Worked examples — Carbon capture, hydrogen economy (electrolysis, fuel cells)

This page is the problem gym for the parent topic. Before touching numbers, we lay out every distinct kind of question this topic can ask, then hit each one with a fully worked example. Nothing here uses a symbol we did not earn in the parent note — but we will re-anchor each tool the first time it appears.

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The scenario matrix

Think of a problem as living in a grid. Each row is a physical quantity you might be asked about; each column is the flavour of the ask — a plain forward calculation, a reverse/backward one, a zero-or-degenerate edge case, or a real-world twist. We must land at least one example in every column.

Case class	What makes it special	Hit by
Charge → mass (Faraday, forward)	Given current & time, find product mass	Ex 1
Mass → charge/time (reverse)	Given product wanted, find current or time	Ex 2
Voltage from thermodynamics	Turn $\Delta G^{\circ }$ into a cell voltage (sign matters!)	Ex 3
Overpotential / real energy cost	Practical voltage $>$ theoretical; energy per kg	Ex 4
Sign-flip / degenerate case	Fuel cell vs electrolysis: same reaction, opposite sign	Ex 5
Efficiency vs Carnot (limiting comparison)	Compare a heat engine's ceiling to a fuel cell	Ex 6
Le Chatelier edge (zero net / equilibrium shift)	Which way does amine capture go at a given $T$?	Ex 7
Real-world word problem (whole-system)	A power plant sizing / emissions question	Ex 8
Exam twist (stoichiometry trap)	${\text{O}}_2$ needs 4 $e^{-}$, ${\text{H}}_2$ needs 2 — mixing them up	Ex 9

We will keep re-using three tools, so pin them down once.

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Ex 1 — Charge → mass (the forward Faraday case)

Forecast: guess an order of magnitude first. A few amps for half an hour — do you expect grams or kilograms? (Answer: a fraction of a gram.)

1. Convert time to seconds, then charge. $t=30.0\times 60=1800\text{s}$, so $Q=It=5.00\times 1800=9000\text{C}$. Why this step? Faraday's law counts charge, not current — current is only "charge per second," so we must multiply by the number of seconds.

2. Charge → moles of electrons. $n_{e^{-}}=Q/F=9000/96485=0.09328\text{mol}$. Why this step? $F$ is the exchange rate "coulombs per mole of electrons." Dividing by it converts our pile of coulombs into a count of electron-moles.

3. Electrons → moles of ${\text{H}}_2$. Cathode: $2{\text{H}}^{+}+2e^{-}\to {\text{H}}_2$, so 2 electrons per ${\text{H}}_2$. $n_{{\text{H}}_2}=0.09328/2=0.04664\text{mol}$. Why this step? The half-reaction is the "recipe": it tells us the electron : molecule ratio. Skipping it is the classic trap (Ex 9).

4. Moles → grams. $M({\text{H}}_2)=2.016\text{g/mol}$, so mass $=0.04664\times 2.016=0.0940\text{g}$. Why this step? The question asked for grams, but stoichiometry only ever gives us moles. Molar mass $M$ is the "grams per mole" price tag that converts a molecule-count into a weighable mass.

Verify: units cancel as $\frac{\text{C}}{\text{C/mol}}=\text{mol}$, then $\text{mol}\times \text{g/mol}=\text{g}$. And $0.094\text{g}$ is indeed "a fraction of a gram" — matches the forecast.

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Ex 2 — Mass → current/time (running Faraday backwards)

Forecast: 1 kg of the lightest molecule is a lot of moles. Expect a very large current — hundreds of amps at least.

1. Mass → moles ${\text{H}}_2$. $n_{{\text{H}}_2}=1000/2.016=496.03\text{mol}$. Why this step? Faraday's law speaks in electrons, and electrons only connect to moles, not grams — so we must first turn the target mass into a mole-count using $M$.
2. Moles ${\text{H}}_2$ → moles electrons. $\times 2=992.06\text{mol }{e}^{-}$. Why? Reverse of Ex 1 step 3: each ${\text{H}}_2$ consumed 2 electrons, so making this many needs twice as many electron-moles.
3. Electrons → charge. $Q=n_{e^{-}}F=992.06\times 96485=9.573\times 10^7\text{C}$. Why this step? We can't set a current from moles of electrons directly. Multiplying by $F$ (coulombs per mole of electrons) converts the electron-count into the total charge that has to flow — the quantity current is built from.
4. Charge → current. Spread over one hour $=3600\text{s}$: $I=Q/t=9.573\times 10^7/3600=2.659\times 10^4\text{A}$. Why this step? Current is charge per second, so to find the steady amps we divide the total charge by the time window it must flow in — the exact inverse of $Q=It$ from Ex 1.

