AP EAPCET (Agriculture & Pharmacy) · Chemistry (JEE & NEET)
Redox Reactions and Electrochemistry
Oxidation states, balancing redox equations, electrochemical cells, electrode potentials, Nernst equation and conductance.
Ten concepts from Class 11 redox through Class 12 electrochemistry: oxidation numbers and balancing, galvanic cells, E° and ΔG°, Nernst, Faraday electrolysis, molar conductivity/Kohlrausch, oxidising strength from E°, then NCERT’s batteries, H₂–O₂ fuel cells, and corrosion as an unwanted galvanic cell.
- AP EAPCET (Agriculture & Pharmacy)
- Medium level
- 10 concepts
- 5 practice questions
1Oxidation numbers and the ion-electron method
Oxidation is loss of electrons — equivalently, an increase in oxidation number. Reduction is the gain. A redox equation is two half-reactions that together conserve both mass and charge: the electrons lost by the reductant equal the electrons gained by the oxidant.
The ion-electron method writes each half in acidic (or basic) medium, balances O with H₂O and H with H⁺, then multiplies so the electron counts match before adding. Oxidation numbers are a bookkeeping device for spotting which species change; they are not physical charges on atoms in a covalent molecule.
Figure. Balance redox by half-reactions: MnO₄⁻ to Mn²⁺ consumes five electrons; Fe²⁺ to Fe³⁺ produces one. Multiply the iron half by five so the electrons cancel, then add the halves.
Balancing a redox equation
- Assign oxidation numbersFind which atoms change. The species that increases ON is oxidised; the one that decreases is reduced.
- Write the two halvesBalance atoms other than O and H, then O with H₂O and H with H⁺ (acidic medium). Add electrons to balance charge.
- Equalise electrons and addMultiply halves so e⁻ cancel, then cancel spectators. Check atoms and net charge on both sides.
Permanganate oxidizing Fe²⁺
Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium, and state how many moles of Fe²⁺ one mole of MnO₄⁻ oxidises.
- Mn half: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OMn +7 → +2, gains 5e⁻
- Fe half: Fe²⁺ → Fe³⁺ + e⁻loses 1e⁻
- ×5 on Fe half, then addMnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
- Charge check: left −1+10+8 = +17; right +2+15 = +17balanced
Pro tip. In acidic medium the electron count on MnO₄⁻ → Mn²⁺ is always 5. That single fact turns most titration stoichiometry into one multiplication.
In the half-reaction Cr₂O₇²⁻ → 2Cr³⁺ (acidic), the number of electrons gained is
- 3
- 6
- 7
Each Cr goes from +6 to +3, so two Cr atoms gain 6 electrons. The balanced half is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
2Galvanic cell: anode, cathode and cell notation
A galvanic (voltaic) cell converts chemical energy into electrical energy. Oxidation runs at the anode and reduction at the cathode. In a spontaneously discharging cell the anode is the negative terminal — electrons leave the cell there — and the cathode is positive.
Cell notation writes anode half on the left and cathode on the right, phases separated by | and the salt bridge by ∥: Zn | Zn²⁺(aq) ∥ Cu²⁺(aq) | Cu. The salt bridge completes the circuit with ions without mixing the two solutions.
Figure. Oxidation at the Zn anode frees electrons that travel through the external wire to the Cu cathode. The dashed salt bridge returns the ionic current; it is not an electron path. Signs shown are for spontaneous discharge.
Reading a Daniell cell
- Find the spontaneous directionThe couple with the more positive reduction potential is reduced (cathode). The other is oxidised (anode).
- Assign signsAnode negative, cathode positive while the cell discharges spontaneously.
- Write the notationAnode | anode ion ∥ cathode ion | cathode, with concentrations if non-standard.
Writing the Daniell cell
For the spontaneous cell with Zn/Zn²⁺ and Cu/Cu²⁺ electrodes, identify anode and cathode and write the cell notation at unit activities.
- E°(Cu²⁺/Cu) = +0.34 V; E°(Zn²⁺/Zn) = −0.76 VCu couple more positive
- Cathode (reduction)Cu²⁺ + 2e⁻ → Cu
- Anode (oxidation)Zn → Zn²⁺ + 2e⁻
- Cell notationZn | Zn²⁺(aq) ∥ Cu²⁺(aq) | Cu
Pro tip. If you reverse the notation you reverse the sign of E_cell — the chemistry did not change, only which way you chose to write it.
