MHT-CET Chemistry · Electrochemistry

Galvanic Cells, EMF, Nernst Equation and Thermodynamics

A galvanic cell turns a spontaneous redox reaction into a voltage: E°cell = E°cathode − E°anode from the electrochemical series, corrected for concentration by the Nernst equation, and tied to ΔG° = −nFE° and to K.

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Why this matters

73 PYQs, 10 HARD — and all ten are Nernst: the electrode potential of M → M(n+) at 0.1 or 0.01 M, or how much the emf moves when one ion's concentration drops tenfold. The rest are E°cell subtractions, ΔG° = −nFE° in kJ, E° from K, and recall of which electrode is positive and which species is the strongest reducing or oxidising agent.

Concept 1 of 6: Reading a Cell: Anode Left, Cathode Right

In cell notation the anode (oxidation) is written on the left and the cathode (reduction) on the right, the double bar being the salt bridge. In a GALVANIC cell the anode is the negative electrode and the cathode the positive one; electrons flow through the wire from anode to cathode. The electrode with the higher reduction potential is the cathode.

Definition

$\text{Zn}(s)\,|\,\text{Zn}^{2+}\,\|\,\text{Ag}^+\,|\,\text{Ag}(s)$ : Zn oxidised at the left (anode, −), Ag⁺ reduced at the right (cathode, +). Net: $\text{Zn} + 2\text{Ag}^+ \to \text{Zn}^{2+} + 2\text{Ag}$ — balance the electrons.
A galvanic (voltaic) cell converts CHEMICAL energy to electrical; an electrolytic cell does the reverse. A dry cell is a voltaic cell.
With SHE: whichever has the higher E° is the cathode. Zn/SHE — zinc is the anode, the positive electrode carries $2\text{H}^+ + 2e^- \to \text{H}_2$ . Cu/SHE — copper is the cathode, net $\text{H}_2 + \text{Cu}^{2+} \to 2\text{H}^+ + \text{Cu}$ .
SHE difficulties: pure H₂, exactly 1 bar, exactly 1 M H⁺ — NOT 'running the reaction in reverse', which is easy.

Cell notation

\text{anode (–)}\ \big|\ \text{anode ion}\ \big\|\ \text{cathode ion}\ \big|\ \text{cathode (+)}

Worked example

Write the cell in which

\text{Ni} + 2\text{Ag}^+ \to \text{Ni}^{2+} + 2\text{Ag}

occurs and name the positive electrode.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q56Moderate

Example 1 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

For a Galvanic cell consisting zinc electrode and standard hydrogen electrode,

E^{\circ}\left( Zn_{(aq)}^{+ 2}\mid Zn_{(s)} \right)= - 0.76\text{ }V

Identify the reaction that takes place at positive electrode during working of cell?

Anode = positive

That is true in an ELECTROLYTIC cell only. In a galvanic cell electrons leave the anode, so it is negative; the cathode, where they arrive, is positive.

Drill 4 more on reading a cell: anode left, cathode right

Concept 2 of 6: E°cell = E°cathode − E°anode

Both tabulated values are REDUCTION potentials. The cathode reduces, so its value is used as is; the anode oxidises, so its reduction potential is subtracted. Subtracting a negative number adds — the Zn/Pb cell with two negative E° still has a positive emf.

Definition

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ (reduction potentials). Cd/Ag: $0.799 - (-0.403) = 1.202$ V. Zn/Pb: $-0.126 - (-0.763) = 0.637$ V.
Against SHE ( $E^\circ = 0$ ): Al anode, $0 - (-1.66) = 1.66$ V.
A known cell gives an unknown electrode: Zn/calomel, $1.007 = 0.242 - E^\circ_{\text{Zn}}$ → $E^\circ_{\text{Zn}} = -0.765$ V.
Coefficients do NOT change E°: $2\text{Al} + 3\text{Ni}^{2+}$ still gives $-0.25 - (-1.66) = 1.41$ V.
Positive E°cell means spontaneous as written; negative means the reverse runs.

