Redox and Electrochemistry Formula Sheet
All key Redox and Electrochemistry formulas, electrode potentials, Nernst equation, Faraday's laws, and cell reactions for JEE Main & Advanced quick revision.
Last-minute revision sheet for Redox Reactions and Electrochemistry: every formula, electrode potential value, key relation, and must-know cell reaction from this chapter, grouped so you can scan it the night before the exam.
Oxidation Number and Redox Basics
Oxidation Is Loss of electrons, Reduction Is Gain of electrons. Alternatively: LEO the lion says GER (Lose Electrons = Oxidation, Gain Electrons = Reduction).
| Process | Electronic definition | Oxidation state |
|---|---|---|
| Oxidation | Loss of electrons | Increases |
| Reduction | Gain of electrons | Decreases |
| Oxidizing agent | Gets reduced (gains $e^-$) | Causes oxidation |
| Reducing agent | Gets oxidized (loses $e^-$) | Causes reduction |
Rules for Assigning Oxidation Numbers
| Species | Oxidation state | Exceptions |
|---|---|---|
| Free element (Na, O$_2$, P$_4$, S$_8$) | $0$ | — |
| Monoatomic ion | $=$ charge | — |
| Oxygen | $-2$ | Peroxides $-1$; superoxides $-\tfrac{1}{2}$; OF$_2$ $+2$ |
| Hydrogen | $+1$ | Metal hydrides $-1$ |
| Fluorine | $-1$ | always |
| Other halogens (Cl, Br, I) | $-1$ | with O or more electronegative halogen |
| Group 1 / Group 2 / Al | $+1$ / $+2$ / $+3$ | — |
Sum of oxidation states $= 0$ (neutral molecule) or $=$ charge (polyatomic ion).
Worked Oxidation States (from chapter)
- S in H$_2$SO$_4$: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$
- Cr in Cr$_2$O$_7^{2-}$: $2x + 7(-2) = -2 \Rightarrow x = +6$
- Average Fe in Fe$_3$O$_4$ $= +\tfrac{8}{3}$ (fractional, mixed +2 and +3)
- C in CH$_3$COOH: methyl C $=-3$, carboxyl C $=+3$, average $=0$
Types of Redox Reactions
| Type | Defining feature | Example |
|---|---|---|
| Combination | Elements/compounds join | C $+$ O$_2$ $\to$ CO$_2$ |
| Decomposition | Single reactant breaks down | 2H$_2$O$_2$ $\to$ 2H$_2$O $+$ O$_2$ |
| Displacement | One element displaces another | Zn $+$ CuSO$_4$ $\to$ ZnSO$_4$ $+$ Cu |
| Disproportionation | Same element oxidized and reduced | 3Cl$_2$ $+$ 6OH$^-$ $\to$ 5Cl$^-$ $+$ ClO$_3^-$ $+$ 3H$_2$O |
Balancing (Ion-Electron Method)
- Split into oxidation and reduction half-reactions.
- Balance atoms except O and H.
- Balance O with H$_2$O, then H with H$^+$.
- Balance charge with electrons.
- Equalize electrons, add, and cancel.
- Basic medium: add OH$^-$ to both sides to neutralize H$^+$.
Standard balanced result (acidic):
$$\boxed{\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}}$$Electrochemical Cells and EMF
| Quantity | Relation | Notes |
|---|---|---|
| Cell EMF | $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ | always cathode minus anode |
| Standard EMF | $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ | at 298 K, 1 bar, 1 M |
| Spontaneity | $E^\circ_{\text{cell}} > 0 \Rightarrow$ spontaneous | galvanic cell |
Cell notation: Anode $|$ anode electrolyte $||$ cathode electrolyte $|$ cathode. Anode (oxidation) on left, cathode (reduction) on right; $|$ = phase boundary, $||$ = salt bridge.
$$\text{Zn}_{(s)} \,|\, \text{Zn}^{2+}_{(1M)} \,||\, \text{Cu}^{2+}_{(1M)} \,|\, \text{Cu}_{(s)} \qquad E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}$$AN OX = Anode is Oxidation; RED CAT = Reduction at Cathode. CROP: Cathode $\to$ Reduction $\to$ Oxidizing agent $\to$ Positive terminal (in a galvanic cell). Anode $\to$ Oxidation $\to$ Reducing agent $\to$ Negative terminal.
