Electrochemical Cells and Cell Potential

Spec Mapping — OCR H432 Module 5.2.3 — Electrochemical cells and cell potential, covering the calculation of standard cell potential from two half-cell potentials via $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$Ecell∘​=Ecathode∘​−Eanode∘​ (where the cathode is the more positive electrode); the construction of overall ionic equations by combining the cathode reduction half-equation with the reversed anode half-equation, balancing electrons by scaling; the prediction of redox feasibility from the sign of $E^\circ_{cell}$Ecell∘​ (positive = thermodynamically feasible); the role of the salt bridge in carrying ionic current to balance charge; the bridge between electrochemistry and thermodynamics via $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​ with the Faraday constant $F = 96\,485$F=96485 C mol$^{-1}$−1; qualitative use of Le Chatelier on half-cell equilibria to predict the effect of concentration changes on $E$E; the principle and operation of hydrogen–oxygen fuel cells (alkaline KOH and acidic PEM variants) with overall $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \to \text{H}_2\text{O}$H2​+21​O2​→H2​O, $E^\circ_{cell} \approx +1.23$Ecell∘​≈+1.23 V; awareness of methanol fuel cells as a portable alternative; the distinction between thermodynamic feasibility and kinetic accessibility; and applications to rechargeable Li-ion and lead-acid batteries (refer to the official OCR H432 specification document for exact wording).

Lesson 8 built the vocabulary — the SHE, the standard electrode potential, the IUPAC cell notation, the electrochemical series. This lesson turns vocabulary into arithmetic: how to combine two $E^\circ$E∘ values into a single $E^\circ_{cell}$Ecell∘​, how to construct the overall ionic equation, how to use $E^\circ_{cell}$Ecell∘​ to predict feasibility, and how to bridge to the Gibbs framework of Lesson 7 via $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​. Six worked examples walk through the Daniell cell, magnesium/copper, copper/silver, acidified MnO$_4^-$4−​/Fe$^{2+}$2+ (the Lesson 10 redox-titration setup), Cl$_2$2​/Fe$^{2+}$2+, and Mg/Ag$^+$+. The lesson then applies the framework to hydrogen and methanol fuel cells, the lithium-ion and lead-acid rechargeable batteries, and the qualitative Le-Chatelier effect of concentration changes on cell emf. Lesson 10 closes the module by applying these ideas to quantitative redox titrations.

Key Equations: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \quad \text{(with cathode = more positive)}$Ecell∘​=Ecathode∘​−Eanode∘​(with cathode = more positive) $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​ with $n$n = moles of electrons transferred per mole of reaction, $F = 96\,485$F=96485 C mol$^{-1}$−1, $E^\circ_{cell}$Ecell∘​ in V, $\Delta G^\circ$ΔG∘ in J mol$^{-1}$−1 (divide by 1000 for kJ). Feasibility criterion: $E^\circ_{cell} > 0 \Leftrightarrow \Delta G^\circ < 0$Ecell∘​>0⇔ΔG∘<0.

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Learning Objectives (OCR Spec 5.2.3)

By the end of this lesson you should be able to:

• Calculate $E^\circ_{cell}$Ecell∘​ from two standard electrode potentials using $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$Ecell∘​=Ecathode∘​−Eanode∘​, with cathode = more positive
• Construct the overall ionic equation by reversing the anode half-equation, scaling to balance electrons, and adding to the cathode half-equation
• Predict the thermodynamic feasibility of a redox reaction from the sign of $E^\circ_{cell}$Ecell∘​
• Apply $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​ to convert cell emf into Gibbs free-energy change and confirm equivalence with the Lesson 7 feasibility criterion $\Delta G^\circ < 0$ΔG∘<0
• Use Le Chatelier's principle on a half-cell equilibrium to predict the qualitative effect of changing the concentration of an oxidised or reduced species on $E$E (Nernst-equation foreshadow)
• Describe the operation of a hydrogen–oxygen fuel cell (both alkaline KOH and acidic PEM variants), state the half-equations at each electrode, and write the overall reaction with $E^\circ_{cell} \approx +1.23$Ecell∘​≈+1.23 V
• Contrast a fuel cell (continuously fed) with a rechargeable battery (closed system, externally regenerated)
• Distinguish thermodynamic feasibility ($E^\circ_{cell} > 0$Ecell∘​>0) from kinetic accessibility (the overpotential / activation-energy barrier)

