. What are the main rules for assigning oxidation number?
Oxidation number rules help calculate the oxidation state of elements.
Key rules:
- Free elements have oxidation number 0.
- Monoatomic ions have oxidation number equal to their charge.
- Oxygen is usually -2.
- Oxygen is -1 in peroxides.
- Hydrogen is usually +1.
- Hydrogen is -1 in metal hydrides.
- Fluorine is always -1.
- The sum of oxidation numbers in a neutral compound is 0.
- The sum of oxidation numbers in an ion equals ionic charge.
Final Result: These rules decide oxidation number in compounds and ions.
Q18. Find the oxidation number of sulphur in H2SO5.
The oxidation number of sulphur in H2SO5 is +6.
Known values:
- H = +1
- Normal O = -2
- Peroxide O = -1
- H2SO5 contains one peroxide linkage.
Equation:
2(+1) + x + 3(-2) + 2(-1) = 0
Calculation:
2 + x - 6 - 2 = 0
x - 6 = 0
x = +6
Final Result: Sulphur oxidation number = +6.
Q19. Find the oxidation number of chromium in Cr2O7^2-.
The oxidation number of chromium in Cr2O7^2- is +6.
Known value:
O = -2
Let chromium be x.
Equation:
2x + 7(-2) = -2
Calculation:
2x - 14 = -2
2x = +12
x = +6
Final Result: Chromium oxidation number = +6.
Q20. Find the oxidation number of nitrogen in NO3^-.
The oxidation number of nitrogen in NO3^- is +5.
Known value:
O = -2
Let nitrogen be x.
Equation:
x + 3(-2) = -1
Calculation:
x - 6 = -1
x = +5
Final Result: Nitrogen oxidation number = +5.
Q21. Calculate the oxidation number of Mn in KMnO4.
The oxidation number of Mn in KMnO4 is +7.
Known values:
- K = +1
- O = -2
Let Mn be x.
Equation:
+1 + x + 4(-2) = 0
Calculation:
1 + x - 8 = 0
x = +7
Final Result: Mn oxidation number = +7.
Q22. Calculate the oxidation number of Mn in K2MnO4.
The oxidation number of Mn in K2MnO4 is +6.
Known values:
- K = +1
- O = -2
Let Mn be x.
Equation:
2(+1) + x + 4(-2) = 0
Calculation:
2 + x - 8 = 0
x = +6
Final Result: Mn oxidation number = +6.
Q23. Calculate the oxidation number of boron in NaBH4.
The oxidation number of boron in NaBH4 is +3.
Known values:
- Na = +1
- H = -1 in metal hydrides
Let B be x.
Equation:
+1 + x + 4(-1) = 0
Calculation:
1 + x - 4 = 0
x = +3
Final Result: Boron oxidation number = +3.
Q24. What is Stock notation?
Stock notation shows oxidation number with Roman numerals in brackets after the metal symbol.
Examples:
Fe(II)O means iron has +2 oxidation state.
Fe2(III)O3 means iron has +3 oxidation state.
Final Result: Stock notation names variable oxidation states of metals.
Q25. Why can oxidation number be fractional?
Fractional oxidation number is only an average value.
Example:
Fe3O4 gives average iron oxidation number +8/3.
Fe3O4 is better written as FeO.Fe2O3.
- FeO contains Fe^2+.
- Fe2O3 contains Fe^3+.
- The average value becomes fractional.
Final Result: Actual oxidation states remain whole-number values.
Redox Reaction Examples Class 11 With Oxidising Agent and Reducing Agent
Redox reaction examples Class 11 often ask students to identify the oxidising agent and reducing agent. The oxidant gets reduced, and the reductant gets oxidised.
Q26. Justify that 2Cu2O + Cu2S → 6Cu + SO2 is a redox reaction.
This is a redox reaction because copper gets reduced and sulphur gets oxidised.
Reaction:
2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g)
Oxidation states:
- Cu in Cu2O = +1
- Cu in Cu2S = +1
- Cu in Cu = 0
- S in Cu2S = -2
- S in SO2 = +4
Changes:
- Cu changes from +1 to 0.
- S changes from -2 to +4.
- Reduction and oxidation occur together.
Final Result: Cu2O acts as oxidant, and Cu2S acts as reductant.
Q27. Identify oxidising agent and reducing agent in CuO + H2 → Cu + H2O.
CuO is the oxidising agent, and H2 is the reducing agent.
Reaction:
CuO(s) + H2(g) → Cu(s) + H2O(g)
- Cu changes from +2 in CuO to 0 in Cu.