Galvanic vs Electrolytic Cell
| Feature | Galvanic | Electrolytic |
|---|---|---|
| Energy | Chemical $\to$ Electrical | Electrical $\to$ Chemical |
| Reaction | Spontaneous | Non-spontaneous |
| $\Delta G$ | Negative | Positive |
| $E^\circ_{\text{cell}}$ | Positive | Negative |
| Anode polarity | Negative ($-$) | Positive ($+$) |
| Cathode polarity | Positive ($+$) | Negative ($-$) |
In both cells: oxidation at anode, reduction at cathode — only the polarity flips.
Daniell Cell Reactions
- Anode: $\text{Zn}_{(s)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2e^-$
- Cathode: $\text{Cu}^{2+}_{(aq)} + 2e^- \rightarrow \text{Cu}_{(s)}$
- Overall: $\text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)}$, $E^\circ_{\text{cell}} = 1.10$ V
Standard Hydrogen Electrode (SHE)
Reference electrode, $E^\circ = 0.00$ V. Conditions: Pt(black), H$_2$ at 1 bar, H$^+$ at 1 M, 298 K.
$$2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2 \qquad E^\circ = 0.00\ \text{V}$$Gibbs Energy, Equilibrium and EMF
$$\boxed{\Delta G^\circ = -nFE^\circ_{\text{cell}}}$$$$\boxed{\Delta G^\circ = -nFE^\circ_{\text{cell}} = -RT\ln K = -2.303\,RT\log K}$$$$\boxed{E^\circ_{\text{cell}} = \frac{0.059}{n}\log K}$$At 298 K: $\Delta G^\circ = -5.706\log K$ kJ/mol.
| $E^\circ_{\text{cell}}$ | $\Delta G^\circ$ | $K$ | Reaction |
|---|---|---|---|
| Positive | Negative | $>1$ | Spontaneous (products favored) |
| Zero | Zero | $=1$ | Equilibrium |
| Negative | Positive | $<1$ | Non-spontaneous |
Constants: $F = 96{,}500$ C/mol ($\approx 96{,}485$), $R = 8.314$ J/(mol·K), $n =$ electrons transferred.
Nernst Equation
$$\boxed{E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q = E^\circ_{\text{cell}} - \frac{2.303\,RT}{nF}\log Q}$$At 298 K, $\dfrac{2.303RT}{F} = \dfrac{2.303 \times 8.314 \times 298}{96500} = 0.0591 \approx 0.059$:
$$\boxed{E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q}$$Single electrode ($\text{M}^{n+} + ne^- \rightarrow \text{M}$):
$$E = E^\circ + \frac{0.059}{n}\log[\text{M}^{n+}]$$Reaction quotient for $aA + bB \rightarrow cC + dD$:
$$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$Solids and pure liquids omitted; gases use partial pressure; solutions use molarity.
$Q > 1 \Rightarrow \log Q > 0 \Rightarrow E_{\text{cell}}$ drops. $Q < 1 \Rightarrow E_{\text{cell}}$ rises. As products build up, the driving force falls. $E$ and $E^\circ$ are intensive — do not multiply by stoichiometric coefficients.
Concentration Cells ($E^\circ_{\text{cell}} = 0$)
$$\boxed{E_{\text{cell}} = \frac{0.059}{n}\log\frac{C_{\text{concentrated}}}{C_{\text{dilute}}}}$$Dilute solution at anode (left), concentrated at cathode (right) for positive EMF. For a gas-electrode (pressure) cell: $E_{\text{cell}} = \dfrac{0.059}{n}\log\dfrac{P_1}{P_2}$ with $P_1 > P_2$ (higher-pressure gas oxidised at the anode).
