This MCQ module is based on: Balancing Redox Electrochemistry
TOPIC 3 OF 13
Balancing Redox Electrochemistry
🎓 Class 11
Chemistry
CBSE
Theory
Ch 7 – Redox Reactions
⏱ ~14 min
🌐 Language: [gtranslate]
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Balancing Redox Equations and Electrochemical Cells
7.3 Balancing of Redox Reactions
A balanced redox equation must satisfy two conservation laws: (i) atoms of every element, and (ii) total electrical charge. Two complementary recipes do the job: the oxidation-number method and the half-reaction (ion-electron) method.
Method 1 — Oxidation Number Method L3 Apply
1Write the unbalanced skeleton equation.
2Assign oxidation numbers to every atom.
3Identify the elements whose oxidation numbers change.
4Compute total increase and total decrease in ON.
5Multiply the oxidised and reduced species by small integers so that total e⁻ lost = total e⁻ gained.
6Balance all other atoms except H and O.
7Balance O by adding H2O; then balance H by H⁺ (acidic medium) or by H2O + OH⁻ (basic medium).
8Check that atoms AND total charge balance on both sides.
Method 2 — Half-Reaction (Ion-Electron) Method
1Write the two half-reactions (oxidation & reduction), each with only the species that change.
2Balance atoms other than H and O.
3Balance O by H2O.
4Balance H by adding H⁺ (acidic) or by adding the same number of H2O to the opposite side with OH⁻ (basic).
5Balance the charge on each half by adding electrons.
6Multiply each half so that electrons cancel when added.
7Add the halves and cancel identical species on both sides.
Worked Examples — Balancing in Acidic Medium
Example 7.17 — Balance MnO4⁻ + Fe2+ → Mn2+ + Fe3+ (acidic)
Halves:
Oxidation: Fe²⁺ → Fe³⁺ + e⁻
Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Equalise electrons: multiply oxidation by 5.
5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Add and cancel electrons:
MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O
Charge check: LHS = (−1) + 5(+2) + 8(+1) = +17; RHS = (+2) + 5(+3) + 0 = +17. ✓
Example 7.18 — Balance Cr2O72− + I⁻ → Cr3+ + I2 (acidic)
Halves:
Oxidation: 2 I⁻ → I₂ + 2 e⁻
Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Multiply oxidation by 3 to give 6 e⁻:
6 I⁻ → 3 I₂ + 6 e⁻
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Net:
Cr₂O₇²⁻ + 6 I⁻ + 14 H⁺ → 2 Cr³⁺ + 3 I₂ + 7 H₂O
Problem 7.7 — Balance Cr2O72− + H2S → Cr3+ + S (acidic)
Halves:
Ox: H₂S → S + 2 H⁺ + 2 e⁻
Red: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Red: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
Multiply oxidation by 3: 3 H₂S → 3 S + 6 H⁺ + 6 e⁻
Add:
Cr₂O₇²⁻ + 3 H₂S + 8 H⁺ → 2 Cr³⁺ + 3 S + 7 H₂O
(14 H⁺ on red side − 6 H⁺ on ox side = net 8 H⁺ on LHS.)
Example 7.19 — Permanganate–Oxalate titration (acidic)
MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in dilute H₂SO₄.
Ox: C₂O₄²⁻ → 2 CO₂ + 2 e⁻
Red: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Red: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Multiply ox × 5 and red × 2:
2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O
This is the stoichiometry of the classic redox titration (warm, ~60 °C, self-indicating).
Worked Examples — Balancing in Basic Medium
Problem 7.6 — Balance MnO4⁻ + S2− → MnO2 + S (basic)
Halves (start neutral):
Ox: S²⁻ → S + 2 e⁻
Red: MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻
Red: MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻
Multiply ox × 3 and red × 2 to match 6 e⁻:
3 S²⁻ → 3 S + 6 e⁻
2 MnO₄⁻ + 4 H₂O + 6 e⁻ → 2 MnO₂ + 8 OH⁻
2 MnO₄⁻ + 4 H₂O + 6 e⁻ → 2 MnO₂ + 8 OH⁻
Net:
2 MnO₄⁻ + 3 S²⁻ + 4 H₂O → 2 MnO₂ + 3 S + 8 OH⁻
Example 7.20 — KMnO4 + Na2SO3 → MnO2 + Na2SO4 (basic)
Core redox: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻.
Ox: SO₃²⁻ + 2 OH⁻ → SO₄²⁻ + H₂O + 2 e⁻
Red: MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻
Red: MnO₄⁻ + 2 H₂O + 3 e⁻ → MnO₂ + 4 OH⁻
Multiply ox × 3, red × 2 (6 e⁻ each side):
3 SO₃²⁻ + 6 OH⁻ + 2 MnO₄⁻ + 4 H₂O → 3 SO₄²⁻ + 3 H₂O + 2 MnO₂ + 8 OH⁻
Cancel 3 H₂O and 6 OH⁻:
2 MnO₄⁻ + 3 SO₃²⁻ + H₂O → 2 MnO₂ + 3 SO₄²⁻ + 2 OH⁻
Interactive: Redox Balancer Walkthrough
Pick a classic equation and watch the half-reaction balancing unfold step by step.