While a Daniell cell discharges, electrons in the external wire travel
- From Cu to Zn
- From Zn to Cu
- Through the salt bridge from Zn²⁺ to Cu²⁺
Oxidation at Zn frees electrons that flow through the wire to the Cu cathode, where Cu²⁺ is reduced. The salt bridge carries ions, not the electron current.
3Standard cell potential, ΔG° and K
Under standard conditions the cell potential is E°_cell = E°_cathode − E°_anode, both written as reduction potentials. A positive E°_cell means the cell reaction as written is spontaneous, because ΔG° = −nFE°_cell is then negative.
The same E° fixes the equilibrium constant through E°_cell = (0.059/n) log K at 298 K. E° is intensive: doubling every coefficient in the balanced equation doubles n and leaves E° unchanged — ΔG° doubles, log K doubles, and K is squared, which is consistent.
Figure. Standard reduction potentials on one scale. More positive means a stronger oxidising agent as the oxidised form; more negative means a stronger reducing agent as the metal. F₂ is the strongest oxidiser here and Li the strongest reducer. Drawn lengths follow the values.
From E° to ΔG° and K
- Subtract reduction potentialsE°_cell = E°(cathode couple) − E°(anode couple). Do not flip a sign by hand and then subtract again.
- Gibbs energyΔG° = −nFE° with F = 96500 C mol⁻¹. Positive E° gives negative ΔG°.
- Equilibrium constantAt 298 K, E° = (0.059/n) log K, so log K = nE°/0.059.
Daniell cell thermodynamics
For Zn | Zn²⁺ ∥ Cu²⁺ | Cu, E°_cell = 1.10 V and n = 2. Find ΔG° and K at 298 K.
- ΔG° = −nFE° = −2 × 96500 × 1.10−212300 J mol⁻¹ = −212.3 kJ mol⁻¹
- log K = nE°/0.059 = 2 × 1.10 / 0.05937.288
- K = 10^37.2881.94 × 10³⁷
- Check: E° = (0.059/2) log K = 0.0295 × 37.2881.10 V ✓
Pro tip. If you double the equation to 2Zn + 2Cu²⁺ → …, E° stays 1.10 V while n becomes 4. ΔG° doubles to −424.6 kJ and K becomes (1.94 × 10³⁷)² — never multiply E° by the stoichiometric factor.
A cell reaction is doubled so every coefficient is ×2. What happens to E°_cell and ΔG°?
- Both double
- E° unchanged, ΔG° doubles
- E° halves, ΔG° unchanged
E° is intensive and does not scale with the equation. ΔG° = −nFE° scales with n, so it doubles when the reaction is doubled.
4Nernst equation under non-standard conditions
When concentrations leave 1 M, the cell potential is E = E° − (0.059/n) log Q at 298 K. Q is the reaction quotient written exactly like K, with solids and pure liquids omitted. Lowering the concentration of a product or raising that of a reactant increases E — the same direction Le Chatelier predicts for a spontaneous forward reaction.
At equilibrium Q = K and E = 0, which recovers E° = (0.059/n) log K. The 0.059 V factor is (2.303 RT/F) at 298 K; at any other temperature use (2.303 RT/nF) log Q with T in kelvin.
Figure. E against log Q for the Daniell cell (n = 2). Slope is −0.0295 V per decade. The vertical axis sits at log Q = 0; the dashed horizontal rule is E° = 1.10 V, not E = 0. The worked example (Q = 0.1) is the marked point above E°.
Applying Nernst
- Write Q for the cell reactionProducts over reactants, each to its stoichiometric power. Omit solids.
- Insert into NernstE = E° − (0.059/n) log Q at 298 K. Watch the sign: subtracting a negative log raises E.
- Sanity-check directionQ < 1 (reactant-rich) gives E > E°; Q > 1 gives E < E°.
EMF of a Daniell cell (Nernst)
For Zn | Zn²⁺(0.1 M) ∥ Cu²⁺(1.0 M) | Cu, E°_cell = 1.10 V. Find the cell EMF at 298 K.