Nernst Applications
| Use | Relation |
|---|---|
| pH via H$_2$ electrode (1 atm) | $E = -0.059\,\text{pH}$, so $\text{pH} = -\dfrac{E}{0.059}$ |
| Solubility ($K_{sp}$) via AgCl electrode | $E = E^\circ - 0.059\log[\text{Cl}^-]$ |
| Equilibrium constant | $E^\circ_{\text{cell}} = \dfrac{0.059}{n}\log K$ |
Electrode Potentials and Electrochemical Series
All electrode potentials are quoted as reduction potentials: $\text{Ox} + ne^- \rightarrow \text{Red}$. To get the oxidation potential, reverse the reaction and flip the sign.
- Higher $E^\circ$ $\Rightarrow$ stronger oxidizing agent (top of series, gets reduced, goes to cathode).
- Lower $E^\circ$ $\Rightarrow$ stronger reducing agent (bottom of series, gets oxidized, goes to anode).
- Spontaneous: $E_{\text{reduction}} > E_{\text{oxidation}}$, i.e. $E^\circ_{\text{cell}} > 0$.
Standard Reduction Potentials to Memorize
| Half-reaction | $E^\circ$ (V) |
|---|---|
| F$_2$ + 2e$^-$ → 2F$^-$ | $+2.87$ |
| MnO$_4^-$ + 8H$^+$ + 5e$^-$ → Mn$^{2+}$ + 4H$_2$O | $+1.51$ |
| Au$^{3+}$ + 3e$^-$ → Au | $+1.50$ |
| Cl$_2$ + 2e$^-$ → 2Cl$^-$ | $+1.36$ |
| Cr$_2$O$_7^{2-}$ + 14H$^+$ + 6e$^-$ → 2Cr$^{3+}$ + 7H$_2$O | $+1.33$ |
| O$_2$ + 4H$^+$ + 4e$^-$ → 2H$_2$O | $+1.23$ |
| Br$_2$ + 2e$^-$ → 2Br$^-$ | $+1.09$ |
| Ag$^+$ + e$^-$ → Ag | $+0.80$ |
| Fe$^{3+}$ + e$^-$ → Fe$^{2+}$ | $+0.77$ |
| I$_2$ + 2e$^-$ → 2I$^-$ | $+0.54$ |
| O$_2$ + 2H$_2$O + 4e$^-$ → 4OH$^-$ | $+0.40$ |
| Cu$^{2+}$ + 2e$^-$ → Cu | $+0.34$ |
| 2H$^+$ + 2e$^-$ → H$_2$ | $0.00$ |
| Pb$^{2+}$ + 2e$^-$ → Pb | $-0.13$ |
| Sn$^{2+}$ + 2e$^-$ → Sn | $-0.14$ |
| Ni$^{2+}$ + 2e$^-$ → Ni | $-0.25$ |
| Fe$^{2+}$ + 2e$^-$ → Fe | $-0.44$ |
| Zn$^{2+}$ + 2e$^-$ → Zn | $-0.76$ |
| 2H$_2$O + 2e$^-$ → H$_2$ + 2OH$^-$ | $-0.83$ |
| Al$^{3+}$ + 3e$^-$ → Al | $-1.66$ |
| Mg$^{2+}$ + 2e$^-$ → Mg | $-2.37$ |
| Na$^+$ + e$^-$ → Na | $-2.71$ |
| Ca$^{2+}$ + 2e$^-$ → Ca | $-2.87$ |
| K$^+$ + e$^-$ → K | $-2.93$ |
| Li$^+$ + e$^-$ → Li | $-3.05$ |
Lock these in: Li$^+$/Li $=-3.05$ (most negative, strongest reducing agent), Zn$^{2+}$/Zn $=-0.76$, H$^+$/H$_2$ $=0.00$, Cu$^{2+}$/Cu $=+0.34$, Ag$^+$/Ag $=+0.80$, F$_2$/F$^-$ $=+2.87$ (most positive, strongest oxidizing agent).
