University Chemistry — Year 2 · Bachelor Year 2

9Thermodynamics of Electrochemical Cells

A hydrogen fuel-cell bus carries a stack of a few hundred thin cells. In each, hydrogen and the oxygen of the air combine to water without a flame, and the energy comes out as an electric current. Two numbers govern such a stack, and both come straight from tables of Gibbs energies: no cell can give more than $1.23\,\mathrm{V}$ , and no fuel cell, however perfect, turns all the energy of its hydrogen into electrical work. This chapter connects the voltage of a cell to the Gibbs energy of its reaction, derives the Nernst equation used since the Year 1 volume, and reads entropies and enthalpies of reaction off a voltmeter and a thermometer.

You already know

The Year 1 volume: cells, half-cells, anode and cathode, salt bridge, cell voltage, electrode potential, standard potential and the standard hydrogen electrode, the Nernst equation (admitted there), equilibrium constants from potentials. Chapter 2: $\Delta_r G$ , $\Delta_r G^\circ$ and $\Delta_r G = \Delta_r G^\circ + RT\ln Q$ ; the evolution criterion. Chapter 3: activities. The school volume: electrolysis.

A hydrogen fuel-cell bus at a stop: the hydrogen tanks are under the fairing on the roof, and the only exhaust is water, dripping under the bus.

9.1 Electrical work and the Faraday constant

Definition 9.1 (Faraday constant)

The Faraday constant is the electric charge of one mole of elementary charges, $F = N_A e = 96\,485.33\,\mathrm{C}/\mathrm{mol}$ .

Proposition 9.2 (Charge passed)

If the reaction of a cell, written with $n$ electrons exchanged, advances by $\Delta\xi$ , the charge that flows through the external circuit is $Q = nF\,\Delta\xi$ .

Proof. An advancement $\Delta\xi$ transfers $n\,\Delta\xi$ moles of electrons from the anode to the cathode through the wire, each of charge $e$ in magnitude: $Q = n\,\Delta\xi\,N_A e$ . ∎

When a charge $Q$ is driven through a circuit by a voltage $U$ , the work received by the circuit is $QU$ . Seen from the cell, which delivers that work, the electrical work it receives is $\delta W_{\text{el}} = -U\,\delta Q = -nFU\, d\xi$ .

A cell as a thermodynamic system at constant temperature and pressure: it exchanges heat with its surroundings and delivers electrical work through the external circuit. With the sign convention of the first law (work and heat received counted positive), W_ el = -nFU\,d. — A cell as a thermodynamic system at constant temperature and pressure: it exchanges heat with its surroundings and delivers electrical work through the external circuit. With the sign convention of the first law (work and heat received counted positive), $\delta W_{\text{el}} = -nFU\,d\xi$ .

9.2 The Gibbs energy of a cell reaction

Theorem 9.3 (Electrical work and Gibbs energy)

A cell working at constant $T$ and $p$ delivers an electrical work at most equal to the decrease of its Gibbs energy: for an advancement $d\xi$ , $nFU\,d\xi \leq -\Delta_r G\,d\xi$ . When it works reversibly (an infinitesimal current), the equality holds, and the cell voltage is the electromotive force

\Delta_r G = -nFE .

Proof. First law at constant pressure, with the pressure work separated: $dH = \delta Q + \delta W_{\text{el}}$ . Second law: $dS \geq \delta Q/T$ . At constant $T$ and $p$ , $dG = dH - T\,dS \leq \delta W_{\text{el}} = -nFU\,d\xi$ . With $dG = \Delta_r G\,d\xi$ , $\Delta_r G\,d\xi \leq -nFU\,d\xi$ , that is $nFU\,d\xi \leq -\Delta_r G\,d\xi$ . For a reversible change the inequality of the second law is an equality, and the voltage then measured, with no current, is $E$ : $\Delta_r G = -nFE$ . ∎

A cell can thus run forward ( $d\xi > 0$ , delivering work) only if $\Delta_r G < 0$ , that is $E > 0$ with the poles chosen so that the reaction written runs from anode to cathode. A real cell, through which a current flows, delivers less work than $-\Delta_r G$ : the difference is dissipated as heat in its resistance and at its electrodes (Chapter 10).