Verify: enormous, as forecast — real industrial cells are stacked in parallel precisely because a single cell can't sink tens of kiloamps. Sanity re-check: scaling Ex 1's $0.094\text{g per }5\text{A}\cdot 0.5\text{h}$ up to $1000\text{g/h}$ predicts a current of order $5\times \frac{1000}{0.094}\times \frac{1}{2}\approx 2.7\times 10^4$ A. ✓

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Ex 3 — Voltage from thermodynamics (sign matters)

Forecast: the parent note already whispered the answer — around $1.2\text{V}$. Confirm it and watch the sign.

1. Set up the bridge. $E^{\circ }=-\Delta G^{\circ }/(nF)$. Why this tool and not another? $\Delta G^{\circ }$ is energy per reaction event; dividing by "charge moved per event" ($nF$) converts joules-per-mole into joules-per-coulomb, which is volts. No other formula turns thermodynamics into a voltage.

2. Plug in. $\Delta G^{\circ }=+474000\text{J}$, $n=4$, $F=96485$: $E^{\circ }=\frac{-474000}{4\times 96485}=-1.228\text{V}.$ Why this step? We do the arithmetic to turn the abstract bridge into a number we can act on: $nF=4\times 96485$ is the total charge (in coulombs) shifted per reaction event, and dividing the energy by it lands us in exact joules-per-coulomb, i.e. volts.

3. Interpret the minus sign. $E_{\text{cell}}^{\circ }$ came out negative → the reaction is non-spontaneous → you must supply at least $1.23\text{V}$ to force it. The magnitude is what we call "the minimum voltage."

Verify: $4\times 96485\times 1.228=4.74\times 10^5\text{J}=474\text{kJ}$, back to $\Delta G^{\circ }$. ✓ See figure below for the energy landscape.

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Ex 4 — Overpotential & the real energy bill

Forecast: efficiency should be below 100% (you're paying extra), and the real energy bill should exceed $474\text{kJ}$.

1. Overpotential = extra volts over the thermodynamic floor. ${\eta }_{\text{over}}=1.90-1.23=0.67\text{V}$. Why? $1.23\text{V}$ is the reversible minimum; everything above it is kinetic "friction" at the electrodes, dissipated as heat.

2. Voltage efficiency = floor / actual. $\epsilon =1.23/1.90=0.647=64.7\%$. Why this ratio? It measures how much of the energy you paid for went into useful water-splitting versus wasted heat.

3. Real energy = charge $\times$ applied volts. Charge for $n=4$ mol $e^{-}$: $Q=4\times 96485=3.859\times 10^5\text{C}$. $E_{\text{elec}}=Q\times E_{\text{applied}}=3.859\times 10^5\times 1.90=7.33\times 10^5\text{J}=733\text{kJ}$. Why this step? A volt is defined as a joule per coulomb, so energy $=$ charge $\times$ voltage. We use the applied $1.90\text{V}$ (not the $1.23\text{V}$ floor) because that is the voltage the power supply actually pushes every coulomb through — the true bill includes the wasted overpotential.

Verify: $733\text{kJ}>474\text{kJ}$ (you paid more than the thermodynamic minimum ✓), and $474/733=0.647=\epsilon$ (energy efficiency equals voltage efficiency because charge cancels ✓).

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Ex 5 — The sign-flip / degenerate case (fuel cell = electrolysis reversed)

Forecast: same $1.23\text{V}$ magnitude, but now positive because the reaction is spontaneous.

1. Apply the bridge. $E^{\circ }=-\Delta G^{\circ }/(nF)=-(-237000)/(2\times 96485)=+1.228\text{V}$. Why positive now? $\Delta G^{\circ }<0$ (energy released), so $-\Delta G^{\circ }>0$, so $E^{\circ }>0$ — the cell delivers voltage.

