Redox Reactions & Electrochemistry Formula Sheet — JEE Main Chemistry

Every key Redox Reactions & Electrochemistry formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 6 sub-topics

Concept of oxidation and reduction
Redox reactions
Oxidation number
Balancing redox reactions
In terms of loss and gain of electron and change in oxidation number
Applications of redox reactions

Redox Reactions & Oxidation Number

Oxidation & reduction (always together)

▸ $Oxidation$ = loss of electrons / increase in oxidation number (O.N.).
▸ $Reduction$ = gain of electrons / decrease in O.N.
▸A $redox$ reaction couples the two — electrons lost = electrons gained.
▸ $Oxidising agent$ (oxidant) is itself reduced; $reducing agent$ (reductant) is itself oxidised.

Type of redox reaction	Example
Combination	$C + O_{2} \to C O_{2}$
Decomposition	$2KCl O_{3} \to 2KCl + 3 O_{2}$
Displacement	$Zn + CuS O_{4} \to ZnS O_{4} + Cu$
Disproportionation	$2C u^{+} \to C u^{2 +} + Cu$ (same element oxidised & reduced)

Redox couple & agents: In

Zn + Cu^{2 +} \to Zn^{2 +} + Cu

Zn

is oxidised (reducing agent),

Cu^{2 +}

is reduced (oxidising agent). The pair

Zn^{2 +} / Zn

is a

redox couple

Oxidation-number rules (in order)

▸Free element (e.g. $O_{2}$ , $Na$ ) $= 0$ ; monatomic ion $=$ its charge.
▸Group 1 $= + 1$ , Group 2 $= + 2$ ; F $= - 1$ always.
▸H $= + 1$ (but $- 1$ in metal hydrides like $NaH$ ).
▸O $= - 2$ (but $- 1$ in peroxides, $+ 2$ in $O F_{2}$ ).
▸ $\sum$ O.N. $= 0$ in a neutral molecule, $=$ charge in an ion.

💡 Tip · Reading O.N. fast

S in

H_{2} S O_{4}

2 (+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 6

. Mn in

KMn O_{4} = + 7

; Cr in

K_{2} C r_{2} O_{7} = + 6

. These high-O.N. species are strong oxidants.

🎯 Exam · Balancing redox equations

Half-reaction (ion-electron) method

: split into oxidation & reduction halves; balance atoms, then O with

H_{2} O

, H with

H^{+}

(add

O H^{-}

in base), balance charge with

e^{-}

; multiply so

e^{-}

cancel and add.

n-factor

= total change in O.N.; equivalent mass

= \frac{molar mass}{n -factor}

🚫 Examiner Trap · Examiner traps

(1) Oxidation

=

LOSS of e

^{-}

(O.N.

↑

); the oxidising agent is itself REDUCED. (2) O is

- 2

except peroxides (

- 1

) and O

F_{2}

(

+ 2

); H is

+ 1

except metal hydrides (

- 1

). (3)

Disproportionation

: same element both oxidised AND reduced (Cu

^{+}

, C

l_{2}

in base). (4) n-factor

=

total O.N. change; equiv. mass

= M / n

-factor.

Galvanic Cells & Electrode Potential

Galvanic (voltaic) cell: Converts the Gibbs energy of a

spontaneous

redox reaction into electrical energy. Two half-cells (each a metal in its ion solution) are joined externally by a wire and internally by a salt bridge.

Oxidation at anode

(

-

reduction at cathode

(

+

Daniell cell: zinc anode in zinc sulphate and copper cathode in copper sulphate joined by a salt bridge and an external voltmeter, with half-reactions and cell notation giving 1.10 V — Daniell cell — electrons flow Zn (anode) to Cu (cathode); $E_{ce ll}^{\circ} = 1.10$ V.

Conventions

▸Electrons flow anode $\to$ cathode in the wire; conventional current is opposite.
▸ $Salt bridge$ (e.g. KCl in agar) completes the circuit and keeps each solution electrically neutral.
▸Cell notation: anode $∣$ anode soln $∥$ cathode soln $∣$ cathode (left = oxidation).

Standard electrode potential (

E^{\circ}

): The potential of an electrode (as a

reduction

, IUPAC) measured against the standard hydrogen electrode when all species are at unit concentration/activity, gases at 1 bar, 298 K.

★ Remember · Standard Hydrogen Electrode (SHE)

Pt (s) ∣ H_{2} (g, 1 bar) ∣ H^{+} (aq, 1 M)

is the reference, assigned

E^{\circ} = 0.00

V at all temperatures for

H^{+} + e^{-} \to \frac{1}{2} H_{2}

Standard cell potential

E_{ce ll}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = E_{r i g h t}^{\circ} - E_{l e f t}^{\circ}

Both are reduction potentials.