Example 9.4 (The Daniell cell)

For $\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$ , $\Delta_r G^\circ = \Delta_f G^\circ(\ce{Zn^2+}) - \Delta_f G^\circ(\ce{Cu^2+}) = -147.06 - 65.49 = -212.55\,\mathrm{kJ}/\mathrm{mol}$ , and $n = 2$ : $E^\circ = 212\,550/(2 \times 96\,485) = 1.10\,\mathrm{V}$ , the voltage measured on a Daniell cell in standard conditions.

9.3 The Nernst equation and standard potentials

The Year 1 volume admitted the Nernst equation; it now follows from $\Delta_r G = \Delta_r G^\circ + RT\ln Q$ .

Theorem 9.5 (The Nernst equation, derived)

For a half-reaction $\alpha\,\mathrm{Ox} + n\,\mathrm e^- \rightleftharpoons \beta\,\mathrm{Red}$ , the potential of the electrode at temperature $T$ is

E = E^\circ + \frac{RT}{nF}\ln\frac{a_{\mathrm{Ox}}^{\alpha}}{a_{\mathrm{Red}}^{\beta}} ,

where the activities include those of every other species of the half-reaction ( $\ce{H+}$ , water excepted), with their stoichiometric powers.

Proof. The potential of an electrode is the voltage of the cell made of the standard hydrogen electrode (anode) and that electrode (cathode). Its reaction is

\alpha\,\mathrm{Ox} + \tfrac n2\,\ce{H2} \longrightarrow \beta\,\mathrm{Red} + n\,\ce{H+} ,

whose reaction quotient, with $a_{\ce{H+}} = 1$ and $p_{\ce{H2}} = p^\circ$ at the standard hydrogen electrode, reduces to $a_{\mathrm{Red}}^\beta/ a_{\mathrm{Ox}}^\alpha$ . From Theorem 9.3, $E = -\Delta_r G/(nF) = -\Delta_r G^\circ/(nF) - (RT/nF)\ln Q$ ; with $E^\circ = -\Delta_r G^\circ/(nF)$ this is the formula. At $298.15\,\mathrm{K}$ , $RT\ln 10/F = 0.0592\,\mathrm{V}$ . ∎

Proposition 9.6 (Standard potentials from Gibbs energies)

With the conventions $\Delta_f G^\circ(\ce{H+}, \text{aq}) = 0$ and $\Delta_f G^\circ(\ce{H2}, \text{g}) = 0$ , the standard potential of a couple is $E^\circ = -\Delta_r G^\circ/(nF)$ , where $\Delta_r G^\circ$ is the standard Gibbs energy of the half-reaction written as a reduction, the electrons counted as having zero Gibbs energy.

Proof. In the reaction against the hydrogen electrode, $\tfrac n2\,\ce{H2}$ and $n\,\ce{H+}$ contribute nothing to $\Delta_r G^\circ$ under those conventions: the standard Gibbs energy of the cell reaction is that of the half-reaction alone. ∎

Method 9.7 (A standard potential from tables)

Write the half-reaction as a reduction, balanced with $\ce{H+}$ and $\ce{H2O}$ , with $n$ electrons.
Compute $\Delta_r G^\circ = \sum\nu_i\Delta_f G^\circ_i$ (products positive), with zero for the elements, $\ce{H+}$ and the electrons.
$E^\circ = -\Delta_r G^\circ/(nF)$ , in volts with $\Delta_r G^\circ$ in joules per mole.

Example 9.8 (The silver chloride electrode)

$\ce{AgCl(s) + e- -> Ag(s) + Cl^-}$ : $\Delta_r G^\circ = -131.228 - (-109.789) = -21.439\,\mathrm{kJ}/\mathrm{mol}$ , $E^\circ = 21\,439/96\,485 = 0.222\,\mathrm{V}$ . Its potential depends on the chloride activity only: in a saturated potassium chloride solution it is a stable reference electrode.

Proposition 9.9 (Combining potentials)

If a half-reaction (3), with $n_3$ electrons, is the sum of half-reactions (1) and (2), with $n_1$ and $n_2$ electrons, then

n_3E_3^\circ = n_1E_1^\circ + n_2E_2^\circ ;

the potentials themselves do not add.

