Redox Reactions — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Redox Reactions" — 8 important questions with detailed answers for CBSE board exam preparation.
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TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Redox Reactions" — 8 important questions with detailed answers for CBSE board exam p…
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Key Questions Covered:
- Define oxidation and reduction. Are they always simultaneous?
- Determine oxidation numbers of all atoms in K2Cr2O7 and KMnO4
- Balance the redox equation: MnO4- + Fe2+ -> Mn2+ + Fe3+ in acidic medium
- Balance redox equation: Cr2O7^2- + I- -> Cr^3+ + I2 in acidic medium
- What is electrochemistry? Explain galvanic and electrolytic cells.
- Calculate number of electrons transferred when 0.5 mol of MnO4- is completely…
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Define oxidation and reduction. Are they always simultane… | ✓ Solved |
| Determine oxidation numbers of all atoms in K2Cr2O7 and K… | ✓ Solved |
| Balance the redox equation: MnO4- + Fe2+ -> Mn2+ + Fe3+ i… | ✓ Solved |
| Balance redox equation: Cr2O7^2- + I- -> Cr^3+ + I2 in ac… | ✓ Solved |
| What is electrochemistry? Explain galvanic and electrolyt… | ✓ Solved |
| Calculate number of electrons transferred when 0.5 mol of… | ✓ Solved |
Showing 6 of 8 questions
Q1: Define oxidation and reduction. Are they always simultaneous?
Oxidation: Loss of electrons or increase in oxidation number.
Reduction: Gain of electrons or decrease in oxidation number.
Redox reaction: A reaction in which both oxidation and reduction occur simultaneously.
They are ALWAYS simultaneous because:
- Electrons lost by one species must be gained by another
- Total electrons lost = Total electrons gained (law of conservation of electrons)
- Oxidizing agent: Species that gets reduced (gains electrons)
- Reducing agent: Species that gets oxidized ...
Q2: Determine oxidation numbers of all atoms in K2Cr2O7 and KMnO4
Oxidation number rules:
- Alkali metals (Group 1) = +1
- Alkaline earth metals (Group 2) = +2
- Oxygen in compounds = -2 (except peroxides)
- Sum = charge on compound (for neutral: sum = 0)
In K2Cr2O7:
K: +1 (Group 1)
Cr: ?
O: -2
2(+1) + 2(Cr) + 7(-2) = 0
2 + 2Cr - 14 = 0
2Cr = 12
Cr = +6
In KMnO4:
K: +1
Mn: ?
O: -2
(+1) + Mn + 4(-2) = 0
1 + Mn - 8 = 0
Mn = +7
Summary:
K2Cr2O7: K = +1, Cr = +6, O = -2
KMnO4: K = +1, Mn = +7, O = -2
Q3: Balance the redox equation: MnO4- + Fe2+ -> Mn2+ + Fe3+ in acidic medium
Step 1: Identify oxidation states
Mn in MnO4-: +7 (reduction, gains 5e-)
Fe2+: +2 (oxidation, loses 1e-)
Step 2: Write half-reactions
Reduction: MnO4- -> Mn2+
Oxidation: Fe2+ -> Fe3+
Step 3: Balance half-reactions (acidic medium)
Reduction: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Oxidation: Fe2+ -> Fe3+ + 1e-
Step 4: Equalize electrons
Multiply reduction by 1, oxidation by 5
Reduction: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
Oxidation: 5Fe2+ -> 5Fe3+ + 5e-
Step 5: Add and cancel electron...
Q4: Balance redox equation: Cr2O7^2- + I- -> Cr^3+ + I2 in acidic medium
Step 1: Identify oxidation states
Cr in Cr2O7^2-: +6 (reduction, gains 3e- per Cr, total 6e-)
I-: -1 (oxidation, loses 1e-)
Step 2: Write half-reactions
Reduction: Cr2O7^2- -> Cr^3+
Oxidation: I- -> I2
Step 3: Balance half-reactions (acidic medium)
Reduction: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O
Oxidation: 2I- -> I2 + 2e-
Step 4: Equalize electrons (LCM of 6 and 2 = 6)
Multiply reduction by 1, oxidation by 3
Reduction: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O
Oxidation: 6I- -&g...
Q5: What is electrochemistry? Explain galvanic and electrolytic cells.
Electrochemistry: Study of chemical reactions involving electron transfer and their relationship with electrical energy.
Galvanic (Voltaic) Cell:
- Spontaneous redox reaction generates electrical energy
- Oxidation at anode (negative electrode), reduction at cathode (positive electrode)
- Electrons flow from anode to cathode externally through wire
- Salt bridge allows ion flow to complete circuit
- Example: Zn-Cu cell (Daniel cell)
Anode: Zn -> Zn2+ + 2e- (oxidation)
Cathode: Cu2+ + 2e-...
Q6: Calculate number of electrons transferred when 0.5 mol of MnO4- is completely reduced to Mn2+
From the half-reaction for permanganate reduction:
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
This shows: 1 mol MnO4- gains 5 mol electrons
For 0.5 mol MnO4-:
Electrons transferred = 0.5 mol × 5 e-/mol = 2.5 mol electrons
In terms of Faraday's constant (F = 96500 C/mol e-):
Charge = 2.5 mol e- × 96500 C/mol = 241,250 C ≈ 241.25 kC
Or in terms of number of electrons:
2.5 mol × 6.022 × 10^23 = 1.5 × 10^24 electrons
Showing 6 of 8 questions. Visit the full page for complete solutions.
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