E_{ce ll}^{\circ} > 0

means the cell reaction is spontaneous as written.

Vertical electrochemical series of standard reduction potentials from fluorine at plus 2.87 volts down to lithium at minus 3.05 volts, with SHE at zero, oxidising power increasing upward and reducing power increasing downward — Electrochemical series: higher $E^{\circ}$ = stronger oxidant; lower = stronger reductant.

🚫 Examiner Trap · Examiner traps

(1) In a

galvanic

cell anode is

-

, cathode

+

; oxidation is ALWAYS at the anode (sign flips in electrolytic). (2)

E_{ce ll}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}

, both as

reduction

potentials. (3) SHE

= 0.00

V by definition. (4) Higher

E^{\circ}

\Rightarrow

stronger oxidant; the salt bridge keeps solutions neutral, not the current path metal.

Nernst Equation, ΔG & Equilibrium

Nernst equation (298 K)

E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.059}{n} lo g Q

General form

E_{ce ll} = E_{ce ll}^{\circ} - \frac{RT}{n F} ln Q

;

n =

electrons transferred,

Q =

reaction quotient. Increasing product-ion or decreasing reactant-ion concentration lowers

E_{ce ll}

✎ Example · Worked Nernst calculation

For

Mg ∣ M g^{2 +} (0.130 M) ∥ A g^{+} (1.0 \times 1 0^{- 4} M) ∣ Ag

with

E_{ce ll}^{\circ} = 3.17

V, find

E_{ce ll}

$n = 2$ ; $Q = \frac{[ M g ^{2 +} ]}{[ A g ^{+} ] ^{2}} = \frac{0.130}{( 1 0 ^{- 4} ) ^{2}}$ .
$E_{ce ll} = 3.17 - \frac{0.059}{2} lo g \frac{0.130}{( 1 0 ^{- 4} ) ^{2}} = 3.17 - 0.21$ .

∴

E_{ce ll} = 2.96 V

Gibbs energy

Δ_{r} G = - nFE_{ce ll}

Standard

Δ_{r} G^{\circ} = - nFE_{ce ll}^{\circ}

Equilibrium

Δ_{r} G^{\circ} = - RT ln K

From

E^{\circ}

lo g K_{c} = \frac{n E _{ce ll}^{\circ}}{0.059}

🎯 Exam · Spontaneity at a glance

E_{ce ll}^{\circ} > 0 \Leftrightarrow Δ_{r} G^{\circ} < 0 \Leftrightarrow K > 1

(reaction favours products).

F = 96500

C mo

l^{- 1}

; use

Δ_{r} G^{\circ}

in joules (

= - nFE^{\circ}

with

E^{\circ}

in volts).

Concentration cell

▸Same electrodes, different ion concentrations: $E_{ce ll}^{\circ} = 0$ .
▸ $E_{ce ll} = \frac{0.059}{n} lo g \frac{[ conc. ]}{[ dil. ]}$ — EMF arises purely from the concentration difference, and falls to zero as concentrations equalise.

🚫 Examiner Trap · Examiner traps

(1) Nernst:

E = E^{\circ} - \frac{0.059}{n} lo g Q

— get n and Q right (products/reactants, powers

=

coefficients). (2)

Δ_{r} G = - nFE_{ce ll}

(use joules, E in volts). (3)

E_{ce ll}^{\circ} > 0 \Leftrightarrow Δ_{r} G^{\circ} < 0 \Leftrightarrow K > 1

. (4) A concentration cell has

E_{ce ll}^{\circ} = 0

but

E_{ce ll} \neq = 0

until concentrations equalise.

Conductance & Molar Conductivity

Conductivity (

κ

): Reciprocal of resistivity: the conductance of a 1 m (or 1 cm) cube of solution. From

R = ρ \frac{l}{A}

κ = \frac{1}{ρ} = \frac{1}{R} \cdot \frac{l}{A}

. SI unit

S m^{- 1}

; the factor

l / A

is the

cell constant

G^{*}

Molar conductivity (

Λ_{m}

): Conductance of all the ions from

one mole

of electrolyte between electrodes 1 unit apart:

Λ_{m} = \frac{κ}{c}

. Rises as concentration falls (ions occupy more volume).

Plot of molar conductivity versus square root of concentration showing a nearly straight line for strong KCl extrapolating to its limiting value and a steeply rising curve for weak acetic acid that cannot be extrapolated, plus the key conductivity relations — $Λ_{m}$ vs $c$ : linear for strong, steeply rising for weak electrolytes.