Calculating $E^\circ_{cell}$Ecell∘​ — the master equation

When two half-cells are joined through a salt bridge and an external wire, the overall standard cell emf is:

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$Ecell∘​=Ecathode∘​−Eanode∘​

where:

• The cathode is where reduction occurs — the half-cell with the more positive $E^\circ$E∘.
• The anode is where oxidation occurs — the half-cell with the more negative $E^\circ$E∘.

This is not an arbitrary convention: it falls out of the IUPAC cell-notation rule (Lesson 8) that the cell is written with the more positive electrode on the right. The subtraction is signed — be careful with negative $E^\circ$E∘ values, because the minus signs propagate. The cell will always give $E^\circ_{cell} > 0$Ecell∘​>0 when the more positive electrode is correctly placed on the right (the cathode position).

Algorithm for any $E^\circ_{cell}$Ecell∘​ problem:

1. Look up both half-cell $E^\circ$E∘ values.
2. Identify the more positive — that is the cathode (reduction).
3. The less positive is the anode (oxidation) — reverse its half-equation.
4. $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$Ecell∘​=Ecathode∘​−Eanode∘​, subtracted as signed numbers.
5. Scale each half-equation so the electrons match, then add. Electrons cancel; the overall reaction has no electrons.
6. If asked about $\Delta G^\circ$ΔG∘: $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​, with $n$n = the electrons per mole of the balanced overall equation.

graph TD
  A[Two half-cells with known E values] --> B[Identify more positive E = CATHODE = reduction]
  A --> C[Identify more negative E = ANODE = oxidation - REVERSE the equation]
  B --> D["E_cell = E_cathode - E_anode (signed subtraction)"]
  C --> D
  D --> E{E_cell positive?}
  E -->|Yes| F[Reaction THERMODYNAMICALLY feasible as written]
  E -->|No| G[Reverse direction feasible]
  F --> H[Scale electrons -- add half-equations -- electrons cancel]
  H --> I["Apply delta G = -nFE_cell if required (n = electrons per mole of overall reaction)"]

Worked Example 1: The Daniell Cell (Zn/Cu, the canonical example)

Half-cells: $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$E∘(Cu2+/Cu)=+0.34 V; $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$E∘(Zn2+/Zn)=−0.76 V.

Step 1 — identify cathode/anode. Cu$^{2+}$2+/Cu is more positive $\to$→ cathode. Zn$^{2+}$2+/Zn is more negative $\to$→ anode.

Step 2 — compute $E^\circ_{cell}$Ecell∘​.

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (+0.34) - (-0.76) = +1.10\;\text{V}$Ecell∘​=Ecathode∘​−Eanode∘​=(+0.34)−(−0.76)=+1.10V

Step 3 — half-equations.

• Cathode (reduction, as written): $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \to \text{Cu(s)}$Cu2+(aq)+2e−→Cu(s)
• Anode (oxidation, reversed): $\text{Zn(s)} \to \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-$Zn(s)→Zn2+(aq)+2e−

Step 4 — overall equation (electrons match at 2, so add directly):

$\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \to \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}, \quad E^\circ_{cell} = +1.10\;\text{V}$Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s),Ecell∘​=+1.10V

Step 5 — $\Delta G^\circ$ΔG∘ via $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​ with $n = 2$n=2:

$\Delta G^\circ = -2 \times 96\,485 \times 1.10 = -212\,267\;\text{J mol}^{-1} = -212\;\text{kJ mol}^{-1}$ΔG∘=−2×96485×1.10=−212267J mol−1=−212kJ mol−1