- n = 2; Q = [Zn²⁺]/[Cu²⁺] = 0.1/1.00.1
- log Q = log(0.1)−1
- E = 1.10 − (0.059/2)(−1) = 1.10 + 0.02951.1295 V ≈ 1.13 V
- Direction check: Q < 1 so E > E°1.13 > 1.10 ✓
Pro tip. Lowering the anode-ion concentration raises the EMF, exactly as Le Chatelier predicts for Zn → Zn²⁺ + 2e⁻.
For a cell with n = 1, raising Q by a factor of 10 at 298 K changes E by
- −0.059 V
- +0.059 V
- −0.0295 V
ΔE = −(0.059/n) Δ(log Q). One decade in Q at n = 1 shifts E by −0.059 V.
5Faraday's laws of electrolysis
The mass of substance deposited or liberated at an electrode is proportional to the charge passed: w = ItM/(nF), with F = 96500 C mol⁻¹. Here M/n is the equivalent mass — one faraday deposits one equivalent of any electrolyte.
Equivalents deposited equal It/96500. Current and time enter only as their product; 2 A for 965 s is the same charge as 1 A for 1930 s, and deposits the same mass.
Figure. Mass of Cu deposited against charge passed, for n = 2 and M = 63.5. The line is through the origin with slope M/(nF) = 3.29 × 10⁻⁴ g C⁻¹. The worked example sits at Q = 1930 C.
Mass from charge
- Charge passedQ = It in coulombs. Convert minutes to seconds before multiplying.
- Electrons per formula unitRead n from the reduction half: Cu²⁺ + 2e⁻ → Cu has n = 2.
- Massw = QM/(nF). Equivalently, moles = Q/(nF), then multiply by M.
Copper deposited by electrolysis
How much copper is deposited when 2 A flows for 965 s through CuSO₄? (M of Cu = 63.5 g mol⁻¹.)
- Q = It = 2 × 9651930 C
- Cu²⁺ + 2e⁻ → Cu, so n = 2n = 2
- w = QM/(nF) = 1930 × 63.5 / (2 × 96500)0.635 g
- Check: Q/F = 1930/96500 = 0.02 equiv; mass = 0.02 × (63.5/2)0.635 g ✓
Pro tip. 96500 C (1 F) deposits one equivalent = M/n grams of any metal. For Cu that is 31.75 g per faraday.
Passing 0.5 F of charge through molten AlCl₃ deposits how many moles of Al? (Al³⁺ + 3e⁻ → Al)
- 0.5 mol
- 1/6 mol
- 1.5 mol
Moles = Q/(nF) = 0.5/3 = 1/6 mol. One faraday deposits 1/3 mol of Al, so half a faraday deposits half of that.
6Molar conductivity and Kohlrausch's law
Molar conductivity is Λ_m = κ × 1000 / C, with κ the conductivity of the solution and C the molarity. Λ_m rises on dilution because ion–ion interactions weaken and, for weak electrolytes, because the degree of dissociation rises.
Kohlrausch's law of independent migration says the limiting molar conductivity is the sum of ionic contributions: Λ_m° = ν₊ λ₊° + ν₋ λ₋°. That sum is how Λ_m° of a weak electrolyte is obtained — from strong-electrolyte limits that can be extrapolated linearly against √C — when the weak electrolyte's own Λ_m never levels off in an accessible concentration window.
Figure. Λm against √c. A strong electrolyte is linear — Kohlrausch's Λm = Λm° − A√c — and extrapolates to Λm° at √c = 0. A weak electrolyte bows upward on dilution and never gives a trustworthy linear intercept in the same window; its Λm° is assembled from ionic conductivities instead.
Using Kohlrausch
- Measure or be given κ and CΛ_m = κ × 1000 / C. Units: S cm² mol⁻¹ when κ is in S cm⁻¹ and C in mol L⁻¹.
- Strong electrolyte → √C plotΛ_m = Λ_m° − A√C is linear; the intercept is Λ_m°.
- Weak electrolyte → ionic sumBuild Λ_m° from λ₊° and λ₋° of the ions, taken from strong salts that share them.