Standard emf

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

Worked example

E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V. Find E°cell for the spontaneous cell and write it.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q56Easy

Example 2 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

Calculate

E°_{cell}

for

\text{Cd}(s)|\text{Cd}^{2+}(1M)\|\text{Ag}^+(1M)|\text{Ag}(s)

E°_{Cd} = -0.403\,\text{V}

;

E°_{Ag} = 0.799\,\text{V}

Adding the two potentials

E^\circ_{\text{anode}} + E^\circ_{\text{cathode}}

is offered as a 'relation' and as a number. Reduction potentials are SUBTRACTED; for Zn/Cd that gives 0.36 V, not −1.17 V.

Drill 8 more on e°cell = e°cathode − e°anode

Concept 3 of 6: The Electrochemical Series: Who Reduces, Who Oxidises, Who Deposits

Arrange reduction potentials from most negative (Li, K) to most positive (F₂). The bottom of that list are the metals most eager to give electrons — the strongest reducing agents; the top are the species most eager to take them — the strongest oxidising agents. A metal displaces from solution any ion above it.

Definition

Most negative E°: Li⁺/Li (−3.04), K⁺/K (−2.93), Mg (−2.37), Al (−1.66), Zn (−0.76), Fe (−0.44), Sn (−0.14), Pb (−0.13), H (0), Cu (+0.34), Ag (+0.80), Cl₂ (+1.36), F₂ (+2.87).
Strongest reducing agent = most negative E° metal (K among K, Al, Mg, Ag). Strongest oxidising agent = most positive E° species (F₂ over Li, Li⁺, F⁻).
Deposition order: higher E° deposits first: Ag > Cu > Sn > Cd.
Spontaneous displacement: the metal LOWER in the series reduces the ion of one higher. Zn + Cu²⁺ (E° = +1.10 V) yes; Cu + Mg²⁺ no.

Spontaneity test

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0 \iff \text{spontaneous}

Worked example

Will Sn displace Cd²⁺ from solution? E°(Sn) = −0.14, E°(Cd) = −0.40 V.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q100Easy

Example 3 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

What is the decreasing order of deposition of metal on electrode if standard reduction potentials are given as

E^0(\text{Ag}^+/\text{Ag}) = 0.80\text{ V}, E^0(\text{Cu}^{2+}/\text{Cu}) = 0.337\text{ V}, E^0(\text{Sn}^{2+}/\text{Sn}) = -0.136\text{ V}, E^0(\text{Cd}^{2+}/\text{Cd}) = -0.403\text{ V}

Picking F⁻ as the strongest oxidising agent

F⁻ is already reduced — it can only be oxidised. The oxidising agent is the species that GETS reduced: F₂. Likewise Li (the metal), not Li⁺, is the reducing agent.

Drill 4 more on the electrochemical series: who reduces, who oxidises, who deposits

Concept 4 of 6: The Nernst Equation: E = E° − (0.0592/n) log Q

Away from 1 M the emf shifts by (0.0592/n) volts per power of ten in Q, where Q puts product ions over reactant ions with their stoichiometric powers. More product ion lowers E; less product ion (or more reactant ion) raises it. Equal 0.1 M on both sides of a 1:1 reaction gives Q = 1 and no shift at all.

Definition

$E = E^\circ - \dfrac{RT}{nF}\ln Q = E^\circ - \dfrac{0.0592}{n}\log_{10} Q$ at 298 K, $Q = \dfrac{[\text{products}]}{[\text{reactants}]}$ (ions only).
Cd | Cd²⁺ || Cu²⁺ | Cu: $E = E^\circ - 0.0296\log\dfrac{[\text{Cd}^{2+}]}{[\text{Cu}^{2+}]}$ . [Cd²⁺] ten times [Cu²⁺]: E is LOWER by 0.0296 V.
Zn | Zn²⁺(1 M) || Ag⁺ | Ag, n = 2, $Q = [\text{Zn}^{2+}]/[\text{Ag}^+]^2$ : [Zn²⁺] → 0.1 M raises E by 0.0296 V; [Ag⁺] → 0.1 M lowers E by 0.0592 V (the square); [Ag⁺] = 10 M raises it by 0.0592 V.
E lower than E° by 0.0592 V means $\log Q = 2$ : $[\text{Zn}^{2+}] = 1$ , $[\text{Ag}^+] = 0.1$ .
Zn | Zn²⁺(0.1) || Cr³⁺(0.1) | Cr: n = 6, $Q = (0.1)^3/(0.1)^2 = 0.1$ , $E = 0.02 + 0.0592/6 = 0.03$ V.
Hydrogen electrode at 1 atm: $E = -0.0592 \times \text{pH}$ ; pH 1 → −0.0592 V.