2. Check the "degenerate" consistency. Electrolysis (Ex 3) was $2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2$, $\Delta G^{\circ }=+474\text{kJ}$ for 2 waters. Per one water that's $+237\text{kJ}$. The fuel cell is exactly this reversed → $-237\text{kJ}$. Same reaction, opposite sign — the "degenerate pair." Why care? This is the whole conceptual unity of the chapter: one reaction, two directions.

Verify: $2\times (-237)=-474\text{kJ}$ is the negative of the electrolysis $\Delta G^{\circ }=+474\text{kJ}$. Signs are exact mirror images. ✓ (See Electrochemistry — Standard Electrode Potentials.)

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Ex 6 — Efficiency vs Carnot (the limiting comparison)

Forecast: the turbine's ceiling is high on paper; but real turbines fall far short, and the fuel cell isn't even playing the same game.

1. Carnot ceiling. ${\eta }_{\text{max}}=1-T_{\text{cold}}/T_{\text{hot}}=1-300/1500=0.800=80\%$. Why this formula and not another? Any heat engine — anything that converts heat into work — is capped by the ratio of temperatures it works between. This is a ceiling, not the real value.

2. Reality vs ceiling. Real turbines reach maybe $35$–$45\%$, well under 80%, because of irreversibilities.

3. Fuel cell is not a heat engine. It converts chemical $\to$ electrical energy directly, never routing energy through "heat at temperature $T$." So Carnot's $1-T_c/T_h$ simply does not apply — 55% is genuine, and it can in principle exceed the practical engine number.

Verify: $1-300/1500=0.8$. And note $55\%>45\%$ (a good real turbine), the point the parent note made about fuel cells "dodging Carnot." ✓

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Ex 7 — Le Chatelier edge case (which way does capture go?)

Forecast: cold → grab ${\text{CO}}_2$ (forward); hot → release it (reverse). Somewhere between lies balance.

1. Cold absorber ($40°\text{C}$). Forward is exothermic (releases heat). By Le Chatelier's Principle, removing heat (staying cold) pushes the system to make more heat → forward → ${\text{CO}}_2$ absorbed. Why? Le Chatelier: a system opposes an imposed stress; low $T$ is a "heat deficit" the exothermic forward direction relieves. Direction: forward (capture).
2. Hot stripper ($120°\text{C}$). Adding heat pushes the reaction to consume heat → reverse → ${\text{CO}}_2$ released, amine regenerated. Why this step? Same Le Chatelier logic, mirrored: heating is a "heat surplus" stress. The system relieves it by running the direction that absorbs heat — and since forward is exothermic, the heat-absorbing direction is the reverse (endothermic) one. That is exactly why the stripper is run hot: to force release. Direction: reverse (release).
3. The degenerate "zero net" point. At some intermediate $T$ the forward and reverse rates match — net reaction $=0$ (equilibrium with no useful transfer). Engineers deliberately operate away from it, swinging between 40 and 120 °C. This $T$-swing is the main energy cost. Why this step? We must cover the degenerate case the matrix demands: at the crossover temperature neither absorption nor release wins, so no ${\text{CO}}_2$ is net moved. Naming it explains why a single fixed temperature is useless — you need the swing between two temperatures on opposite sides of this balance point to shuttle ${\text{CO}}_2$ in one vessel and out the other.

Verify: consistency check — the parent note states "low $T$ favours absorption, high $T$ favours release." Our two directions match exactly, and the acid–base driver (a base grabbing acidic ${\text{CO}}_2$) is in Acids, Bases and Anhydrides. ✓

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Ex 8 — Real-world whole-system word problem

Forecast: megawatts for an hour is gigajoules of energy → expect tens of kilograms.