Λ_m° of acetic acid from Kohlrausch
Given λ°(H⁺) = 349.6, λ°(Na⁺) = 50.1, λ°(CH₃COO⁻) = 40.9 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹, find Λ_m°(CH₃COOH).
- Λ_m°(HCl) = λ°(H⁺) + λ°(Cl⁻) = 349.6 + 76.3425.9
- Λ_m°(NaCH₃COO) = 50.1 + 40.991.0
- Λ_m°(NaCl) = 50.1 + 76.3126.4
- Λ_m°(CH₃COOH) = Λ_m°(HCl) + Λ_m°(NaAc) − Λ_m°(NaCl)425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹
Pro tip. The same answer is λ°(H⁺) + λ°(CH₃COO⁻) = 349.6 + 40.9 = 390.5 — Kohlrausch's combination of three salts is just a way to assemble those two ionic values when you were not given them directly.
On dilution of a weak electrolyte, Λ_m rises mainly because
- κ rises sharply
- The degree of dissociation α rises
- The ions move faster through denser solvent
κ actually falls on dilution (fewer ions per cm³), but α rises enough that Λ_m = κ×1000/C still increases, approaching Λ_m° as α → 1.
7Oxidising and reducing strength from E°
A more positive standard reduction potential means a stronger oxidising agent in its oxidised form. F₂ (E° ≈ +2.87 V) is the strongest common oxidiser; the corresponding F⁻ is an extremely weak reducer. Conversely, a large negative E° means a strong reducing agent in its reduced form — Li metal, not Li⁺.
Comparing two couples: the oxidised form of the higher-E° couple oxidises the reduced form of the lower-E° couple. That is the same statement as E°_cell = E°_high − E°_low > 0.
Figure. The oxidised form of a high-E° couple is a strong oxidiser; the reduced form of a low-E° couple is a strong reducer. Pair them and the cell runs spontaneously — F₂ oxidises Li metal, never the reverse under standard conditions.
Who oxidises whom
- List the couplesWrite each as ox/red with its E°.
- Higher E° is reducedThat couple's oxidised form is the oxidising agent.
- Lower E° is oxidisedThat couple's reduced form is the reducing agent. E°_cell is their difference.
Will Fe³⁺ oxidise I⁻?
E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V. Predict whether Fe³⁺ oxidises I⁻ under standard conditions, and find E°_cell.
- Higher E° couple: Fe³⁺/Fe²⁺Fe³⁺ reduced
- Lower E° couple: I₂/I⁻I⁻ oxidised
- E°_cell = 0.77 − 0.540.23 V > 0
- Spontaneous cell reaction2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Pro tip. Flip the argument for the reverse: I₂ cannot oxidise Fe²⁺ under standard conditions, because that cell would have E° = −0.23 V.
Among F₂, Cl₂, Br₂ and I₂, the strongest oxidising agent is
- I₂
- Cl₂
- F₂
Oxidising power of the halogens tracks their reduction potentials: F₂ > Cl₂ > Br₂ > I₂. F₂ has the most positive E°.
8Primary and secondary batteries
A practical battery is a galvanic cell (or several in series) engineered so voltage stays usable and the pack is compact. Primary batteries discharge once and cannot be recharged: the dry cell (Leclanché) uses a Zn anode (the can), a graphite cathode in MnO₂/C, and a moist NH₄Cl/ZnCl₂ paste — about 1.5 V. The mercury cell (Zn–Hg amalgam anode, HgO cathode, KOH/ZnO electrolyte) sits near 1.35 V and stays unusually flat because the overall reaction Zn(Hg) + HgO → ZnO + Hg involves no solution ion whose concentration drifts.
Secondary batteries are recharged by driving current backward. The lead storage battery (automobiles, inverters) has a Pb anode, PbO₂ on a lead grid as cathode, and ~38% H₂SO₄. Discharge: Pb + PbO₂ + 2 H₂SO₄ → 2 PbSO₄ + 2 H₂O; charging reverses that. Nickel–cadmium is another secondary cell with longer life but higher cost; NCERT quotes Cd + 2 Ni(OH)₃ → CdO + 2 Ni(OH)₂ + H₂O on discharge.