Proof. Standard Gibbs energies add, $\Delta_r G_3^\circ = \Delta_r G_1^\circ + \Delta_r G_2^\circ$ , and the electron counts add, $n_3 = n_1 + n_2$ . Divide each $\Delta_r G^\circ$ by $-F$ . ∎

Example 9.10 (Iron in its two oxidation states)

$\ce{Fe^3+ + e- -> Fe^2+}$ has $E^\circ = (-78.90 + 4.7)/(-96.485) = 0.77\,\mathrm{V}$ and $\ce{Fe^2+ + 2e- -> Fe}$ has $E^\circ = -78.90/(2 \times 96.485) = -0.41\,\mathrm{V}$ . Hence $\ce{Fe^3+ + 3e- -> Fe}$ : $E^\circ = (0.77 - 0.82)/3 = -0.02\,\mathrm{V}$ , not the sum of the two.

9.4 Temperature, entropy and the efficiency of a fuel cell

Definition 9.11 (Temperature coefficient)

The temperature coefficient of a cell is the derivative $(\partial E/\partial T)_p$ of its electromotive force with respect to temperature.

Proposition 9.12 (Entropy and enthalpy from cell data)

For the standard cell reaction,

\Delta_r S^\circ = nF\,\frac{dE^\circ}{dT}, \qquad \Delta_r H^\circ = -nF\Big(E^\circ - T\frac{dE^\circ}{dT}\Big),

and a cell working reversibly at temperature $T$ receives from its surroundings the heat $T\Delta_r S$ per mole of reaction.

Proof. $\Delta_r S^\circ = -d\Delta_r G^\circ/dT$ (from $dG = -S\,dT + V\,dp$ , differentiated with respect to $\xi$ ) $= nF\,dE^\circ/dT$ ; then $\Delta_r H^\circ = \Delta_r G^\circ + T\Delta_r S^\circ$ . For the reversible change, $\delta Q = T\,dS = T\Delta_r S\, d\xi$ . ∎

Method 9.13 (From cell data to the thermodynamics of the reaction)

Measure $E$ at several temperatures; the slope is $dE/dT$ .
$\Delta_r G = -nFE$ , $\Delta_r S = nF\,dE/dT$ , $\Delta_r H = \Delta_r G + T\Delta_r S$ .
Heat exchanged per mole when the cell works reversibly: $T\Delta_r S$ ; when it delivers a current at a voltage $U < E$ , the heat released is $-\Delta_r H - nFU$ per mole.

For the hydrogen–oxygen cell producing liquid water, the CODATA key values give $\Delta_r H^\circ = -285.83\,\mathrm{kJ}/\mathrm{mol}$ and

\Delta_r S^\circ = 69.95 - 130.68 - \tfrac12 \times 205.15 = -163.3\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol}) ,

so $\Delta_r G^\circ = -237.14\,\mathrm{kJ}/\mathrm{mol}$ and $E^\circ = 1.229\,\mathrm{V}$ ; the temperature coefficient is $\Delta_r S^\circ/(2F) = -0.85\,\mathrm{mV}/\mathrm{K}$ . A hydrogen fuel cell gives less voltage hot than cold.

The hydrogen–oxygen cell from the JANAF tables. Left: standard voltage - _r G /(2F) with liquid water (up to 500\, K, the liquid kept under its standard state) and with water vapour. Right: maximum efficiency _r G / _r H of the cell producing vapour, against the Carnot efficiency of a heat engine working between T and 298\, K; the engine wins only above about 1150\, K. — The hydrogen–oxygen cell from the JANAF tables. Left: standard voltage $-\Delta_r G^\circ/(2F)$ with liquid water (up to $500\,\mathrm{K}$ , the liquid kept under its standard state) and with water vapour. Right: maximum efficiency $\Delta_r G^\circ/\Delta_r H^\circ$ of the cell producing vapour, against the Carnot efficiency of a heat engine working between $T$ and $298\,\mathrm{K}$ ; the engine wins only above about $1150\,\mathrm{K}$ .

Definition 9.14 (Thermodynamic efficiency of a fuel cell)

The thermodynamic efficiency of a fuel cell is the ratio of the electrical work it delivers to the heat its fuel would release by combustion at the same temperature, $-\Delta_r H$ per mole.