Predictions from the Series
- Displacement: a metal displaces another from solution if it has lower $E^\circ$ (e.g. Zn displaces Cu).
- Reaction with dilute acids: metals with $E^\circ < 0$ (below H) liberate H$_2$ (Zn, Fe, Al, Mg, Na). Cu, Ag, Au ($E^\circ > 0$) do not.
- Sacrificial protection: metals with lower $E^\circ$ corrode first (Zn, Mg protect Fe).
- Maximum EMF: pair lowest-$E^\circ$ anode with highest-$E^\circ$ cathode.
Conductance and Kohlrausch’s Law
| Quantity | Formula | Notes |
|---|---|---|
| Resistance | $R = \rho\,\dfrac{l}{A}$ | $\rho$ = resistivity |
| Conductance | $G = \dfrac{1}{R} = \kappa\,\dfrac{A}{l}$ | $\kappa$ = conductivity (S/m) |
| Molar conductivity | $\Lambda_m = \dfrac{\kappa \times 1000}{M}$ | $M$ = molarity |
| Kohlrausch’s law | $\Lambda_m^\circ = \lambda^\circ_+ + \lambda^\circ_-$ | at infinite dilution |
Kohlrausch’s law lets you find $\Lambda_m^\circ$ of weak electrolytes, the degree of dissociation, and the solubility of sparingly soluble salts.
Electrolysis and Faraday’s Laws
$$\boxed{Q = I \times t}$$First law — mass deposited $\propto$ charge:
$$\boxed{m = Z\,Q = Z\,I\,t}$$Electrochemical equivalent:
$$\boxed{Z = \frac{E}{96500} = \frac{M}{nF}} \qquad E = \frac{M}{n}\ (\text{equivalent weight})$$Working forms:
$$\boxed{m = \frac{M \times I \times t}{nF} = \frac{M \times Q}{nF}} \qquad \text{moles} = \frac{Q}{nF} = \frac{It}{nF}$$Gas volume at STP:
$$V = \frac{22.4 \times I \times t}{nF}\ \text{litres}$$Second law — same charge through different cells:
$$\boxed{\frac{m_1}{m_2} = \frac{E_1}{E_2} = \frac{M_1/n_1}{M_2/n_2}}$$1 Faraday $= 96{,}500$ C $=$ charge of 1 mole of electrons.
(1) $Q = I\,t$ → (2) moles of $e^- = Q/F$ → (3) moles of substance $= (\text{moles } e^-)/n$ → (4) mass $=$ moles $\times M$, or volume $=$ moles $\times 22.4$ L. Always convert time to seconds.
Preferential Discharge
| Electrode | Rule | Order |
|---|---|---|
| Cathode (reduction) | Higher $E^\circ$ discharged first | Ag$^+$ > Cu$^{2+}$ > H$^+$ > Zn$^{2+}$ > Na$^+$ |
| Anode (oxidation) | Lower $E^\circ$ (oxidation) first | I$^-$ > Br$^-$ > Cl$^-$ > OH$^-$ > SO$_4^{2-}$, NO$_3^-$ |
Concentrated solutions favour discharge of the salt ion over water (overpotential effect — e.g. conc. NaCl gives Cl$_2$, not O$_2$, at the anode).
Key Electrolysis Results
| Electrolyte (Pt electrodes) | Net reaction |
|---|---|
| Molten NaCl | $2\text{NaCl}_{(l)} \to 2\text{Na} + \text{Cl}_2$ |
| Molten Al$_2$O$_3$ (Hall–Héroult) | $2\text{Al}_2\text{O}_3 \to 4\text{Al} + 3\text{O}_2$ |
| CuSO$_4$ (aq) | $2\text{CuSO}_4 + 2\text{H}_2\text{O} \to 2\text{Cu} + 2\text{H}_2\text{SO}_4 + \text{O}_2$ |
| NaCl (aq), conc. (chlor-alkali) | $2\text{NaCl} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 + \text{Cl}_2$ |
| H$_2$SO$_4$ (aq) / acidified water | $2\text{H}_2\text{O} \to 2\text{H}_2 + \text{O}_2$ |
Batteries and Fuel Cells
| Cell | Type | Voltage |
|---|---|---|
| Dry cell (Leclanché) / Alkaline | Primary | 1.5 V |
| Mercury (button) cell | Primary | 1.35 V |
| Lead-acid (per cell) | Secondary | 2.0 V |
| NiCd / NiMH | Secondary | 1.2 V |
| Lithium-ion | Secondary | 3.7 V |
| H$_2$–O$_2$ fuel cell | Fuel cell | ~1.0 V |
- Primary: irreversible, non-rechargeable. Secondary: reversible, rechargeable (recharged by electrolysis). Fuel cell: continuous external fuel supply.