Nernst equation (298 K)

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n}\log_{10} Q

Worked example

For Ni | Ni²⁺(0.01 M) || Cu²⁺(1 M) | Cu with E° = 0.59 V, find E at 298 K.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q90Hard

Example 4 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

For the cell,

\ ^{\ominus}Zn_{(s)}\left| Zn^{+ 2}(lM) \right|\left| Ag^{+ 1}(lM) \right|Ag_{(s)}\ ^{\oplus}

If concentration of

Zn^{+ 2}

decreases to 0.1 M at 298 K , then emf of cell

Forgetting the square on [Ag⁺]

Two Ag⁺ per Zn, so Q carries

[\text{Ag}^+]^2

. Dropping Ag⁺ to 0.1 M multiplies Q by 100, a two-decade shift of 0.0592 V — twice the 0.0296 V that dropping Zn²⁺ gives.

Drill 9 more on the nernst equation: e = e° − (0.0592/n) log q

Concept 5 of 6: Potential of One Electrode at a Given Concentration

For the OXIDATION M → M(n+)(c) + ne⁻, the standard oxidation potential is the negative of the tabulated reduction value, and the Nernst term is −(0.0592/n) log c. Since c < 1 makes the log negative, the oxidation potential comes out MORE positive than E°ox by (0.0592/n) per decade. Doubling the equation changes nothing — potential is intensive.

Definition

$E_{\text{ox}} = E^\circ_{\text{ox}} - \dfrac{0.0592}{n}\log[\text{M}^{n+}]$ , with $E^\circ_{\text{ox}} = -E^\circ_{\text{red}}$ .
Mg → Mg²⁺(0.01 M): $+2.37 + 0.0592 = +2.4292$ V. At 0.1 M: $+2.37 + 0.0296 = +2.3996$ V.
Zn → Zn²⁺(0.01): $+0.76 + 0.0592 = +0.8192$ V. Al → Al³⁺(0.1): $+1.66 + 0.0197 = +1.679$ V.
Metals with POSITIVE E°red: Cu → Cu²⁺(0.1): $-0.34 + 0.0296 = -0.3104$ V. Ag → Ag⁺(0.01): $-0.80 + 0.1184 = -0.6816$ V.
Intensive: $2\text{Cu} \to 2\text{Cu}^{2+} + 4e^-$ has $E^\circ = -0.34$ V, the same as for one Cu. $2\text{Zn} \to 2\text{Zn}^{2+} + 4e^-$ : $+0.76$ V.

Oxidation electrode potential

E_{\text{ox}} = -E^\circ_{\text{red}} - \frac{0.0592}{n}\log[\text{M}^{n+}]

Worked example

E°(Ni²⁺/Ni) = −0.25 V. Find the potential for Ni → Ni²⁺(0.001 M) + 2e⁻ at 298 K.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q100Hard

Example 5 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

E^{\circ}\left( Mg_{(aq)}^{+ 2}\mid Mg_{(s)} \right)= - 2.37\text{ }V

. What is potential for

Mg_{(s)}\longrightarrow Mg^{+ 2}(0.01M) + 2e^{-}

at 298 K ?

Doubling E° when the equation is doubled

Electrode potential is intensive — it does not scale with the amount. 2Zn → 2Zn²⁺ + 4e⁻ is still +0.76 V; +1.52 V and −1.52 V are the planted options.