Proposition 9.15 (Maximum efficiency of a fuel cell)

The thermodynamic efficiency of a fuel cell is at most $\Delta_r G/\Delta_r H$ , reached when it works reversibly, and equal to $(nFU)/(-\Delta_r H)$ at a working voltage $U$ . Unlike a heat engine, it is not limited by the Carnot efficiency $1 - T_0/T$ .

Proof. By Theorem 9.3 the work per mole is $nFU \leq -\Delta_r G$ ; divide by $-\Delta_r H$ . The cell converts chemical energy directly into work at a single temperature: no heat is taken from a hot source and partly given to a cold one, so the Carnot bound, which concerns such cycles, does not apply. Its own bound is set by $T\Delta_r S$ , the heat the reversible cell must exchange. ∎

Where the energy of one mole of hydrogen goes in a fuel cell at 298\, K (liquid water). Top: the heat a flame would release. Middle: the reversible cell, which must give up -T _r S = 48.7\, kJ as heat whatever its design. Bottom: a real cell working at 0.70\, V. — Where the energy of one mole of hydrogen goes in a fuel cell at $298\,\mathrm{K}$ (liquid water). Top: the heat a flame would release. Middle: the reversible cell, which must give up $-T\Delta_r S^\circ = 48.7\,\mathrm{kJ}$ as heat whatever its design. Bottom: a real cell working at $0.70\,\mathrm{V}$ .

History — Faraday’s laws of electrolysis

In 1834 Michael Faraday, measuring the products of electrolyses with a voltameter of his own design, stated that the mass of a substance produced at an electrode is proportional to the quantity of electricity passed, and that for a given quantity of electricity the masses of different substances are in the ratio of their chemical equivalents. He coined the words electrode, anode, cathode, ion, anion and cation. The constant that bears his name turns those laws into $Q = nF\,\Delta\xi$ . (Portrait by Thomas Phillips, 1842, public domain, Wikimedia Commons.)

9.5 Concentration cells

Definition 9.16 (Concentration cell)

A concentration cell is made of two half-cells of the same couple that differ only in the concentration (or pressure) of one species.

Proposition 9.17 (Voltage of a concentration cell)

Two electrodes of a metal M dipping into solutions of $\ce{M^{n+}}$ at concentrations $c_1 < c_2$ , joined by a salt bridge, give a voltage

U = \frac{RT}{nF}\ln\frac{c_2}{c_1} ,

the positive pole being the more concentrated side. The cell works until the two concentrations are equal.

Proof. By the Nernst equation each electrode has $E_i = E^\circ + (RT/nF)\ln c_i$ (ideal dilute solutions); $E^\circ$ cancels in the difference. The reaction is $\ce{M^{n+}}$ (side 2) $\to$ $\ce{M^{n+}}$ (side 1): metal is deposited on the concentrated side (cathode) and dissolved on the dilute side (anode), which evens the concentrations out; $\Delta_r G = RT\ln(c_1/c_2) < 0$ until they are equal. ∎

A silver concentration cell at 25 C. The dilute side is the anode: silver dissolves there and is deposited on the concentrated side, until the two concentrations are equal. U = 0.0592 (0.10/0.010) = 59\, mV. — A silver concentration cell at $25{}^{\circ}\mathrm{C}$ . The dilute side is the anode: silver dissolves there and is deposited on the concentrated side, until the two concentrations are equal. $U = 0.0592\log(0.10/0.010) = 59\,\mathrm{mV}$ .

The same principle measures concentrations. A glass electrode, whose thin membrane separates a reference solution from the sample, develops a membrane potential of $(RT/F)\ln 10 = 59\,\mathrm{mV}$ per unit of pH difference at $25{}^{\circ}\mathrm{C}$ : the pH meter of the Year 1 volume is a concentration cell for $\ce{H+}$ . Across the membrane of a living cell, the unequal concentrations of $\ce{K+}$ create in the same way a potential of several tens of millivolts.

9.6 Exercises

Exercise 9.1 ★

From $\Delta_f G^\circ(\ce{H2O}, \text{l}) = -237.14\,\mathrm{kJ}/\mathrm{mol}$ , compute the standard voltage of the hydrogen–oxygen cell.

Solution

Solution of Exercise 9.1.