Headline Cell Reactions
Dry cell (cathode): $2\text{MnO}_2 + 2\text{NH}_4^+ + 2e^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{NH}_3 + \text{H}_2\text{O}$
Lead-acid (discharge):
$$\boxed{\text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \xrightarrow{\text{discharge}} 2\text{PbSO}_4 + 2\text{H}_2\text{O}}$$Charging reverses this. H$_2$SO$_4$ is consumed on discharge $\Rightarrow$ electrolyte density falls (hydrometer test of charge state).
Li-ion (discharge): $\text{LiCoO}_2 + 6\text{C} \rightarrow \text{Li}_{1-x}\text{CoO}_2 + \text{Li}_x\text{C}_6$ (Li$^+$ “rocking chair”).
H$_2$–O$_2$ fuel cell (overall):
$$\boxed{2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}} \qquad \Delta G^\circ = -237\ \text{kJ/mol H}_2\text{O}$$Alkaline: anode $2\text{H}_2 + 4\text{OH}^- \to 4\text{H}_2\text{O} + 4e^-$; cathode $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^-$.
The mercury cell holds a steady ~1.35 V because the KOH electrolyte concentration doesn’t change during discharge (ZnO and H$_2$O formed don’t alter [OH$^-$]). The dry cell’s voltage drops as NH$_4$Cl is consumed.
Corrosion
Corrosion is spontaneous electrochemical oxidation of a metal — a galvanic cell on the metal surface. Rusting needs both O$_2$ and H$_2$O (electrolyte accelerates it).
Anode (iron corrodes):
$$\boxed{\text{Fe}_{(s)} \rightarrow \text{Fe}^{2+}_{(aq)} + 2e^-}$$Cathode (neutral/basic, common):
$$\boxed{\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-}$$Cathode (acidic): $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$
Overall rust formation:
$$\boxed{4\text{Fe} + 3\text{O}_2 + 2x\,\text{H}_2\text{O} \rightarrow 2\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}}$$Protection Methods
| Method | Principle | Example |
|---|---|---|
| Galvanization | Zn coating (sacrificial — protects even if scratched) | Iron sheets, pipes |
| Tin plating | Sn barrier only (NOT sacrificial; Fe corrodes if scratched) | Tin cans |
| Sacrificial anode | Attach lower-$E^\circ$ metal (Zn, Mg) → it corrodes | Ships, pipelines |
| Impressed current | External DC forces structure to be cathode | Tanks, offshore rigs |
| Passivation | Adherent, non-porous oxide layer | Al$_2$O$_3$, stainless steel Cr$_2$O$_3$ |
| Alloying / Painting | Resistant elements / barrier coat | Stainless steel / cars |
Zn ($E^\circ = -0.76$ V) is more reactive than Fe ($-0.44$ V), so it sacrifices itself and protects iron even when the coating breaks. Sn ($-0.14$ V) is less reactive, so a scratched tin coating makes Fe the anode and speeds up corrosion.
Corrosion-rate electrode ladder:
$$\text{Mg}(-2.37) < \text{Zn}(-0.76) < \text{Fe}(-0.44) < \text{Sn}(-0.14) < \text{Cu}(+0.34)\ \text{V}$$More negative $\Rightarrow$ corrodes first $\Rightarrow$ good sacrificial anode.