Cell constant & conductivity

κ = G^{*} \times G = \frac{G ^{*}}{R} Λ_{m} = \frac{κ \times 1000}{c}

c in mol

L^{- 1}

gives

Λ_{m}

in S c

m^{2}

l^{- 1}

G^{*} = l / A

is found from a standard KCl solution of known

κ

Quantity	SI unit	Common unit
Conductivity $κ$	$S m^{- 1}$	$S c m^{- 1}$
Molar cond. $Λ_{m}$	$S m^{2} mo l^{- 1}$	$S c m^{2} mo l^{- 1}$
Conversion	$1 S m^{2} mo l^{- 1}$	$= 1 0^{4} S c m^{2} mo l^{- 1}$

Variation with concentration

▸ $κ$ $always decreases$ on dilution (fewer ions per unit volume).
▸ $Λ_{m}$ $increases$ on dilution for both types.
▸ $Strong$ electrolytes: $Λ_{m} = Λ_{m}^{\circ} - A c$ (gentle, linear in $c$ ).
▸ $Weak$ electrolytes: $Λ_{m}$ rises steeply near $c \to 0$ , so $Λ_{m}^{\circ}$ cannot be found by extrapolation.

🚫 Examiner Trap · Examiner traps

(1) On dilution

κ

DECREASES (fewer ions/volume) but

Λ_{m}

INCREASES — opposite directions. (2)

Λ_{m} = κ \times 1000/ c

with c in mol/L gives S c

m^{2}

/mol. (3) Strong electrolytes:

Λ_{m} = Λ_{m}^{\circ} - A c

(extrapolate to

Λ_{m}^{\circ}

); weak: rises steeply, CANNOT extrapolate. (4) Find the cell constant from a standard KCl solution.

Kohlrausch's Law & Its Applications

Kohlrausch's law of independent migration: At infinite dilution each ion migrates independently, so the limiting molar conductivity of an electrolyte is the sum of the limiting contributions of its ions.

Limiting molar conductivity

Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}

ν_{+}, ν_{-}

= number of cations/anions per formula unit;

λ^{\circ}

= limiting molar conductivity of each ion. E.g.

Λ_{m}^{\circ} (CaC l_{2}) = λ_{C a^{2 +}}^{\circ} + 2 λ_{C l^{-}}^{\circ}

Cation	$λ^{\circ}$	Anion	$λ^{\circ}$
$H^{+}$	349.6	$O H^{-}$	199.1
$N a^{+}$	50.1	$C l^{-}$	76.3
$K^{+}$	73.5	$B r^{-}$	78.1
$C a^{2 +}$	119.0	$C H_{3} CO O^{-}$	40.9
$M g^{2 +}$	106.0	$S O_{4}^{2 -}$	160.0

★ Remember · Why

H^{+}

O H^{-}

are huge

H^{+}

(349.6) and

O H^{-}

(199.1) conduct far better than other ions via the

Grotthuss

proton-hopping mechanism, not bodily migration. Units: S c

m^{2}

l^{- 1}

Applications

▸ $Λ_{m}^{\circ}$ of a $weak$ electrolyte (unreachable by extrapolation) is built from strong-electrolyte data.
▸ $Degree of dissociation$ : $α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$ .
▸ $Dissociation constant$ : $K_{a} = \frac{c α ^{2}}{1 - α} = \frac{c Λ _{m}^{2}}{Λ _{m}^{\circ} ( Λ _{m}^{\circ} - Λ _{m} )}$ .

✎ Example ·

Λ_{m}^{\circ}

of a weak acid

Given

Λ_{m}^{\circ}

: HCl

= 425.9

, NaAc

= 91.0

, NaCl

= 126.4

S c

m^{2}

l^{- 1}

. Find

Λ_{m}^{\circ}

of acetic acid (HAc).

$Λ_{m}^{\circ} (HAc) = Λ_{m}^{\circ} (HCl) + Λ_{m}^{\circ} (NaAc) - Λ_{m}^{\circ} (NaCl)$ .
$= 425.9 + 91.0 - 126.4$ .

∴

Λ_{m}^{\circ} (HAc) = 390.5 S c m^{2} mo l^{- 1}

💡 Tip · Then get

α

and

K_{a}

Λ_{m} = 48.15

at some c, then

α = \frac{48.15}{390.5} = 0.123

; feed

α

and c into

K_{a} = \frac{c α ^{2}}{1 - α}

🚫 Examiner Trap · Examiner traps

(1)

Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ}

— multiply by the number of each ion (CaC

l_{2}

has

2

^{-}

). (2) Get a WEAK electrolyte's

Λ_{m}^{\circ}

from strong-electrolyte data (HAc

=

HCl

+

NaAc

-

NaCl). (3)

α = Λ_{m} / Λ_{m}^{\circ}

. (4) H

^{+}

/OH

^{-}

conduct anomalously high (Grotthuss hopping).