$\ce{H2 + 1/2O2 -> H2O(l)}$ , $n = 2$ : $E^\circ = 237\,140/(2 \times 96\,485) = 1.229\,\mathrm{V}$ .

Exercise 9.2 ★

A battery is rated $1.0\,\mathrm{A}\,\mathrm{h}$ . Compute the charge it holds, in coulombs, and the mass of zinc consumed at its anode ( $\ce{Zn -> Zn^2+ + 2e-}$ ) when it is fully discharged.

Solution

Solution of Exercise 9.2.

$Q = 1.0 \times 3600 = 3.6 \times 10^{3}\,\mathrm{C}$ ; $n(\ce{Zn}) = 3600/(2 \times 96\,485) = 1.87 \times 10^{-2}\,\mathrm{mol}$ , that is $1.22\,\mathrm{g}$ .

Exercise 9.3 ★

A Daniell cell delivers $1.10\,\mathrm{V}$ with no current. Compute $\Delta_r G$ for one mole of zinc and the maximum electrical work for $10.0\,\mathrm{g}$ of zinc.

Solution

Solution of Exercise 9.3.

$\Delta_r G = -2 \times 96\,485 \times 1.10 = -212\,\mathrm{kJ}/\mathrm{mol}$ . For $10.0\,\mathrm{g}$ , $10.0/65.38 = 0.153\,\mathrm{mol}$ : at most $32.5\,\mathrm{kJ}$ .

Exercise 9.4 ★

Two copper electrodes dip into copper(II) sulfate solutions of $1.0 \times 10^{-3}\,\mathrm{mol}/\mathrm{L}$ and $0.10\,\mathrm{mol}/\mathrm{L}$ . Compute the voltage at $25{}^{\circ}\mathrm{C}$ and name the poles.

Solution

Solution of Exercise 9.4.

$U = (0.0592/2)\log(0.10/1.0 \times 10^{-3}) = 0.059\,\mathrm{V}$ ; the positive pole (cathode) is the electrode in the $0.10\,\mathrm{mol}/\mathrm{L}$ solution.

Exercise 9.5 ★★

A cell ( $n = 2$ ) has $E^\circ = 1.015\,\mathrm{V}$ and $dE^\circ/dT = -4.0 \times 10^{-4}\,\mathrm{V}/\mathrm{K}$ at $298\,\mathrm{K}$ (exercise data). Compute $\Delta_r G^\circ$ , $\Delta_r S^\circ$ and $\Delta_r H^\circ$ .

Solution

Solution of Exercise 9.5.

$\Delta_r G^\circ = -2 \times 96\,485 \times 1.015 = -195.9\,\mathrm{kJ}/\mathrm{mol}$ ; $\Delta_r S^\circ = 2 \times 96\,485 \times (-4.0 \times 10^{-4}) = -77.2\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ; $\Delta_r H^\circ = -195.9 + 298 \times (-0.0772) = -218.9\,\mathrm{kJ}/\mathrm{mol}$ .

Exercise 9.6 ★★

The cell of Exercise 9.5 discharges one mole of reaction (a) reversibly; (b) at $0.90\,\mathrm{V}$ . Compute in each case the electrical work and the heat exchanged with the surroundings.

Solution

Solution of Exercise 9.6.

(a) Work delivered $195.9\,\mathrm{kJ}$ ; heat $T\Delta_r S = -23.0\,\mathrm{kJ}$ : $23.0\,\mathrm{kJ}$ released. (b) Work $2 \times 96\,485 \times 0.90 = 173.7\,\mathrm{kJ}$ ; heat released $-\Delta_r H - 173.7 = 218.9 - 173.7 = 45.2\,\mathrm{kJ}$ . The extra $22.2\,\mathrm{kJ}$ is the work lost to the current.

Exercise 9.7 ★★

Compute $E^\circ(\ce{Fe^3+}/\ce{Fe})$ from the Gibbs energies of formation ( $\ce{Fe^2+}$ $-78.90\,\mathrm{kJ}/\mathrm{mol}$ , $\ce{Fe^3+}$ $-4.7\,\mathrm{kJ}/\mathrm{mol}$ ) and check that it equals $(E_1^\circ + 2E_2^\circ)/3$ .

Solution

Solution of Exercise 9.7.