Strongly negative — consistent with the strongly positive $E^\circ_{cell}$Ecell∘​ — confirming the Daniell cell reaction is thermodynamically feasible. This is the canonical demonstration of the framework: the cell is built from Zn rod in ZnSO$_4$4​(aq) and Cu rod in CuSO$_4$4​(aq), connected by a salt bridge of saturated KNO$_3$3​, and a voltmeter reads $+1.10$+1.10 V at 298 K with [Zn$^{2+}$2+] = [Cu$^{2+}$2+] = 1.00 mol dm$^{-3}$−3.

graph TD
  A[Zn rod in 1.0 mol/dm3 ZnSO4 -- anode -- negative terminal] --> B[Zn loses electrons]
  B --> C[External wire carries electrons to Cu electrode]
  C --> D[Cu rod in 1.0 mol/dm3 CuSO4 -- cathode -- positive terminal]
  D --> E[Cu2+ gains electrons -- Cu deposits on rod]
  F[Salt bridge -- saturated KNO3 in agar] --> G[K+ migrates to Cu cell -- NO3- migrates to Zn cell]
  G --> H[Charge balance maintained -- no net build-up]

The salt bridge — why ionic flow matters

The salt bridge is the unsung hero of the electrochemical cell. Its job is to carry ionic current between the two half-cells to balance the charge that would otherwise build up as electrons leave one half-cell and enter the other. Without it, the cell would polarise within microseconds — the anode compartment would accumulate positive charge from each Zn$^{2+}$2+ released, the cathode compartment would accumulate negative charge from each Cu$^{2+}$2+ consumed, and the cell emf would collapse.

In a typical salt bridge of saturated KNO$_3$3​ in agar gel, K$^+$+ ions migrate towards the cathode compartment (where positive charge is being consumed) and NO$_3^-$3−​ ions migrate towards the anode compartment (where positive charge is being released). Both KNO$_3$3​ ions are electrochemically silent — they are not oxidised or reduced at either electrode in the typical potential range, so they don't contribute to the cell reaction. The choice of KNO$_3$3​ also avoids the issues of Cl$^-$− (which would be oxidised by Cl$_2$2​/Cl$^-$−-active half-cells) or SO$_4^{2-}$42−​ (which can precipitate Pb$^{2+}$2+ or Ba$^{2+}$2+ if those are in either compartment).

Critical point: the salt bridge carries ions, not electrons. Electrons flow through the external wire; ions flow through the salt bridge. Saying "the salt bridge completes the circuit" is correct only in the sense that ionic flow closes the electrical loop.

Worked Example 2: Mg|Mg$^{2+}$2+‖Cu$^{2+}$2+|Cu

Half-cells: $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$E∘(Cu2+/Cu)=+0.34 V; $E^\circ(\text{Mg}^{2+}/\text{Mg}) = -2.37$E∘(Mg2+/Mg)=−2.37 V.

Cu$^{2+}$2+/Cu is more positive $\to$→ cathode. Mg$^{2+}$2+/Mg is more negative $\to$→ anode.

$E^\circ_{cell} = (+0.34) - (-2.37) = +2.71\;\text{V}$Ecell∘​=(+0.34)−(−2.37)=+2.71V

A huge cell emf — over twice the Daniell value — driven by the very negative magnesium half-cell. Half-equations (electrons already balanced at 2):

• Cathode: Cu$^{2+}$2+(aq) + 2e$^-$− $\to$→ Cu(s)
• Anode: Mg(s) $\to$→ Mg$^{2+}$2+(aq) + 2e$^-$−

Overall: Mg(s) + Cu$^{2+}$2+(aq) $\to$→ Mg$^{2+}$2+(aq) + Cu(s), $E^\circ_{cell} = +2.71$Ecell∘​=+2.71 V.

$\Delta G^\circ$ΔG∘ with $n = 2$n=2: $\Delta G^\circ = -2 \times 96\,485 \times 2.71 = -523\,029$ΔG∘=−2×96485×2.71=−523029 J mol$^{-1} = -523$−1=−523 kJ mol$^{-1}$−1.

This is the chemistry behind magnesium-air primary batteries and sacrificial-anode corrosion protection (a block of Mg attached to a steel ship's hull preferentially oxidises, sparing the iron from rusting).