Electrolysis & Faraday's Laws

Electrolytic cell: An external DC source drives a

non-spontaneous

redox reaction (

E_{ce ll}^{\circ} < 0

). The

anode

is now

+

(oxidation) and the

cathode

-

(reduction) — opposite signs to a galvanic cell, but oxidation is still at the anode.

Faraday's first law

m = Z I t = \frac{M}{n F} Q, Q = I t

Mass deposited

\propto

charge passed.

Z = M / n F

is the electrochemical equivalent; n = electrons per ion;

F = 96500

C mo

l^{- 1}

Faraday's second law

\frac{m _{1}}{m _{2}} = \frac{E _{1}}{E _{2}} (same charge)

For the same Q, masses liberated are in the ratio of equivalent masses

E = M / n

. One mole of electrons (

1 F

) deposits one

gram-equivalent

Products of electrolysis

▸Depend on the species present and their $E^{\circ}$ (and on $overpotential$ ).
▸ $Molten$ NaCl $\to$ Na (cathode) $+$ $C l_{2}$ (anode).
▸ $Aqueous$ NaCl $\to$ $H_{2}$ (cathode) $+$ $C l_{2}$ (anode, due to overpotential) $+$ NaOH.
▸Used industrially: refining of Cu, extraction of Na/Mg/Al.

✎ Example · Mass deposited

CuS O_{4}

is electrolysed for 10 min at 1.5 A. Mass of Cu deposited? (

C u^{2 +} + 2 e^{-} \to Cu

M = 63

)

$Q = I t = 1.5 \times 600 = 900$ C.
$2 F$ (= $2 \times 96500$ C) deposit 63 g, so $m = \frac{63 \times 900}{2 \times 96500}$ .

∴

m = 0.294 g of Cu

🚫 Examiner Trap · Examiner traps

(1) In an

electrolytic

cell anode is

+

, cathode

-

(OPPOSITE to galvanic) — but oxidation is still at the anode. (2)

m = MQ / n F

;

Q = I t

F = 96500

. (3)

Aqueous

NaCl gives

H_{2}

(not Na) at the cathode and C

l_{2}

(overpotential) at the anode;

molten

NaCl gives Na. (4)

1 F

deposits one gram-equivalent.

Batteries, Fuel Cells & Corrosion

Two kinds of batteries

▸ $Primary$ : reaction occurs once, cannot be recharged (dry cell, mercury cell).
▸ $Secondary$ : reversible — recharged by passing current backwards (lead storage, Ni-Cd).
▸A battery is one or more galvanic cells in series, made light, compact and steady in voltage.

Three panels: a Leclanche dry cell with zinc case anode and carbon rod cathode at 1.5 volts, a lead storage battery with Pb and PbO2 plates in sulphuric acid at 2 volts per cell, and a hydrogen-oxygen fuel cell about 70 percent efficient — Dry cell (primary), lead storage (secondary) and $H_{2}$ - $O_{2}$ fuel cell.

Lead anode

Pb + S O_{4}^{2 -} \to PbS O_{4} + 2 e^{-}

Lead cathode

Pb O_{2} + S O_{4}^{2 -} + 4 H^{+} + 2 e^{-} \to PbS O_{4} + 2 H_{2} O

Fuel cell

2 H_{2} + O_{2} \to 2 H_{2} O

Cell	Type	EMF
Dry (Leclanche)	primary	$\sim 1.5$ V
Mercury cell	primary	$1.35$ V (steady)
Lead storage	secondary	$\sim 2$ V/cell
$H_{2}$ - $O_{2}$ fuel	continuous	$\sim 70%$ eff.

Corrosion (rusting of iron): An

electrochemical

process: a moist iron surface acts as tiny galvanic cells. Anode:

Fe \to F e^{2 +} + 2 e^{-}

; cathode:

O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O

F e^{2 +}

is further oxidised to hydrated rust

F e_{2} O_{3} \cdot x H_{2} O

🎯 Exam · Preventing corrosion

Barrier

(paint, oil, grease);

galvanising

(Zn coat);

sacrificial anode

/ cathodic protection — attach a more active metal (Mg, Zn) that corrodes preferentially and saves the object.

🚫 Examiner Trap · Examiner traps

(1)

Primary

cells can't be recharged (dry, mercury);

secondary

can (lead-acid, Ni-Cd). (2) Lead storage uses Pb/Pb

O_{2}

H_{2} S O_{4}

(

\sim 2

V/cell). (3) Corrosion is

electrochemical

(mini galvanic cells) — needs both

O_{2}

and water. (4) Sacrificial anode (Mg/Zn) protects by corroding FIRST (cathodic protection).

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