$\ce{Fe^3+ + 3e- -> Fe}$ : $\Delta_r G^\circ = +4.7\,\mathrm{kJ}/\mathrm{mol}$ , $E^\circ = -4700/(3 \times 96\,485) = -0.016\,\mathrm{V}$ . With $E_1^\circ = 0.769$ and $E_2^\circ = -0.409\,\mathrm{V}$ : $(0.769 - 0.818)/3 = -0.016\,\mathrm{V}$ .

Exercise 9.8 ★★

Compute the standard potential of the couple $\ce{AgCl}$ / $\ce{Ag}$ , then the potential of a silver chloride electrode dipping into $0.10\,\mathrm{mol}/\mathrm{L}$ potassium chloride.

Solution

Solution of Exercise 9.8.

$E^\circ = 0.222\,\mathrm{V}$ (as in the chapter). $E = 0.222 - 0.0592\log 0.10 = 0.281\,\mathrm{V}$ .

Exercise 9.9 ★★

Zinc–air cells use $\ce{2Zn + O2 -> 2ZnO}$ ( $\Delta_f G^\circ(\ce{ZnO}) = -318.30\,\mathrm{kJ}/\mathrm{mol}$ ). Compute the standard voltage and the maximum electrical energy, in $\mathrm{kW}\,\mathrm{h}$ , per kilogram of zinc.

Solution

Solution of Exercise 9.9.

$\Delta_r G^\circ = 2(-318.30) = -636.6\,\mathrm{kJ}$ for $n = 4$ : $E^\circ = 636\,600/(4 \times 96\,485) = 1.65\,\mathrm{V}$ . Per kilogram of zinc, $1000/65.38 = 15.3\,\mathrm{mol}$ , $15.3 \times 318.3 = 4.87\,\mathrm{MJ}$ , that is $1.35\,\mathrm{kW}\,\mathrm{h}$ .

Exercise 9.10 ★★★

A direct methanol fuel cell runs $\ce{CH3OH(l) + 3/2O2 -> CO2 + 2H2O(l)}$ . From $\Delta_f H^\circ$ ( $\mathrm{kJ}/\mathrm{mol}$ ): $\ce{CH3OH(l)}$ $-239$ , $\ce{CO2}$ $-393.51$ , $\ce{H2O(l)}$ $-285.83$ ; and $S^\circ$ ( $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ): $\ce{CH3OH(l)}$ 127.2, $\ce{CO2}$ 213.8, $\ce{H2O(l)}$ 69.95, $\ce{O2}$ 205.15, compute $n$ , $E^\circ$ and the maximum efficiency.

Solution

Solution of Exercise 9.10.

Carbon goes from $-2$ to $+4$ : $n = 6$ . $\Delta_r H^\circ = -393.51 + 2(-285.83) + 239 = -726.2\,\mathrm{kJ}/\mathrm{mol}$ ; $\Delta_r S^\circ = 213.8 + 2(69.95) - 127.2 - 1.5 \times 205.15 = -81.2\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ; $\Delta_r G^\circ = -726.2 + 298.15 \times 0.0812 = -702.0\,\mathrm{kJ}/\mathrm{mol}$ . $E^\circ = 702\,000/(6 \times 96\,485) = 1.21\,\mathrm{V}$ ; maximum efficiency $702.0/726.2 = 0.967$ . (Real methanol cells work far below this voltage.)

Exercise 9.11 ★★★

A cell ( $n = 2$ ) has $E = 0.500\,\mathrm{V}$ and $dE/dT = +3.0 \times 10^{-4}\,\mathrm{V}/\mathrm{K}$ at $298\,\mathrm{K}$ (exercise data). Show that, working reversibly, it delivers more electrical work than $-\Delta_r H$ , and explain where the extra energy comes from.

Solution

Solution of Exercise 9.11.

$\Delta_r G = -2 \times 96\,485 \times 0.500 = -96.5\,\mathrm{kJ}$ ; $\Delta_r S = +57.9\,\mathrm{J}/\mathrm{K}$ ; $\Delta_r H = -96.5 + 298 \times 0.0579 = -79.2\,\mathrm{kJ}$ . The work, $96.5\,\mathrm{kJ}$ , exceeds $-\Delta_r H = 79.2\,\mathrm{kJ}$ : the cell absorbs $T\Delta_r S = 17.3\,\mathrm{kJ}$ of heat from its surroundings and turns it into work as well. The second law is not violated, since the entropy of the reaction increases by as much.

