JEE Mains Chemistry · Electrochemistry

Galvanic Cells and Electrode Potentials

Reading a table of standard reduction potentials: which couple is reduced, which species is the stronger oxidising or reducing agent, and how a cell is set up and written.

All Electrochemistry notes JEE Mains Chemistry strategy

Why this matters

Fourteen PYQs, eleven of them multiple choice, and two from 2026. Six hand you a list of E° values and ask you to rank oxidising or reducing agents. Eight ask how a cell is built: which electrode is the anode, what E°cell is, and which cell runs a given reaction. Two ideas cover the page.

Concept 1 of 2: The electrochemical series

A standard reduction potential measures how badly a couple wants to be reduced. The higher the

E^\circ

, the stronger the oxidised form is as an oxidising agent. The lower (more negative) the

E^\circ

, the stronger the reduced form is as a reducing agent. Every ranking question is this one table, read up or down.

Definition

Each couple is written as a reduction: oxidised form $+\,ne^-\to$ reduced form.
Higher $E^\circ$ : the oxidised form is a stronger oxidising agent ( $\mathrm{F_2}$ , $\mathrm{MnO_4^-}$ , $\mathrm{Cl_2}$ ).
Lower $E^\circ$ : the reduced form is a stronger reducing agent ( $\mathrm{Li}$ , $\mathrm{Na}$ , $\mathrm{Mg}$ ).
An oxidant oxidises every reduced form whose couple lies BELOW it in the table.
For a couple like $\mathrm{Fe^{3+}/Fe^{2+}}$ , the reducing agent is the reduced form, $\mathrm{Fe^{2+}}$ , not the metal.
$E^\circ(\mathrm{M^+/M})$ is built from a cycle: sublimation of the solid, ionisation of the GASEOUS atom, and hydration of the gaseous ion. Ionisation of the solid is not a step.
$\mathrm{Li^+/Li}$ sits below $\mathrm{Na^+/Na}$ because the small $\mathrm{Li^+}$ ion has a very large hydration energy.

Couple	E° at 298 K (V)	What it tells you
$\mathrm{Li^+/Li}$	$-3.05$	Li is the strongest reducing agent in water
$\mathrm{Na^+/Na}$	$-2.71$	Na ionises more easily than Li, yet its E° is higher
$\mathrm{Mg^{2+}/Mg}$	$-2.37$	Mg displaces almost every metal ion from water
$\mathrm{Al^{3+}/Al}$	$-1.66$	Al is a strong reducing agent
$\mathrm{Zn^{2+}/Zn}$	$-0.76$	Zn is the anode of the Daniell cell
$\mathrm{Cr^{3+}/Cr}$	$-0.74$	Cr is a reducing agent close to Zn
$\mathrm{Fe^{2+}/Fe}$	$-0.44$	Fe dissolves in dilute acid and gives H₂
$\mathrm{H^+/\tfrac12 H_2}$	$0.00$	The zero of the scale, by definition
$\mathrm{Cu^{2+}/Cu}$	$+0.34$	Cu does not release H₂ from dilute acid
$\mathrm{I_2/I^-}$	$+0.54$	I⁻ is a fairly good reducing agent
$\mathrm{Fe^{3+}/Fe^{2+}}$	$+0.77$	Fe³⁺ oxidises I⁻ to I₂
$\mathrm{Ag^+/Ag}$	$+0.80$	Ag is oxidised by nitric acid
$\mathrm{NO_3^-/NO}$	$+0.97$	Nitrate in acid oxidises Ag but not Au
$\mathrm{Cr_2O_7^{2-}/Cr^{3+}}$	$+1.33$	Dichromate in acid oxidises Ag and Fe²⁺
$\mathrm{Cl_2/Cl^-}$	$+1.36$	Cl₂ oxidises Br⁻ and I⁻
$\mathrm{Au^{3+}/Au}$	$+1.40$	Au resists every common oxidant here
$\mathrm{MnO_4^-/Mn^{2+}}$	$+1.51$	Permanganate in acid oxidises Cl⁻
$\mathrm{F_2/F^-}$	$+2.87$	F₂ is the strongest oxidising agent

Read down the table for stronger oxidising agents (the left-hand species); read up for stronger reducing agents (the right-hand species).

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q42Moderate

Example 1 · Electrochemistry · Galvanic Cells and Electrode Potentials

Consider the following reduction processes :

\begin{matrix} Al^{3+} + 3e^{-} \rightarrow Al(s), & E^{\circ} = -1.66\ V \\ Fe^{3+} + e^{-} \rightarrow Fe^{2+}, & E^{\circ} = +0.77\ V \\ Co^{3+} + e^{-} \rightarrow Co^{2+}, & E^{\circ} = +1.81\ V \\ Cr^{3+} + 3e^{-} \rightarrow Cr(s), & E^{\circ} = -0.74\ V \end{matrix}

The tendency to act as reducing agent decreases in the order :

Picking the oxidised form as the reducing agent

From

\mathrm{Co^{3+}/Co^{2+}}

the candidate reducing agent is

\mathrm{Co^{2+}}

, not

\mathrm{Co^{3+}}

. Always read the right-hand species of a reduction couple when ranking reducing agents.

Reading the sign backwards

A more negative

E^\circ

means a STRONGER reducing agent, not a weaker one. Lithium, at

-3.05

V, is the strongest of all.