Pick a reaction and click Show steps.
7.4 Redox Reactions as the Basis of Titrations
Many volumetric analyses are driven by a redox reaction between a standard oxidising solution (KMnO4, K2Cr2O7) and a reducing analyte (Fe2+, C2O42−, I⁻). End-point detection uses either an external indicator or a self-indicator.
| Titrant | Indicator | Colour change at end-point |
|---|---|---|
| KMnO4 in dil. H2SO4 | Self-indicator | Colourless → permanent pale pink |
| K2Cr2O7 | Diphenylamine (internal) | Green Cr³⁺ → violet at end |
| I2 / thiosulphate | Starch (added near end) | Blue-black → colourless |
7.5 Limitations of the Oxidation-Number Concept
Although very useful, oxidation numbers are an accounting fiction. They miss the driving force of a reaction: some redox reactions that "look fine" on paper simply do not proceed because the reducing agent is too weak, or the oxidising agent too mild. A quantitative treatment needs electrode potentials (Class 12, Electrochemistry), measured in volts on a universally calibrated scale.
7.6 Redox Reactions and Electrode Processes — a Preview
If we could physically separate the two halves of a redox reaction and force the electrons to travel through an external wire, we would have an electrochemical cell — a battery. The most famous textbook example is the Daniell cell.
Cell conventions:
- Anode: oxidation site — Zn(s) → Zn²⁺(aq) + 2 e⁻. Marked negative in a galvanic cell.
- Cathode: reduction site — Cu²⁺(aq) + 2 e⁻ → Cu(s). Marked positive.
- Salt bridge (e.g. KCl in agar gel) maintains electrical neutrality by carrying ions between the half-cells without mixing the solutions.
- Cell notation: Zn(s) | Zn²⁺(aq, 1 M) ‖ Cu²⁺(aq, 1 M) | Cu(s).
EMF of the Cell
The electromotive force (EMF) is the potential difference when no current flows. At standard conditions (1 M, 1 bar, 298 K) it is written E°cell and calculated from the standard electrode potentials of the two half-cells:
E°cell = E°cathode − E°anode
For the Daniell cell, E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V, so
E°cell = 0.34 − (−0.76) = +1.10 V
| Half-cell (reduction form) | E° / V (vs SHE) |
|---|---|
| F₂ + 2 e⁻ → 2 F⁻ | +2.87 (strongest oxidant) |
| Cl₂ + 2 e⁻ → 2 Cl⁻ | +1.36 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Cu²⁺ + 2 e⁻ → Cu | +0.34 |
| 2 H⁺ + 2 e⁻ → H₂ | 0.00 (reference SHE) |
| Fe²⁺ + 2 e⁻ → Fe | −0.44 |
| Zn²⁺ + 2 e⁻ → Zn | −0.76 |
| Na⁺ + e⁻ → Na | −2.71 |
| Li⁺ + e⁻ → Li | −3.04 (strongest reductant) |
Example 7.21 — EMF of a Zn / Ag⁺ cell
Given E°(Zn²⁺/Zn) = −0.76 V, E°(Ag⁺/Ag) = +0.80 V.
Zn is more negative, so it is the anode.
E°cell = E°cathode − E°anode = 0.80 − (−0.76) = +1.56 V.
Positive E° ⇒ the spontaneous cell reaction is Zn + 2 Ag⁺ → Zn²⁺ + 2 Ag.
Example 7.22 — Will Cu reduce Ag⁺?
E°(Ag⁺/Ag) = +0.80, E°(Cu²⁺/Cu) = +0.34. Cu is the better reducer (lower E°), so Cu reduces Ag⁺:
Cu + 2 Ag⁺ → Cu²⁺ + 2 Ag; E°cell = +0.46 V — positive, so spontaneous. This explains the silver deposit on copper wire dipped in AgNO3.
Activity 7.3 — Build a Daniell cell with a lemon? L5 Evaluate
Materials: two small beakers, Zn and Cu strips, ZnSO4 and CuSO4 (0.1 M each), salt-bridge strip (filter paper soaked in KCl), digital multimeter.
- Fill one beaker with ZnSO4, the other with CuSO4.
- Insert the Zn strip into ZnSO4 and Cu strip into CuSO4.
- Dip the salt-bridge paper into both beakers to bridge them.
- Connect the multimeter between the strips and read the voltage.
Predict: what voltage will you measure?
Close to 1.0–1.1 V. The exact value is below 1.10 V (theoretical) because the solutions are dilute and internal resistance adds up. Reverse the leads — the display flips sign, confirming electron direction from Zn to Cu.
Competency-Based Questions
A student prepares a galvanic cell with Mg (E° = −2.37 V) and Fe (E° = −0.44 V) electrodes dipped in 1 M solutions of their respective sulphates, joined by a KNO3 salt bridge.
Q1. Which electrode is the anode?
B. Mg. Lower (more negative) E° ⇒ greater tendency to lose electrons ⇒ anode.
Q2. Compute E°cell.
E° = (−0.44) − (−2.37) = +1.93 V.