Exercise 9.12 ★★★

Two hydrogen electrodes ( $1\,\mathrm{bar}$ of $\ce{H2}$ ) dip into a reference solution of pH 7.00 and into a sample. The voltage is $0.177\,\mathrm{V}$ at $25{}^{\circ}\mathrm{C}$ , the reference being the positive pole. Find the pH of the sample, and explain why the result does not depend on any standard potential.

Solution

Solution of Exercise 9.12.

Each hydrogen electrode has $E = -0.0592\,\mathrm{pH}$ : $U = 0.0592\,(\mathrm{pH}_s - 7.00) = 0.177$ , $\mathrm{pH}_s = 10.0$ . The two electrodes are of the same couple: the standard potential cancels, as in any concentration cell.

9.7 Problem: The Fuel-Cell Bus

Problem 9.1

Weekend problem — the standard voltage of the hydrogen–oxygen cell from Gibbs energies, its temperature coefficient, the efficiency and heat of a real stack, and the energy in a tank of hydrogen

A fuel-cell stack has 400 cells in series, each with a proton-exchange membrane (acidic electrolyte). Data at $298.15\,\mathrm{K}$ (CODATA, JANAF): $\Delta_f H^\circ(\ce{H2O}, \text{l}) = -285.83\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_f G^\circ(\ce{H2O}, \text{l}) = -237.14\,\mathrm{kJ}/\mathrm{mol}$ , $\Delta_f G^\circ(\ce{H2O}, \text{g}) = -228.58\,\mathrm{kJ}/\mathrm{mol}$ ; $S^\circ$ ( $\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ ): $\ce{H2O(l)}$ 69.95, $\ce{H2}$ 130.68, $\ce{O2}$ 205.15. $F = 96\,485\,\mathrm{C}/\mathrm{mol}$ ; $M(\ce{H2}) = 2.016\,\mathrm{g}/\mathrm{mol}$ .

Part I — The standard voltage.

Write the half-reactions at the anode and at the cathode, and the cell reaction.
How many electrons are exchanged per molecule of hydrogen?
Compute $E^\circ$ when the water is formed as a liquid.
Compute it when the water leaves as vapour.
Explain the difference with the Gibbs energy of vaporisation of water.
Compute $\Delta_r H^\circ$ and the maximum efficiency at $298\,\mathrm{K}$ .

Part II — Temperature.

Compute $\Delta_r S^\circ$ for liquid water.
Deduce $dE^\circ/dT$ .
Estimate $E^\circ$ at $80{}^{\circ}\mathrm{C}$ , the working temperature of the stack.
Compare your method with the JANAF value at $400\,\mathrm{K}$ , $\Delta_f G^\circ(\ce{H2O}, \text{l}) = -221.01\,\mathrm{kJ}/\mathrm{mol}$ .
What heat does the reversible cell exchange per mole of hydrogen at $298\,\mathrm{K}$ ? Is it received or released?
Estimate the maximum efficiency at $80{}^{\circ}\mathrm{C}$ .

Part III — The real stack at $0.70\,\mathrm{V}$ per cell.

Compute the electrical work per mole of hydrogen.
Compute the efficiency referred to $\Delta_r H^\circ$ and to $\Delta_r G^\circ$ .
Compute the heat released per mole of hydrogen.
How much of it is unavoidable, and how much is due to the current?
The stack delivers $300\,\mathrm{A}$ . Compute the hydrogen consumption of one cell, then of the stack, in moles and grams per second.
Compute the electrical power of the stack.
Compute the heat power the cooling circuit must remove.

Part IV — The tank.

What amount of hydrogen is in $5.0\,\mathrm{kg}$ ?
What charge passes through the cells, counted over all of them, when this hydrogen is oxidised?
Compute the electrical energy delivered by the stack, with each cell at $0.70\,\mathrm{V}$ .
Convert it to kilowatt-hours.
For how long can the stack deliver its power of question 18?
State the electrical energy delivered by $5.0\,\mathrm{kg}$ of hydrogen at $0.70\,\mathrm{V}$ per cell.