Drill 5 more on the electrochemical series

Concept 2 of 2: Setting up a galvanic cell

Put two couples together and the one with the higher

E^\circ

wins the electrons. It is reduced at the cathode; the other couple runs backwards and is oxidised at the anode. The cell voltage is the gap between the two potentials, never their sum.

Definition

Anode: oxidation, the negative terminal, written on the LEFT. Cathode: reduction, the positive terminal, written on the RIGHT.
Notation: anode $|$ anode solution $\|$ cathode solution $|$ cathode. The double line is the salt bridge.
$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ , both taken as REDUCTION potentials.
$E^\circ$ is intensive: multiplying a half-reaction by 2 does NOT double its $E^\circ$ .
A negative $E^\circ_{cell}$ for a reaction as written means the REVERSE reaction is spontaneous.
Apply an external potential GREATER than $E_{cell}$ , in opposition, and the current reverses: the cell becomes electrolytic.
Electrode types: gas electrode ( $\mathrm{Pt|H_2|H^+}$ ), metal–metal ion ( $\mathrm{Zn|Zn^{2+}}$ ), redox ( $\mathrm{Pt|Fe^{3+},Fe^{2+}}$ ), and metal–insoluble salt–anion (calomel $\mathrm{Hg|Hg_2Cl_2|Cl^-}$ ; $\mathrm{Ag|AgCl|Cl^-}$ ).
A good reference electrode keeps its potential steady, so it has the smallest $(\partial E/\partial T)_P$ .
To choose the cell for a reaction, check that its electrolyte supplies every ion in the equation.

Standard cell potential

E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}

$E^\circ_{cathode}$ reduction potential of the couple that is reduced (higher)
$E^\circ_{anode}$ reduction potential of the couple that is oxidised (lower)

Worked example

A cell is made from

\mathrm{Ni^{2+}/Ni}

(

E^\circ=-0.25

V) and

\mathrm{Ag^+/Ag}

(

E^\circ=+0.80

V). Name the anode, write the cell and its reaction, and find

E^\circ_{cell}

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q32Moderate

Example 2 · Electrochemistry · Galvanic Cells and Electrode Potentials

An electrochemical cell is constructed using half cells in the direction of spontaneous change:

Fe(OH)_{2}(\text{ }s) + 2e^{-}\rightarrow Fe(s) + 2OH^{-}(aq)\ E^{0}= - 0.88\text{ }V

and

AgBr(s) +e^{-}\rightarrow Ag(s) +Br^{-}(aq)\ E^{0}= + 0.07\text{ }V

Which of the following option is correct?

Adding the two potentials

E^\circ_{cell}

is a difference of two reduction potentials. Flipping the anode's sign and then adding gives the same number, but adding the two tabulated values does not.

Scaling E° with the equation

Balancing electrons may double a half-reaction. Its

\Delta G^\circ

doubles; its

E^\circ

does not, because

E^\circ

is intensive.

Drill 7 more on setting up a galvanic cell

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Setting up a galvanic cell
Standard cell potential
$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$

Reference tables (1)

The electrochemical series18 rows

Couple	E° at 298 K (V)	What it tells you
$\mathrm{Li^+/Li}$	$-3.05$	Li is the strongest reducing agent in water
$\mathrm{Na^+/Na}$	$-2.71$	Na ionises more easily than Li, yet its E° is higher
$\mathrm{Mg^{2+}/Mg}$	$-2.37$	Mg displaces almost every metal ion from water
$\mathrm{Al^{3+}/Al}$	$-1.66$	Al is a strong reducing agent
$\mathrm{Zn^{2+}/Zn}$	$-0.76$	Zn is the anode of the Daniell cell
$\mathrm{Cr^{3+}/Cr}$	$-0.74$	Cr is a reducing agent close to Zn
$\mathrm{Fe^{2+}/Fe}$	$-0.44$	Fe dissolves in dilute acid and gives H₂
$\mathrm{H^+/\tfrac12 H_2}$	$0.00$	The zero of the scale, by definition
$\mathrm{Cu^{2+}/Cu}$	$+0.34$	Cu does not release H₂ from dilute acid
$\mathrm{I_2/I^-}$	$+0.54$	I⁻ is a fairly good reducing agent
$\mathrm{Fe^{3+}/Fe^{2+}}$	$+0.77$	Fe³⁺ oxidises I⁻ to I₂
$\mathrm{Ag^+/Ag}$	$+0.80$	Ag is oxidised by nitric acid
$\mathrm{NO_3^-/NO}$	$+0.97$	Nitrate in acid oxidises Ag but not Au
$\mathrm{Cr_2O_7^{2-}/Cr^{3+}}$	$+1.33$	Dichromate in acid oxidises Ag and Fe²⁺
$\mathrm{Cl_2/Cl^-}$	$+1.36$	Cl₂ oxidises Br⁻ and I⁻
$\mathrm{Au^{3+}/Au}$	$+1.40$	Au resists every common oxidant here
$\mathrm{MnO_4^-/Mn^{2+}}$	$+1.51$	Permanganate in acid oxidises Cl⁻
$\mathrm{F_2/F^-}$	$+2.87$	F₂ is the strongest oxidising agent

Read down the table for stronger oxidising agents (the left-hand species); read up for stronger reducing agents (the right-hand species).

Watch out for (4)

Picking the oxidised form as the reducing agent
→ The electrochemical series
Reading the sign backwards
→ The electrochemical series
Adding the two potentials
→ Setting up a galvanic cell
Scaling E° with the equation
→ Setting up a galvanic cell

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