Drill 7 more on potential of one electrode at a given concentration

Concept 6 of 6: ΔG° = −nFE° and the Bridge to K

The electrical work a cell can do is the fall in Gibbs energy: ΔG° = −nFE°, in joules when F = 96500 and E in volts. At equilibrium E = 0, which turns the Nernst equation into E° = (0.0592/n) log K. E° is intensive; ΔG°, carrying n, is extensive.

Definition

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$ . Mg/Sn, E° = 2.23: $-2 \times 96500 \times 2.23 = -430.4$ kJ. Sn/Ag, 0.90 V: −173.7 kJ. Zn/Ni, 0.5 V: −96.5 kJ.
Reverse: $E^\circ = \dfrac{-\Delta G^\circ}{nF}$ . 2Al + 3Cu²⁺, ΔG° = −1158 kJ, n = 6: $1158000/(6 \times 96500) = 2$ V. A + B²⁺, −386 kJ, n = 2: 2 V. Dry cell: n = 2, $E^\circ = -\Delta G^\circ/2F$ .
$E^\circ = \dfrac{0.0592}{n}\log_{10} K$ . K = 10⁴, n = 2: $0.0296 \times 4 = 0.1184$ V. (Not $\dfrac{0.0592}{nF}$ , not $\ln K$ with 0.0592.)
Maximum electrical work $= -\Delta G^\circ = nFE^\circ$ ; work done BY the cell is reported negative: Zn/Ag at 1.55 V, −299.15 kJ.
Intensive: E°cell, electrode potential. Extensive: ΔG°. Electrode potential DOES depend on concentration.

Gibbs energy and K

\Delta G^\circ = -nFE^\circ,\qquad E^\circ = \frac{0.0592}{n}\log_{10} K

Worked example

For Cu | Cu²⁺ || Ag⁺ | Ag, E° = 0.46 V. Find ΔG° and K at 298 K.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q98Moderate

Example 6 · Electrochemistry · Galvanic Cells, EMF, Nernst Equation and Thermodynamics

For cell reaction,

2Al_{(s)}+ 3Cu^{+ 2}\ _{(aq)}\longrightarrow 2Al^{+ 3}\ _{(aq)}+ 3Cu_{(s)}

G^{\circ}= - 1158\text{ }kJ

, what is

E_{\text{cell~}}^{\circ}

Losing the minus sign

E^\circ = \Delta G^\circ/nF

without the minus is the planted FALSE relation, and −ΔG° written as ΔG° flips the sign of a spontaneous cell's 'work'. A positive E° always goes with a negative ΔG°.

Drill 9 more on δg° = −nfe° and the bridge to k

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 (6)

Reading a Cell: Anode Left, Cathode Right
Cell notation
$\text{anode (–)}\ \big|\ \text{anode ion}\ \big\|\ \text{cathode ion}\ \big|\ \text{cathode (+)}$
E°cell = E°cathode − E°anode
Standard emf
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
The Electrochemical Series: Who Reduces, Who Oxidises, Who Deposits
Spontaneity test
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} > 0 \iff \text{spontaneous}$
The Nernst Equation: E = E° − (0.0592/n) log Q
Nernst equation (298 K)
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n}\log_{10} Q$
Potential of One Electrode at a Given Concentration
Oxidation electrode potential
$E_{\text{ox}} = -E^\circ_{\text{red}} - \frac{0.0592}{n}\log[\text{M}^{n+}]$
ΔG° = −nFE° and the Bridge to K
Gibbs energy and K
$\Delta G^\circ = -nFE^\circ,\qquad E^\circ = \frac{0.0592}{n}\log_{10} K$

Watch out for (6)

Anode = positive
→ Reading a Cell: Anode Left, Cathode Right
Adding the two potentials
→ E°cell = E°cathode − E°anode
Picking F⁻ as the strongest oxidising agent
→ The Electrochemical Series: Who Reduces, Who Oxidises, Who Deposits
Forgetting the square on [Ag⁺]
→ The Nernst Equation: E = E° − (0.0592/n) log Q
Doubling E° when the equation is doubled
→ Potential of One Electrode at a Given Concentration
Losing the minus sign
→ ΔG° = −nFE° and the Bridge to K

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