Carbon capture, hydrogen economy (electrolysis, fuel cells): Exercises

Level 1 — Recognition

L1·Q1 — Name the route

A power plant burns coal, then passes the exhaust ("flue gas") through a cold tank of liquid amine; the $CO_{2}$ sticks to the amine and is stripped off later by heating. Which of the three capture routes is this — post-combustion, pre-combustion, or direct air capture — and why?

Recall Solution

Post-combustion. The clue is the word after: the coal is already burned, and we scrub $CO_{2}$ out of the exhaust that combustion produced. Pre-combustion would convert the fuel before burning; direct air capture works on ordinary outdoor air (~ $0.04%$ $CO_{2}$ ), not on a concentrated flue stream.

L1·Q2 — Colour the hydrogen

Classify each as green, grey, or blue hydrogen: (a) Made by electrolysing water with solar electricity. (b) Made from methane by steam reforming, with the $CO_{2}$ vented to air. (c) Made from methane by steam reforming, with the $CO_{2}$ captured and stored.

Recall Solution

(a) Green — electrolysis + renewable power, truly low-carbon. (b) Grey — steam reforming $CH_{4} + 2 H_{2} O \to CO_{2} + 4 H_{2}$ , $CO_{2}$ released. (c) Blue — grey chemistry plus carbon capture bolted on. See Steam Reforming and Industrial H2 for the reforming details.

L1·Q3 — Which electrode?

In water electrolysis, at which electrode does each event happen? (i) $H_{2}$ is released; (ii) oxidation occurs; (iii) electrons are given to the water.

Recall Solution

(i) Cathode — $2 H^{+} + 2 e^{-} \to H_{2}$ is a reduction (gain of electrons), and reduction is always at the cathode. (ii) Anode — oxidation defines the anode. (iii) Cathode — "electrons given to water" = reduction = cathode.

The picture below labels both electrodes and where each gas leaves — keep it in mind for every Faraday problem.

Figure — Carbon capture, hydrogen economy (electrolysis, fuel cells)

Level 2 — Application

L2·Q1 — Grams of hydrogen from charge

You pass a current of $I = 5.00 A$ for $t = 1930 s$ through acidic water. How many grams of $H_{2}$ form at the cathode? (Cathode: $2 H^{+} + 2 e^{-} \to H_{2}$ .)

Recall Solution

Step 1 — charge (WHAT/WHY). Charge is current × time, because current is defined as charge per second; multiply the flow rate by how long it flowed: $Q = I t = 5.00 \times 1930 = 9650 C$ . Step 2 — moles of electrons (WHY divide by $F$ ). We divide by $F$ because $F$ is the charge carried by exactly one mole of electrons ( $96485 C/mol$ ); dividing coulombs by coulombs-per-mole leaves moles: $n_{e^{-}} = 9650/96485 = 0.1000 mol$ . Step 3 — moles of $H_{2}$ (WHY divide by 2). We divide by $2$ because the half-reaction shows $2 e^{-}$ are consumed for every one $H_{2}$ molecule made — the electron count is the recipe ratio: $n_{H_{2}} = 0.1000/2 = 0.0500 mol$ . Step 4 — grams (WHY multiply by $M$ ). We multiply by the molar mass because mass = moles × grams-per-mole ( $M_{H_{2}} = 2.0 g/mol$ ): $0.0500 \times 2.0 = 0.10 g H_{2}$ .

The flowchart below is the fixed pipeline every Faraday problem walks through — memorise the four boxes, not the numbers.

This is Faraday's Laws of Electrolysis in action: mass $\propto$ charge, divided by the electron count of the half-reaction.

L2·Q2 — Oxygen at the other electrode, at the same time

For the same run as L2·Q1, how many grams of $O_{2}$ form at the anode? (Anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ .)

Recall Solution

Step 1 — same electrons. The same $0.1000 mol$ of electrons flows through both electrodes, because they are wired in series — one and the same current passes through each in turn. Step 2 — moles of $O_{2}$ (WHY divide by 4). We divide by $4$ because the anode half-reaction shows $4 e^{-}$ are released per one $O_{2}$ molecule: $n_{O_{2}} = 0.1000/4 = 0.0250 mol$ . Step 3 — grams (WHY multiply by $M$ ). Mass = moles × molar mass, with $M_{O_{2}} = 32.0 g/mol$ : $0.0250 \times 32.0 = 0.80 g O_{2}$ . Sanity check: $n_{H_{2}} : n_{O_{2}} = 0.0500 : 0.0250 = 2 : 1$ , exactly the $2 H_{2} O \to 2 H_{2} + O_{2}$ ratio. Electrons are the shared currency.

L2·Q3 — Amine and Le Chatelier

Industrial capture uses MEA (monoethanolamine), an amine written $RNH_{2}$ , where R is a placeholder for the rest of the organic molecule — an alkyl group (here $R = HOCH_{2} CH_{2} -$ ). We keep it as "R" because only the reactive $- NH_{2}$ end matters chemically. The capture equilibrium is $CO_{2} + 2 RNH_{2} ⇌ RNHCOO^{-} + RNH_{3}^{+}$ where $RNHCOO^{-}$ is the carbamate (the amine now bonded to $CO_{2}$ ) and $RNH_{3}^{+}$ is a protonated amine. The forward (capture) reaction is exothermic. In the absorber we want maximum capture; in the stripper we want maximum release. What temperature does each need, and why?

Recall Solution

Absorber → cold. For an exothermic forward reaction, lowering $T$ shifts equilibrium forward (the system "adds heat back" by favouring the heat-releasing direction). Cold ⇒ more $CO_{2}$ grabbed. Stripper → hot. Heating shifts equilibrium backward, releasing pure $CO_{2}$ and regenerating the amine $RNH_{2}$ . This temperature swing is the whole engineering trick — and its main energy cost. See Le Chatelier's Principle.