- Hydrogen changes from 0 in H2 to +1 in H2O.
- Cu gets reduced, and hydrogen gets oxidised.
Final Result: Oxidising agent = CuO, reducing agent = H2.
Q28. Identify oxidising agent and reducing agent in Fe2O3 + 3CO → 2Fe + 3CO2.
Fe2O3 is the oxidising agent, and CO is the reducing agent.
Reaction:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
- Fe changes from +3 to 0.
- Carbon changes from +2 in CO to +4 in CO2.
- Fe gets reduced, and carbon gets oxidised.
Final Result: Oxidising agent = Fe2O3, reducing agent = CO.
Q29. Why is AgF2 a strong oxidising agent?
AgF2 is a strong oxidising agent because Ag(II) tends to reduce to stable Ag(I).
Fluorine helps stabilise the higher oxidation state. Still, Ag(II) accepts electrons strongly during redox reactions.
Final Result: AgF2 acts as a strong electron acceptor.
Types of Redox Reactions Class 11 Important Questions
Types of redox reactions Class 11 questions test classification. Students should check whether the reaction shows combination, decomposition, displacement, or disproportionation.
Q30. What is a combination redox reaction?
A combination redox reaction forms one product from two or more reactants with oxidation number change.
General form:
A + B → C
Examples:
C(s) + O2(g) → CO2(g)
3Mg(s) + N2(g) → Mg3N2(s)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Final Result: Reactants combine and show redox change.
Q31. What is a decomposition redox reaction?
A decomposition redox reaction breaks one compound into simpler substances with oxidation number change.
Examples:
2H2O(l) → 2H2(g) + O2(g)
2NaH(s) → 2Na(s) + H2(g)
2KClO3(s) → 2KCl(s) + 3O2(g)
Final Result: One compound decomposes and shows redox change.
Q32. Why is CaCO3 → CaO + CO2 not a redox reaction?
CaCO3 → CaO + CO2 is not redox because no oxidation number changes.
Reaction:
CaCO3(s) → CaO(s) + CO2(g)
Oxidation states:
- Ca = +2 on both sides.
- C = +4 on both sides.
- O = -2 on both sides.
Final Result: No oxidation or reduction occurs.
Q33. What is a displacement redox reaction?
A displacement redox reaction occurs when one element replaces another element from its compound.
General form:
X + YZ → XZ + Y
Example:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Final Result: Zinc displaces copper due to stronger reducing power.
Q34. What is a disproportionation reaction Class 11?
A disproportionation reaction Class 11 answer states that the same element gets oxidised and reduced.
Example:
2H2O2(aq) → 2H2O(l) + O2(g)
- Oxygen in H2O2 has oxidation number -1.
- Oxygen in H2O has oxidation number -2.
- Oxygen in O2 has oxidation number 0.
Final Result: Oxygen gets reduced and oxidised in the same reaction.
Q35. Why does ClO4^- not show disproportionation?
ClO4^- does not show disproportionation because chlorine already has its highest oxidation state.
In ClO4^-, chlorine has +7 oxidation state. It cannot increase its oxidation number beyond +7.
Final Result: ClO4^- cannot undergo oxidation further.
Q36. Classify N2 + O2 → 2NO.
This is a combination redox reaction.
Reaction:
N2(g) + O2(g) → 2NO(g)
- Nitrogen changes from 0 to +2.
- Oxygen changes from 0 to -2.
- Both elements combine to form nitric oxide.
Final Result: Combination redox reaction.
Q37. Classify 2Pb(NO3)2 → 2PbO + 4NO2 + O2.
This is a decomposition redox reaction.
Reaction:
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g)
One compound breaks into lead oxide, nitrogen dioxide, and oxygen. Oxidation number changes occur during decomposition.
Final Result: Decomposition redox reaction.
Q38. Classify 2NO2 + 2OH^- → NO2^- + NO3^- + H2O.
This is a disproportionation redox reaction.
Reaction:
2NO2(g) + 2OH^-(aq) → NO2^-(aq) + NO3^-(aq) + H2O(l)
- Nitrogen in NO2 is +4.
- Nitrogen in NO2^- is +3.
- Nitrogen in NO3^- is +5.
Final Result: Nitrogen gets reduced and oxidised.
Balancing Redox Reactions Class 11 by Oxidation Number Method and Half Reaction Method
Balancing redox reactions Class 11 needs atom balance, charge balance, and electron balance. Use oxidation number method Class 11 or half reaction method Class 11 as the question demands.
Q39. What are the two methods used to balance redox reactions?