Solution

Solution of Problem 9.1.

1. Anode: $\ce{H2 -> 2H+ + 2e-}$ ; cathode: $\ce{1/2O2 + 2H+ + 2e- -> H2O}$ ; cell: $\ce{H2 + 1/2O2 -> H2O}$ . 2. Two. 3. $E^\circ = 237\,140/(2 \times 96\,485) = 1.229\,\mathrm{V}$ . 4. $228\,580/192\,970 = 1.185\,\mathrm{V}$ . 5. $\Delta_r G^\circ$ differs by $\Delta_{\text{vap}}G^\circ(\ce{H2O}) = -228.58 + 237.14 = 8.56\,\mathrm{kJ}/\mathrm{mol}$ at $298\,\mathrm{K}$ : vapour is less stable than liquid there, and the voltage is lower by $8560/192\,970 = 0.044\,\mathrm{V}$ . 6. $\Delta_r H^\circ = -285.83\,\mathrm{kJ}/\mathrm{mol}$ ; $237.14/285.83 = 0.830$ . 7. $\Delta_r S^\circ = 69.95 - 130.68 - 102.58 = -163.3\,\mathrm{J}/(\mathrm{K}\,\mathrm{mol})$ . 8. $dE^\circ/dT = -163.3/192\,970 = -8.46 \times 10^{-4}\,\mathrm{V}/\mathrm{K}$ . 9. $1.229 - 8.46 \times 10^{-4} \times 55 = 1.182\,\mathrm{V}$ . 10. The same estimate at $400\,\mathrm{K}$ gives $1.229 - 8.46 \times 10^{-4} \times 101.85 = 1.143\,\mathrm{V}$ ; JANAF gives $221\,010/192\,970 = 1.145\,\mathrm{V}$ : treating $\Delta_r S^\circ$ as constant costs only $2\,\mathrm{mV}$ . 11. $T\Delta_r S^\circ = 298.15 \times (-163.3) = -48.7\,\mathrm{kJ}$ per mole: $48.7\,\mathrm{kJ}$ released. 12. $\Delta_r G^\circ(353) \approx -237.14 + 55 \times 0.1633 = -228.2\,\mathrm{kJ}$ ; with $\Delta_r H^\circ$ nearly unchanged, $228.2/285.8 = 0.80$ . 13. $2 \times 96\,485 \times 0.70 = 135.1\,\mathrm{kJ}$ . 14. $135.1/285.83 = 0.47$ ; $135.1/237.14 = 0.57$ . 15. $285.83 - 135.1 = 150.7\,\mathrm{kJ}$ . 16. $48.7\,\mathrm{kJ}$ would be released even by a reversible cell; the other $102.0\,\mathrm{kJ}$ come from the losses due to the current (resistance and electrode overpotentials). 17. One cell: $300/(2 \times 96\,485) = 1.55 \times 10^{-3}\,\mathrm{mol}/\mathrm{s}$ ; the stack: $400 \times 1.55 \times 10^{-3} = 0.622\,\mathrm{mol}/\mathrm{s}$ , $1.25\,\mathrm{g}/\mathrm{s}$ . 18. $400 \times 0.70 \times 300 = 84\,\mathrm{kW}$ . 19. $0.622 \times 150.7 = 93.7\,\mathrm{kW}$ : more heat than electricity. 20. $5000/2.016 = 2.48 \times 10^{3}\,\mathrm{mol}$ . 21. $2 \times 2480 \times 96\,485 = 4.79 \times 10^{8}\,\mathrm{C}$ . 22. Each coulomb passing through a cell gives $0.70\,\mathrm{J}$ : $4.79 \times 10^8 \times 0.70 = 3.35 \times 10^{8}\,\mathrm{J}$ . 23. $3.35 \times 10^8/3.6 \times 10^6 = 93\,\mathrm{kW}\,\mathrm{h}$ . 24. $93/84 = 1.1\,\mathrm{h}$ at full power. 25. $5.0\,\mathrm{kg}$ of hydrogen at $0.70\,\mathrm{V}$ per cell deliver $\boldsymbol{\approx 93\,\mathrm{kW}\,\mathrm{h}}$ of electrical energy.

Terms defined in this chapter

See all 852 terms in the glossary