1. Electrical energy in one hour. $E=Pt=2.00\times 10^6\text{W}\times 3600\text{s}=7.20\times 10^9\text{J}$. Why this step? Power is energy per second (a watt is a joule per second), so to get the total energy delivered we multiply the steady power by the number of seconds it flows — here one full hour. We need a total energy, not a rate, before any efficiency can be applied.
2. Useful chemical energy captured as ${\text{H}}_2$. $E_{\text{chem}}=0.65\times 7.20\times 10^9=4.68\times 10^9\text{J}$. Why multiply by 0.65? Overpotential and resistance (the losses of Ex 4) mean only $65\%$ of the electricity survives as stored fuel energy. We must strip out the wasted $35\%$ before converting to ${\text{H}}_2$, or we would overcount the hydrogen.
3. Chemical energy → moles ${\text{H}}_2$. Each mol stores $237000\text{J}$: $n=4.68\times 10^9/237000=1.975\times 10^4\text{mol}$. Why this step? $\Delta G^{\circ }$ is the energy banked per mole of ${\text{H}}_2$, so dividing the total captured energy by it counts how many moles that energy corresponds to.
4. Moles → mass. $m=1.975\times 10^4\times 2.016=3.98\times 10^4\text{g}\approx 39.8\text{kg}$. Why this step? The question wants kilograms; molar mass $M$ converts the mole-count into a weighable mass, then $÷1000$ to reach kg.

Verify: order of magnitude "tens of kg," as forecast ✓. Is it green? Yes — the electrons came from wind (no ${\text{CO}}_2$). Had the power come from a gas plant, the same ${\text{H}}_2$ would be grey. The steam-reforming alternative that would emit ${\text{CO}}_2$ is in Steam Reforming and Industrial H2; running the Green Chemistry — 12 Principles lens, this route scores on "renewable feedstock."

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Ex 9 — Exam twist: the stoichiometry trap (${\text{O}}_2$ vs ${\text{H}}_2$)

Forecast: by the overall reaction $2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2$, you get half as many molecules of ${\text{O}}_2$ as ${\text{H}}_2$ — but ${\text{O}}_2$ is heavier. Guess: is the ${\text{O}}_2$ mass bigger or smaller than the $0.094\text{g}$ of ${\text{H}}_2$?

1. Electrons available. $n_{e^{-}}=9000/96485=0.09328\text{mol}$ (same charge as Ex 1). Why this step? The anode and cathode share the same current, so the same charge — and therefore the same electron-mole count — passes both. We reuse Ex 1's electron pile rather than recomputing it.
2. Use the anode recipe. Anode: $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}$ → 4 electrons per ${\text{O}}_2$. $n_{{\text{O}}_2}=0.09328/4=0.02332\text{mol}$. Why /4 not /2? This is the trap: ${\text{O}}_2$ demands 4 electrons, twice what ${\text{H}}_2$ needs. Same charge therefore makes half the moles of ${\text{O}}_2$ as of ${\text{H}}_2$ — consistent with the 2:1 ratio in $2{\text{H}}_2:{\text{O}}_2$. Using the wrong divisor (2) is precisely the exam mistake this example exists to inoculate against.
3. Moles → grams. $M({\text{O}}_2)=32.00\text{g/mol}$: $m=0.02332\times 32.00=0.746\text{g}$. Why this step? The question wants grams; molar mass converts the ${\text{O}}_2$ mole-count into mass. Note oxygen's $M$ is $\sim 16\times$ that of ${\text{H}}_2$, which is why fewer molecules still weigh far more.

Verify: mole ratio $n_{{\text{H}}_2}:n_{{\text{O}}_2}=0.04664:0.02332=2:1$ ✓ (matches the overall equation). And $0.746\text{g of }{\text{O}}_2>0.094\text{g of }{\text{H}}_2$ despite fewer molecules — because oxygen is $\sim 16\times$ heavier. Mass ratio $0.746/0.094\approx 7.9\approx 8$, exactly $\frac{32}{2}\times \frac{1}{2}$. ✓

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Recall

Recall Same charge, why less

${\text{O}}_2$ than ${\text{H}}_2$ by moles? ${\text{O}}_2$ needs 4 electrons per molecule, ${\text{H}}_2$ needs 2, so equal charge makes half the moles of ${\text{O}}_2$.

Recall Why does the fuel-cell voltage have the opposite sign to electrolysis?

Because $\Delta G^{\circ }$ flips sign when the reaction is reversed, and $E^{\circ }=-\Delta G^{\circ }/nF$ carries that sign through: spontaneous ($\Delta G<0$) ⇒ positive delivered voltage.