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.

All Electrochemistry notes MHT-CET Chemistry strategy

Why this matters

49 PYQs, 12 HARD — every HARD row in the chapter is here, and all twelve 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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1ElectrochemistryMODERATE

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?

[Q56 · 19 April Shift II · 2025]

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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2ElectrochemistryEASY

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}

[Q56 · 14th May Shift 2 · 2024]

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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3ElectrochemistryEASY

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}

[Q100 · 15th May Shift 2 · 2023]

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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4ElectrochemistryHARD

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

[Q90 · 19 April Shift II · 2025]

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

Concept 5 of 6

Potential of One Electrode at a Given Concentration

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5ElectrochemistryHARD

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 ?

[Q100 · 22 April Shift I · 2025]

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

Intuition

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.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6ElectrochemistryMODERATE

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}

[Q98 · 21 April Shift II · 2025]

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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