Figure. Primary packs discharge once (dry cell); secondary packs reverse on charge (lead–acid swaps Pb/PbO₂ with PbSO₄ and restores acid density). The chemistry is the recharge arrow, not the can drawing.
Classify the battery
- Primary vs secondaryDead after one use → primary (dry cell, mercury). Rechargeable by reverse current → secondary (lead storage, Ni–Cd).
- Lead storage netDischarge consumes H₂SO₄ and forms PbSO₄ on both electrodes; density of acid falls — a practical state-of-charge cue.
- Mercury flat voltageNo aqueous ion concentration change in the net reaction → voltage stays near 1.35 V.
| Cell | Class | Net / cue |
|---|---|---|
| Dry cell (Leclanché) | Primary | ~1.5 V; Zn / MnO₂ |
| Mercury cell | Primary | ~1.35 V, flat; Zn(Hg) + HgO |
| Lead storage | Secondary | Pb + PbO₂ + 2 H₂SO₄ ⇌ 2 PbSO₄ + 2 H₂O |
| Ni–Cd | Secondary | Longer life; costlier |
Lead storage discharge
Write the net discharge reaction of a lead storage battery and state what happens to the sulfuric acid.
- AnodePb + SO₄²⁻ → PbSO₄ + 2e⁻
- CathodePbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
- NetPb + PbO₂ + 2 H₂SO₄ → 2 PbSO₄ + 2 H₂O
- Acidconsumed — concentration/density falls
Pro tip. Charging is the reverse arrow: PbSO₄ on both plates returns to Pb and PbO₂ while H₂SO₄ is regenerated.
A lead storage battery is classified as secondary because
- It uses sulfuric acid, which only secondary cells may use
- After discharge it can be recharged by reversing the cell reaction with an external current
- Its voltage is higher than any primary cell
Secondary means rechargeable by driving current opposite to discharge. Acid presence and voltage magnitude do not define the class.
9Fuel cells: continuous galvanic combustion
A fuel cell is a galvanic cell that converts the energy of combustion of a continuously supplied fuel (H₂, CH₄, CH₃OH, …) directly into electricity, with products removed as they form. The classic H₂–O₂ cell used in the Apollo programme bubbles the gases through porous carbon electrodes into concentrated aqueous NaOH; Pt or Pd catalysts speed the electrode reactions.
Cathode: O₂ + 2 H₂O + 4 e⁻ → 4 OH⁻. Anode: 2 H₂ + 4 OH⁻ → 4 H₂O + 4 e⁻. Net: 2 H₂ + O₂ → 2 H₂O. NCERT contrasts ~70% electrical efficiency with ~40% for thermal power plants, and notes the cell is pollution-free while reactants last. Water vapour from Apollo cells was even condensed for drinking water.
Figure. Flow sketch of a continuous H₂–O₂ fuel cell — not a hardware cutaway. Gases enter electrodes; water and current leave. Efficiency numbers live in the prose, not in this graph.
Recognise a fuel cell
- Continuous feedFuel and oxidant supplied; products removed — not a sealed primary battery that goes dead when internal reagents finish.
- Write H₂–O₂ net2 H₂ + O₂ → 2 H₂O in alkaline medium via OH⁻ at both electrodes.
- Efficiency cueDirect chemical → electrical (~70%) beats heat → steam → turbine (~40%) in NCERT’s comparison.
H₂–O₂ electrode balance
Show that the NCERT anode and cathode half-reactions for the alkaline H₂–O₂ fuel cell sum to 2 H₂ + O₂ → 2 H₂O.
- Cathode ×1O₂ + 2 H₂O + 4e⁻ → 4 OH⁻
- Anode ×12 H₂ + 4 OH⁻ → 4 H₂O + 4e⁻
- Add; cancel 4 OH⁻, 4e⁻, 2 H₂O2 H₂ + O₂ → 2 H₂O
Pro tip. OH⁻ is a catalyst-like intermediate in the alkaline write-up — it appears on both sides and cancels in the net combustion.