Worked Example 3: Cu|Cu$^{2+}$2+‖Ag$^+$+|Ag

Half-cells: $E^\circ(\text{Ag}^+/\text{Ag}) = +0.80$E∘(Ag+/Ag)=+0.80 V; $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$E∘(Cu2+/Cu)=+0.34 V.

Ag$^+$+/Ag is more positive $\to$→ cathode. Cu$^{2+}$2+/Cu is more negative $\to$→ anode.

$E^\circ_{cell} = (+0.80) - (+0.34) = +0.46\;\text{V}$Ecell∘​=(+0.80)−(+0.34)=+0.46V

Half-equations — note the electron mismatch (Ag$^+$+/Ag is 1-electron; Cu$^{2+}$2+/Cu is 2-electron) — scale the Ag half-equation by 2:

• Cathode (×2): $2\text{Ag}^+(\text{aq}) + 2\text{e}^- \to 2\text{Ag(s)}$2Ag+(aq)+2e−→2Ag(s)
• Anode: $\text{Cu(s)} \to \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-$Cu(s)→Cu2+(aq)+2e−

Overall: Cu(s) + 2Ag$^+$+(aq) $\to$→ Cu$^{2+}$2+(aq) + 2Ag(s), $E^\circ_{cell} = +0.46$Ecell∘​=+0.46 V.

$\Delta G^\circ = -2 \times 96\,485 \times 0.46 = -88\,766$ΔG∘=−2×96485×0.46=−88766 J mol$^{-1} = -89$−1=−89 kJ mol$^{-1}$−1. This is the "silver tree" demonstration (Lesson 8). Note that scaling a half-equation does NOT change $E^\circ$E∘ — $E^\circ$E∘ is an intensive property, like temperature. We scale the equation, not the potential.

Worked Example 4: MnO$_4^-$4−​ vs Fe$^{2+}$2+ (the redox-titration setup)

Half-cells: $E^\circ(\text{MnO}_4^-/\text{Mn}^{2+}) = +1.51$E∘(MnO4−​/Mn2+)=+1.51 V (in 1.0 mol dm$^{-3}$−3 H$^+$+); $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$E∘(Fe3+/Fe2+)=+0.77 V.

MnO$_4^-$4−​/Mn$^{2+}$2+ is more positive $\to$→ cathode. Fe$^{3+}$3+/Fe$^{2+}$2+ is more negative $\to$→ anode.

$E^\circ_{cell} = (+1.51) - (+0.77) = +0.74\;\text{V}$Ecell∘​=(+1.51)−(+0.77)=+0.74V

Half-equations — MnO$_4^-$4−​ is 5-electron, Fe$^{3+}$3+/Fe$^{2+}$2+ is 1-electron — scale Fe by 5:

• Cathode: $\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \to \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O(l)}$MnO4−​(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2​O(l)
• Anode (×5): $5\text{Fe}^{2+}(\text{aq}) \to 5\text{Fe}^{3+}(\text{aq}) + 5\text{e}^-$5Fe2+(aq)→5Fe3+(aq)+5e−

Overall:

$\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \to \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O(l)} + 5\text{Fe}^{3+}(\text{aq})$MnO4−​(aq)+8H+(aq)+5Fe2+(aq)→Mn2+(aq)+4H2​O(l)+5Fe3+(aq)

$n = 5$n=5, so $\Delta G^\circ = -5 \times 96\,485 \times 0.74 = -357\,000$ΔG∘=−5×96485×0.74=−357000 J mol$^{-1} = -357$−1=−357 kJ mol$^{-1}$−1. Strongly negative — the reaction goes essentially to completion, which is exactly the precondition for using it as the basis of a quantitative titration (Lesson 10). The 1 MnO$_4^-$4−​ : 5 Fe$^{2+}$2+ stoichiometry is set by the electron-balance constraint, and is the foundation of the iron-tablet and iron-ore titration calculations of Lesson 10.

Worked Example 5: Cl$_2$2​ oxidising Fe$^{2+}$2+

Half-cells: $E^\circ(\text{Cl}_2/\text{Cl}^-) = +1.36$E∘(Cl2​/Cl−)=+1.36 V; $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$E∘(Fe3+/Fe2+)=+0.77 V.