The two methods are oxidation number method and half-reaction method.
- Oxidation number method tracks increase and decrease in oxidation number.
- Half-reaction method separates oxidation and reduction steps.
- Both methods produce the same balanced equation.
Final Result: Use either method based on the question.
Q40. Balance Cr2O7^2- + SO3^2- → Cr^3+ + SO4^2- in acidic medium.
The balanced equation is Cr2O7^2- + 3SO3^2- + 8H^+ → 2Cr^3+ + 3SO4^2- + 4H2O.
- Skeleton equation:
Cr2O7^2- + SO3^2- → Cr^3+ + SO4^2-
- Oxidation number change:
Cr changes from +6 to +3.
S changes from +4 to +6.
- Balance Cr and S:
Cr2O7^2- + 3SO3^2- → 2Cr^3+ + 3SO4^2-
- Balance charge with H^+:
Cr2O7^2- + 3SO3^2- + 8H^+ → 2Cr^3+ + 3SO4^2-
- Balance H and O with water:
Cr2O7^2- + 3SO3^2- + 8H^+ → 2Cr^3+ + 3SO4^2- + 4H2O
Final Result: Cr2O7^2- + 3SO3^2- + 8H^+ → 2Cr^3+ + 3SO4^2- + 4H2O.
Q41. Balance MnO4^- + Br^- → MnO2 + BrO3^- in basic medium.
The balanced equation is 2MnO4^- + Br^- + H2O → 2MnO2 + BrO3^- + 2OH^-.
- Skeleton equation:
MnO4^- + Br^- → MnO2 + BrO3^-
- Oxidation number change:
Mn changes from +7 to +4.
Br changes from -1 to +5.
- Balance electron change:
2MnO4^- + Br^- → 2MnO2 + BrO3^-
- Balance charge with OH^-:
2MnO4^- + Br^- → 2MnO2 + BrO3^- + 2OH^-
- Balance hydrogen with water:
2MnO4^- + Br^- + H2O → 2MnO2 + BrO3^- + 2OH^-
Final Result: 2MnO4^- + Br^- + H2O → 2MnO2 + BrO3^- + 2OH^-.
Q42. Balance MnO4^- + I^- → MnO2 + I2 in basic medium.
The balanced equation is 2MnO4^- + 6I^- + 4H2O → 2MnO2 + 3I2 + 8OH^-.
- Oxidation half-reaction:
2I^- → I2 + 2e^-
- Reduction half-reaction:
MnO4^- + 2H2O + 3e^- → MnO2 + 4OH^-
- Equalise electrons:
6I^- → 3I2 + 6e^-
2MnO4^- + 4H2O + 6e^- → 2MnO2 + 8OH^-
- Add half-reactions:
2MnO4^- + 6I^- + 4H2O → 2MnO2 + 3I2 + 8OH^-
Final Result: 2MnO4^- + 6I^- + 4H2O → 2MnO2 + 3I2 + 8OH^-.
Q43. Balance Fe^2+ + Cr2O7^2- → Fe^3+ + Cr^3+ in acidic medium.
The balanced equation is 6Fe^2+ + Cr2O7^2- + 14H^+ → 6Fe^3+ + 2Cr^3+ + 7H2O.
- Oxidation half-reaction:
Fe^2+ → Fe^3+ + e^-
- Reduction half-reaction:
Cr2O7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H2O
- Equalise electrons:
6Fe^2+ → 6Fe^3+ + 6e^-
- Add half-reactions:
6Fe^2+ + Cr2O7^2- + 14H^+ → 6Fe^3+ + 2Cr^3+ + 7H2O
Final Result: 6Fe^2+ + Cr2O7^2- + 14H^+ → 6Fe^3+ + 2Cr^3+ + 7H2O.
Q44. Balance H2O2 + Fe^2+ → Fe^3+ + H2O in acidic medium.
The balanced equation is H2O2 + 2Fe^2+ + 2H^+ → 2Fe^3+ + 2H2O.
- Oxidation half-reaction:
Fe^2+ → Fe^3+ + e^-
- Reduction half-reaction:
H2O2 + 2H^+ + 2e^- → 2H2O
- Equalise electrons:
2Fe^2+ → 2Fe^3+ + 2e^-
- Add half-reactions:
H2O2 + 2Fe^2+ + 2H^+ → 2Fe^3+ + 2H2O
Final Result: H2O2 + 2Fe^2+ + 2H^+ → 2Fe^3+ + 2H2O.
Redox Titration Class 11 Important Questions
Redox titration Class 11 questions test oxidant, reductant, indicator, and end point. Students should remember that KMnO4 acts as a self-indicator.