Unlike a dry cell, a hydrogen–oxygen fuel cell
- Is electrolytic rather than galvanic
- Runs as long as H₂ and O₂ are supplied, converting combustion energy directly to electricity
- Must be recharged overnight like a lead storage battery
Fuel cells are galvanic and continuous-feed. They are not electrolysers, and they are not secondary batteries awaiting reverse charging.
10Corrosion as an electrochemical cell
Corrosion coats metals with oxides or other salts — rusting of iron, tarnish on silver, green patina on copper. Rusting needs air and water and is electrochemical: one spot on iron acts as anode (Fe → Fe²⁺ + 2 e⁻, E°(Fe²⁺/Fe) = −0.44 V) while another spot acts as cathode, reducing oxygen in acidified moisture (O₂ + 4 H⁺ + 4 e⁻ → 2 H₂O, E° = 1.23 V). Net: 2 Fe + O₂ + 4 H⁺ → 2 Fe²⁺ + 2 H₂O with E°(cell) = 1.67 V. Fe²⁺ is further oxidised to hydrated Fe₂O₃·xH₂O (rust), regenerating H⁺.
Prevention blocks the cell: paint or coatings keep air/water off; tin or zinc plating covers the surface; sacrificial anodes (Mg, Zn) corrode instead of the protected object. The exam move is to recognise rusting as a galvanic process with known E° pieces, not a mysterious surface stain.
Figure. Two spots on one iron object form a shorted galvanic cell. Labels are process names, not to-scale oxide thicknesses. E° = 1.67 V is in the ledger.
Read a rusting stem
- Find anode/cathodeIron oxidation spot = anode. O₂/H⁺ reduction spot = cathode. Electrons travel through the metal.
- E° cell1.23 − (−0.44) = 1.67 V for the NCERT Fe/O₂ couple under standard writing.
- Prevention classBarrier coating, metal cladding, or sacrificial anode — all interrupt the electrochemical loop.
Standard cell potential for rusting write-up
Using E°(Fe²⁺/Fe) = −0.44 V and E°(O₂/H⁺/H₂O) = 1.23 V as in NCERT, compute E° for 2 Fe + O₂ + 4 H⁺ → 2 Fe²⁺ + 2 H₂O.
- E°_cathode (O₂)1.23 V
- E°_anode (Fe)−0.44 V
- E°_cell = E°_cathode − E°_anode1.23 − (−0.44) = 1.67 V
Pro tip. Same subtraction rule as any galvanic E°_cell — corrosion is just an unwanted galvanic cell on one piece of metal.
Rusting of iron is best described as
- A purely chemical hydration of Fe that needs no electron transfer
- An electrochemical process with anodic Fe oxidation and cathodic O₂ reduction in moist air
- Electrolysis forced by stray household currents only
NCERT treats rusting as a galvanic couple on the metal surface. It is not mere hydration and does not require external electrolysis.
Notes
- Redox basics: oxidation is loss of electrons (increase in oxidation number) and reduction is gain; redox equations are balanced by the ion-electron (half-reaction) method conserving mass and charge.
- Galvanic cell: converts chemical energy to electrical energy; the anode is negative (oxidation) and the cathode is positive (reduction), written anode \parallel cathode.
- Cell potential: E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}; a positive E^\circ_{cell} means a spontaneous reaction since \Delta G^\circ=-nFE^\circ.
- Nernst equation gives EMF under non-standard conditions: E=E^\circ-\frac{0.059}{n}\log Q at 298 K.
- Conductance: molar conductivity \Lambda_m increases on dilution, and Kohlrausch's law gives \Lambda_m^\circ as the sum of independent ionic conductivities.
Formulas
- E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}
- \Delta G^\circ=-nFE^\circ_{cell},\quad E^\circ_{cell}=\frac{0.059}{n}\log K
- E=E^\circ-\frac{0.059}{n}\log Q (Nernst, 298 K)
- Faraday's law: w=\frac{ItM}{nF},\quad F=96500 C/mol
- \Lambda_m=\frac{\kappa\times1000}{C},\quad \Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ (Kohlrausch)
Exam traps & shortcuts
- A more positive standard reduction potential means a stronger oxidising agent; F_2 is the strongest oxidiser and Li the strongest reducer.
- Equivalents deposited in electrolysis =\frac{It}{96500}; mass is proportional to charge passed.