$E^\circ_{cell} = 1.36 - 0.77 = +0.59\;\text{V}$Ecell∘​=1.36−0.77=+0.59V

Half-equations (both 2-electron when Cl$_2$2​ is balanced; Fe scaled by 2):

• Cathode: Cl$_2$2​(g) + 2e$^-$− $\to$→ 2Cl$^-$−(aq)
• Anode (×2): 2Fe$^{2+}$2+(aq) $\to$→ 2Fe$^{3+}$3+(aq) + 2e$^-$−

Overall: Cl$_2$2​(g) + 2Fe$^{2+}$2+(aq) $\to$→ 2Cl$^-$−(aq) + 2Fe$^{3+}$3+(aq), $E^\circ_{cell} = +0.59$Ecell∘​=+0.59 V.

$\Delta G^\circ = -2 \times 96\,485 \times 0.59 = -114\,000$ΔG∘=−2×96485×0.59=−114000 J mol$^{-1} = -114$−1=−114 kJ mol$^{-1}$−1. Negative — chlorine water can oxidise iron(II) salts to iron(III), the chemistry behind chlorine bleaching of iron-stained fabrics and a school demonstration of feasibility via $E^\circ_{cell}$Ecell∘​.

Worked Example 6: Cu vs Zn$^{2+}$2+ — predicted reverse direction

Can Cu(s) reduce Zn$^{2+}$2+(aq) to Zn(s)? Proposed reaction: Cu(s) + Zn$^{2+}$2+(aq) $\to$→ Cu$^{2+}$2+(aq) + Zn(s).

For this to be feasible as written, Cu would have to be the anode (oxidised) and Zn$^{2+}$2+/Zn the cathode. So $E^\circ_{cell}$Ecell∘​ for the proposed direction is:

$E^\circ_{cell} = E^\circ(\text{Zn}^{2+}/\text{Zn}) - E^\circ(\text{Cu}^{2+}/\text{Cu}) = -0.76 - 0.34 = -1.10\;\text{V}$Ecell∘​=E∘(Zn2+/Zn)−E∘(Cu2+/Cu)=−0.76−0.34=−1.10V

Negative — NOT feasible. The reverse direction (the Daniell cell, Worked Example 1) is feasible with $E^\circ_{cell} = +1.10$Ecell∘​=+1.10 V.

This worked example shows the diagnostic step: if you propose a direction and get a negative $E^\circ_{cell}$Ecell∘​, then the proposed direction is infeasible and the reverse is what spontaneously occurs.

The Gibbs bridge: $\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​

Lesson 7 introduced $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘ as the criterion of feasibility ($\Delta G^\circ < 0$ΔG∘<0). For electrochemical cells the same criterion appears in a different dress:

$\Delta G^\circ = -nFE^\circ_{cell}$ΔG∘=−nFEcell∘​

where:

• $n$n = moles of electrons transferred per mole of reaction (read from the balanced overall equation)
• $F = 96\,485$F=96485 C mol$^{-1}$−1 — the Faraday constant, charge per mole of electrons
• $E^\circ_{cell}$Ecell∘​ in V

$F$F has units of coulombs per mole; $E^\circ_{cell}$Ecell∘​ in V is equivalently J C$^{-1}$−1 (joules per coulomb of charge). So $nFE^\circ_{cell}$nFEcell∘​ has units of joules per mole — the same units as $\Delta G^\circ$ΔG∘. The minus sign ensures that positive $E^\circ_{cell}$Ecell∘​ (feasible) gives negative $\Delta G^\circ$ΔG∘ (feasible).

Equivalences:

$E^\circ_{cell}$Ecell∘​	$\Delta G^\circ$ΔG∘	Reaction status
$> 0$>0	$< 0$<0	Feasible as written
$= 0$=0	$= 0$=0	At equilibrium ($K = 1$K=1)
$< 0$<0	$> 0$>0	Infeasible as written; reverse is feasible

These two criteria are exactly equivalent. Electrochemistry simply expresses Lesson 7's Gibbs criterion in volts rather than in kJ mol$^{-1}$−1.