Q3. Balance in acidic medium: MnO4⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.
MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O.
Q4. True/False: In a salt bridge, electrons physically cross over between the two beakers.
False. Ions move through the salt bridge; electrons travel only through the external wire.
Q5. (HOT) Why does KMnO4 act as a self-indicator but K2Cr2O7 does not?
KMnO4 is intensely purple; its reduction product Mn²⁺ is colourless. A tiny excess drop after equivalence gives a sharp, persistent pink. K2Cr2O7 and its product Cr³⁺ are both coloured (orange → green), so the transition is not sharp; an external indicator (diphenylamine) is required.
Assertion–Reason Questions
Options: A both true, R explains A · B both true, R does not explain A · C A true R false · D A false R true.
A: In the Daniell cell, Zn is the negative electrode.
R: Zn is oxidised at the anode, releasing electrons that make the Zn electrode negative.
A. Both true and R explains A.
A: The reaction 2 Ag + Cu²⁺ → 2 Ag⁺ + Cu is spontaneous.
R: E°(Ag⁺/Ag) = +0.80 V is greater than E°(Cu²⁺/Cu) = +0.34 V.
D. Assertion is false — Ag cannot reduce Cu²⁺ (wrong direction). R is a true statement but it actually forbids the assertion, not supports it.
A: Balancing redox equations requires conservation of both atoms and charge.
R: In ionic equations the charge on the two sides must be equal.
A. Both true; R explains A.
Frequently Asked Questions — Balancing Redox Equations and Electrochemical Cells
How do you balance redox equations by the oxidation number method?
The oxidation number method in NCERT Class 11 Chemistry Chapter 7: (1) write the unbalanced equation; (2) assign oxidation numbers to all atoms; (3) identify atoms whose oxidation numbers change; (4) calculate increase (oxidation) and decrease (reduction) in oxidation numbers; (5) multiply species by small integers so total increase equals total decrease; (6) balance other atoms by inspection; (7) balance O by adding H₂O and H by adding H⁺ (acidic medium) or OH⁻ (basic medium). Example: balance Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O step by step.
How do you balance redox equations by the half-reaction method?
The half-reaction (ion-electron) method in NCERT Class 11 Chemistry Chapter 7: (1) write skeletal ionic equation; (2) split into oxidation and reduction half-reactions; (3) balance atoms other than H and O in each half; (4) balance O by adding H₂O; (5) balance H by adding H⁺ (acidic) or OH⁻ (basic); (6) balance charge by adding electrons; (7) multiply each half-reaction so electrons cancel; (8) add the two half-reactions. Example: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium. This method is more systematic and is preferred for complex aqueous redox reactions.
What is a galvanic cell?
A galvanic cell (or voltaic cell) is an electrochemical device that converts the energy of a spontaneous redox reaction into electrical energy. NCERT Class 11 Chemistry Chapter 7 describes the Daniell cell: zinc in ZnSO₄ solution and copper in CuSO₄ solution connected by a salt bridge and an external wire. Zn (anode) is oxidised to Zn²⁺ and Cu²⁺ is reduced to Cu (cathode). Electrons flow through the external wire from anode to cathode; ions flow through the salt bridge to maintain electrical neutrality. Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Anode is negative; cathode is positive.
What is electrode potential and how is it measured?
Electrode potential is the tendency of an electrode to lose or gain electrons in contact with its ion solution. It cannot be measured for a single electrode in isolation — only the difference between two electrodes (the cell potential) is measurable. To assign individual values, NCERT Class 11 Chemistry Chapter 7 uses a reference: the standard hydrogen electrode (SHE) with E° = 0.00 V by convention. Standard electrode potentials (E°) are measured against SHE at 1 M concentration, 1 bar pressure and 298 K. The electrochemical series ranks elements by E°, predicting reactivity, displacement and spontaneity of redox.
What is the standard hydrogen electrode (SHE)?
The standard hydrogen electrode (SHE) is the universal reference electrode for measuring standard electrode potentials. It consists of a platinum wire coated with platinum black dipped in 1 M HCl solution, with pure H₂ gas at 1 bar bubbled around it. The half-reaction is: 2H⁺(aq, 1 M) + 2e⁻ ⇌ H₂(g, 1 bar), assigned E° = 0.00 V by convention at 298 K. NCERT Class 11 Chemistry Chapter 7 uses SHE as the reference against which all other half-cell potentials are reported. Although SHE is difficult to use practically, secondary reference electrodes (calomel) are calibrated against it.
How does cell potential predict spontaneity of redox reactions?
The standard cell potential is E°_cell = E°_cathode − E°_anode (both as reduction potentials). NCERT Class 11 Chemistry Chapter 7 rules: (1) if E°_cell > 0, the cell reaction is spontaneous as written; (2) if E°_cell < 0, the reverse reaction is spontaneous; (3) the larger E°_cell, the more thermodynamically favoured the reaction. Cell potential is related to Gibbs free energy: ΔG° = −nFE°_cell. For the Daniell cell, E°_cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = +0.34 − (−0.76) = +1.10 V, confirming spontaneity. The electrochemical series enables prediction of any redox reaction direction.
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