- E^\circ_{cell} is intensive: do not multiply it when you scale the balanced equation.
Reference tables
Signs reverse for an electrolytic cell forced by an external battery; every row below is for a spontaneously discharging galvanic cell.
| Item | Rule | Watch for |
|---|---|---|
| Anode | Oxidation; negative terminal while discharging | In electrolysis the anode is positive — external supply, not chemistry, sets the sign |
| Cathode | Reduction; positive terminal while discharging | Cations still migrate to the cathode in both cell types |
| Cell notation | Anode | anode ion ∥ cathode ion | cathode | Reversing the written order flips the sign of E_cell |
| E°_cell | E°_cathode − E°_anode (both reductions) | Never multiply E° when you scale the equation |
Read every row as a statement about the species named, not about its partner in the couple.
| Observation | Means | Example |
|---|---|---|
| Large positive E° | Oxidised form is a strong oxidiser | F₂, MnO₄⁻/H⁺ |
| Large negative E° | Reduced form is a strong reducer | Li, K, Ca |
| E°_A > E°_B | ox_A oxidises red_B | Fe³⁺ oxidises I⁻ (0.77 > 0.54) |
| E°_cell > 0 | Cell reaction as written is spontaneous | ΔG° = −nFE° < 0 |
Every line should be reconstructible from the concept it came from.
| Relation | Reads as | Watch for |
|---|---|---|
| E°_cell = E°_cathode − E°_anode | Standard cell potential | Both potentials are reductions |
| ΔG° = −nFE°_cell | Link to thermodynamics | F = 96500 C mol⁻¹; E° intensive, ΔG° extensive |
| E° = (0.059/n) log K | Equilibrium constant at 298 K | log is base 10; at other T use 2.303RT/F |
| E = E° − (0.059/n) log Q | Nernst equation at 298 K | Q uses the same form as K |
| w = ItM/(nF) | Faraday's first law | n from the electrode half-reaction |
| Λ_m = κ×1000/C | Molar conductivity | κ in S cm⁻¹, C in mol L⁻¹ |
| Λ_m° = ν₊λ₊° + ν₋λ₋° | Kohlrausch | Route to Λ_m° of weak electrolytes |
Batteries, fuel cells and corrosion close the electrochemistry unit.
| Section | Concept |
|---|---|
| 2.6 Batteries | primary_and_secondary_batteries |
| 2.7 Fuel cells | fuel_cells |
| 2.8 Corrosion | corrosion_as_electrochemistry |
Recap
Read only this the night before.
- Anode / cathode
- Oxidation at the anode, reduction at the cathode. In a discharging galvanic cell the anode is negative. Electrolytic signs flip because the battery forces the direction.
- E° and scaling
- E°_cell = E°_cathode − E°_anode. Doubling the equation leaves E° alone and doubles ΔG°. Never multiply E° by a stoichiometric factor.
- Nernst
- E = E° − (0.059/n) log Q at 298 K. One decade in Q shifts E by 0.059/n volts. Q < 1 raises E above E°.
- Faraday
- w = ItM/(nF) with F = 96500 C mol⁻¹. One faraday deposits one equivalent = M/n grams.
- Kohlrausch
- Λ_m° is the sum of ionic λ° values. Strong electrolytes extrapolate linearly against √C; weak ones need the ionic sum.
- Who oxidises whom
- More positive E° wins as the oxidiser. F₂ is the strongest common oxidiser; Li metal is the strongest common reducer.
- Batteries
- Primary = once (dry cell ~1.5 V; mercury ~1.35 V flat). Secondary = rechargeable (lead storage; Ni–Cd).
- Lead storage
- Pb + PbO₂ + 2 H₂SO₄ → 2 PbSO₄ + 2 H₂O on discharge; reverse on charge.
- Fuel cell
- Continuous H₂/O₂ → H₂O galvanic combustion; ~70% vs ~40% thermal plants.
- Corrosion
- Anodic Fe oxidation + cathodic O₂ reduction; E°=1.67 V. Paint, cladding, sacrificial Zn/Mg.
Practise Redox Reactions and Electrochemistry
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 5 exam-style questions on this topic, with explanations
- A 8-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device