Q45. What is redox titration?
Redox titration is a titration based on oxidation-reduction reaction between an oxidant and a reductant.
It helps determine the strength of one solution using another solution. The end point appears through a clear colour change.
Q46. Why is KMnO4 called a self-indicator?
KMnO4 is called a self-indicator because MnO4^- has an intense purple colour.
In acidic medium, MnO4^- gets reduced to nearly colourless Mn^2+. A faint permanent pink colour marks the end point.
Final Result: KMnO4 needs no external indicator.
Q47. Why is diphenylamine used in dichromate titration?
Diphenylamine is used because dichromate does not act as a self-indicator.
Dichromate oxidises diphenylamine near the end point. The intense blue colour signals completion.
Final Result: Diphenylamine acts as an external redox indicator.
Q48. What is iodometric titration?
Iodometric titration uses iodine liberation and reaction with thiosulphate ions.
Key reactions:
2Cu^2+(aq) + 4I^-(aq) → Cu2I2(s) + I2(aq)
I2(aq) + 2S2O3^2-(aq) → 2I^-(aq) + S4O6^2-(aq)
Starch gives blue colour with iodine. The blue colour disappears at the end point.
Electrode Potential Class 11 and Redox Electrode Processes
Electrode potential Class 11 questions connect redox reactions with electron flow. This concept builds the base for electrochemistry in Class 12.
Q49. What is a redox couple?
A redox couple contains the oxidised and reduced forms of the same species.
Examples:
Zn^2+/Zn
Cu^2+/Cu
Fe^3+/Fe^2+
The oxidised form appears before the reduced form.
Final Result: A redox couple shows one oxidation-reduction pair.
Q50. What is Daniell cell?
Daniell cell is an electrochemical cell based on zinc oxidation and copper ion reduction.
Cell reaction:
Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s)
At zinc electrode:
Zn(s) → Zn^2+(aq) + 2e^-
At copper electrode:
Cu^2+(aq) + 2e^- → Cu(s)
Final Result: Zinc acts as anode, and copper acts as cathode.
Q51. What is electrode potential?
Electrode potential is the potential difference between an electrode and its ion solution.
It shows the tendency of a redox couple to gain or lose electrons. A stronger reducing species loses electrons more easily.
Q52. What is standard electrode potential?
Standard electrode potential is electrode potential measured under standard conditions.
Standard conditions:
- Ion concentration = 1 M
- Gas pressure = 1 atm
- Temperature = 298 K
The standard hydrogen electrode has E° = 0.00 V.
Final Result: Standard electrode potential compares redox strength.
Q53. What does negative standard electrode potential mean?
Negative E° means the redox couple acts as a stronger reducing agent than H^+/H2.
Example:
Zn^2+/Zn has E° = -0.76 V.
Zinc loses electrons more readily than hydrogen.
Final Result: Negative E° indicates stronger reducing tendency.
Q54. What does positive standard electrode potential mean?
Positive E° means the redox couple acts as a weaker reducing agent than H^+/H2.
Example:
Cu^2+/Cu has E° = +0.34 V.
Copper ions accept electrons more readily than zinc ions.
Final Result: Positive E° indicates stronger reduction tendency.
NCERT Class 11 Chemistry Chapter 7 Board Exam Pattern Questions
NCERT Class 11 Chemistry Chapter 7 questions usually test rules through reactions. Students should revise every oxidation number and charge balance step.
Q55. Which areas are most repeated from Redox Reactions in CBSE 2026?
Oxidation number, oxidant-reductant identification, and redox equation balancing are the most repeated areas.
Common question formats:
- Assign oxidation number.
- Identify oxidised and reduced species.
- Identify oxidising agent and reducing agent.
- Classify redox reactions.
- Balance equations in acidic medium.
- Balance equations in basic medium.
- Explain disproportionation.
- Compare oxidising power of halogens.
- Explain electrode processes.
Final Result: Balancing and oxidation number carry the highest practice value.
Q56. Where do students lose marks in Class 11 Chemistry Chapter 7 questions and answers?
Students lose marks by using wrong oxidation number signs and skipping charge balance.
Common errors:
- Treating peroxide oxygen as -2.
- Forgetting hydrogen is -1 in metal hydrides.
- Calling fractional oxidation number an actual state.
- Balancing atoms but not charge.
- Adding H^+ in basic medium without conversion.
- Confusing oxidising agent with oxidised species.
Final Result: Most redox errors come from signs, charges, and medium.
Class 11 Chemistry Important Links