Effect of non-standard conditions — Le Chatelier on the half-cell

The $E^\circ$E∘ values assume [solute] = 1.00 mol dm$^{-3}$−3, $p$p(gas) = 100 kPa, $T = 298$T=298 K. When conditions deviate, $E$E (the actual electrode potential) shifts from $E^\circ$E∘. Treating each half-equation as an equilibrium and applying Le Chatelier: increasing [oxidised species] (LHS) shifts equilibrium right and makes $E$E more positive; decreasing [oxidised species] makes $E$E more negative; increasing [reduced species] makes $E$E more negative. For H$^+$+-dependent half-cells (e.g. MnO$_4^-$4−​ + 8H$^+$+ + 5e$^-$−), increasing [H$^+$+] shifts the reduction right, raising $E$E and making MnO$_4^-$4−​ a stronger oxidising agent. Increasing temperature also shifts $E$E via the half-cell $K_{eq}$Keq​. The Nernst equation $E = E^\circ - (RT/nF)\ln Q$E=E∘−(RT/nF)lnQ makes these qualitative arguments quantitative (undergraduate footnote); at A-Level, qualitative Le-Chatelier reasoning suffices.

Hydrogen Fuel Cells — Continuous Chemistry, Steady Power

A fuel cell is an electrochemical cell in which the reactants are fed continuously from external tanks and the products are removed, giving a steady output of electrical power. Unlike a battery, a fuel cell does not "run down" — it keeps generating power as long as H$_2$2​ (fuel) and O$_2$2​ (oxidant) are supplied. The hydrogen–oxygen fuel cell is the workhorse of low-emission transport and space-mission power generation (the Apollo command module ran on alkaline H$_2$2​/O$_2$2​ cells supplied by Pratt & Whitney).

Alkaline hydrogen–oxygen fuel cell (electrolyte: aqueous KOH, ~6 mol dm$^{-3}$−3):

• Anode (oxidation of H$_2$2​): $\text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \to 2\text{H}_2\text{O(l)} + 2\text{e}^-$H2​(g)+2OH−(aq)→2H2​O(l)+2e−
• Cathode (reduction of O$_2$2​): $\tfrac{1}{2}\text{O}_2(\text{g}) + \text{H}_2\text{O(l)} + 2\text{e}^- \to 2\text{OH}^-(\text{aq})$21​O2​(g)+H2​O(l)+2e−→2OH−(aq)
• Overall: $\text{H}_2(\text{g}) + \tfrac{1}{2}\text{O}_2(\text{g}) \to \text{H}_2\text{O(l)}, \quad E^\circ_{cell} \approx +1.23\;\text{V}$H2​(g)+21​O2​(g)→H2​O(l),Ecell∘​≈+1.23V

Acidic PEM (proton-exchange membrane) hydrogen–oxygen fuel cell (electrolyte: solid acidic polymer membrane, typically Nafion — used in fuel-cell vehicles):

• Anode: $\text{H}_2(\text{g}) \to 2\text{H}^+(\text{aq}) + 2\text{e}^-$H2​(g)→2H+(aq)+2e−
• Cathode: $\tfrac{1}{2}\text{O}_2(\text{g}) + 2\text{H}^+(\text{aq}) + 2\text{e}^- \to \text{H}_2\text{O(l)}$21​O2​(g)+2H+(aq)+2e−→H2​O(l)
• Overall: same as alkaline, with $E^\circ_{cell} \approx +1.23$Ecell∘​≈+1.23 V.

Both versions have the same net chemistry — water from hydrogen and oxygen, with all the redox energy released as electrical work rather than heat. The choice of electrolyte (alkaline vs acidic) affects operating temperature, longevity, and electrolyte handling but not the thermodynamics.

graph TD
  A[H2 fuel tank] --> B[Anode: H2 oxidised to H+ or H2O]
  B --> C[Electrons flow through external circuit -- ELECTRICAL WORK]
  D[O2 from air] --> E[Cathode: O2 reduced to OH- or H2O]
  C --> E
  F[Electrolyte -- acidic PEM or alkaline KOH] --> G[H+ or OH- migration completes circuit]
  E --> H[Water exhaust]
  B --> H