Level 3 — Analysis

L3·Q1 — Where does 1.23 V come from?

The Gibbs free energy to form water from its elements, for the reaction $2 H_{2} + O_{2} \to 2 H_{2} O$ , is $Δ G^{\circ} = - 474 kJ$ (so splitting water costs $+ 474 kJ$ for those $2$ mol of water). Using $E^{\circ} = - Δ G^{\circ} / (n F)$ with $n = 4$ electrons, verify the reversible splitting voltage is $1.23 V$ . Then explain why $n = 4$ here.

Recall Solution

The formula (WHY this tool). $Δ G^{\circ} = - n F E^{\circ}$ is the bridge from Gibbs energy to voltage: it says the electrical work a cell can do equals its free-energy change. Rearranged, $E^{\circ} = - Δ G^{\circ} / (n F)$ . Plug in (WHY the numbers go where they do). Splitting reverses the sign, so $Δ G^{\circ} = + 474000 J$ ; we put charge-per-mole $F$ and electron count $n$ in the denominator because they convert energy-per-mole into energy-per-coulomb, i.e. volts: $E^{\circ} = \frac{474000}{4 \times 96485} = \frac{474000}{385940} = 1.228 \approx 1.23 V .$ Why $n = 4$ ? Balance the overall reaction $2 H_{2} O \to 2 H_{2} + O_{2}$ : the cathode makes $2 H_{2}$ using $4 e^{-}$ , and the anode makes $1 O_{2}$ releasing $4 e^{-}$ . Four electrons move per turnover of the balanced equation, so $Δ G^{\circ} = + 474 kJ$ (for $2$ mol water) pairs with $n = 4$ . If you'd written the reaction for one water molecule, both $Δ G^{\circ}$ and $n$ halve, and $E^{\circ}$ is unchanged — as it must be (voltage is intensive).

L3·Q2 — Efficiency lost to overpotential

Real electrolysis of this cell runs at an applied voltage of $V_{app} = 1.85 V$ , while the thermodynamic floor is $1.23 V$ . Define voltage efficiency as $η_{V} = E_{cell}^{\circ} / V_{app}$ . Compute it, and say in words where the "lost" voltage went.

Recall Solution

We take the ratio of the useful minimum to what we actually paid, because efficiency asks "how much of my input was strictly necessary?": $η_{V} = \frac{1.23}{1.85} = 0.665 = 66.5% .$ The extra $1.85 - 1.23 = 0.62 V$ is overpotential plus ohmic (resistance) losses — the kinetic "friction" of forcing electrons across the electrode surfaces. It appears as heat, not as stored chemical energy. So about $\frac{1}{3}$ of the electrical input is wasted before any $H_{2}$ is banked.

L3·Q3 — Fuel cell vs Carnot

A hydrogen–oxygen fuel cell delivers electricity at an ideal efficiency near $83%$ (from $Δ G /Δ H$ ). A steam turbine draws heat from a boiler at $T_{hot} = 800 K$ and rejects it at $T_{cold} = 320 K$ . Compute the turbine's Carnot ceiling and compare.

Recall Solution

Carnot ceiling (WHY this tool). Any heat engine is capped by $η_{Carnot} = 1 - T_{cold} / T_{hot}$ — the Carnot limit — because you cannot turn all heat into work. $η_{Carnot} = 1 - \frac{320}{800} = 1 - 0.40 = 0.60 = 60% .$ Compare. The fuel cell's $83%$ ceiling beats the turbine's $60%$ ceiling because a fuel cell is not a heat engine — it converts chemical energy directly to electricity and never has to obey Carnot. That is the structural advantage, not just a better engine.

The bar chart below puts the two ceilings side by side — the fuel cell's bar simply is not bound by the Carnot line.

Level 4 — Synthesis

L4·Q1 — Sizing a green-hydrogen plant

A solar-powered electrolyser must produce $1.00 kg$ of $H_{2}$ per hour. (a) How many moles of electrons per hour are needed? (b) What steady current (in A) does that require? (Recall $1 h = 3600 s$ .)

Recall Solution

(a) Step 1 — moles of $H_{2}$ (WHY divide by $M$ ). Moles = mass ÷ molar mass, because molar mass is grams-per-mole: $1000 g /2.0 g/mol = 500 mol$ . Step 2 — moles of electrons (WHY multiply by 2). Each $H_{2}$ needs $2 e^{-}$ (the half-reaction recipe), so $n_{e^{-}} = 500 \times 2 = 1000 mol e^{-} per hour$ . (b) Step 1 — charge (WHY multiply by $F$ ). Charge = moles of electrons × $F$ , because $F$ is coulombs-per-mole: $Q = 1000 \times 96485 = 9.6485 \times 1 0^{7} C$ per hour. Step 2 — current (WHY divide by $t$ ). Current = charge ÷ time, because current is charge per second: $I = \frac{9.6485 \times 1 0 ^{7}}{3600} = 2.680 \times 1 0^{4} A \approx 2.68 \times 1 0^{4} A (26.8 kA) .$ That huge current is why real electrolysers stack many cells: same current, many parallel plates.

L4·Q2 — Energy budget of the loop

Splitting $2$ mol of water stores $Δ G^{\circ} = 474 kJ$ (thermodynamic minimum). But you supplied it at $1.85 V$ instead of $1.23 V$ . Then the fuel cell gives it back at an ideal $83%$ of the same $Δ G$ . (a) Actual electrical energy input for those $2$ mol of water? (b) Electrical energy recovered by the fuel cell? (c) Round-trip efficiency = output ÷ input.

Recall Solution

(a) Input (WHY $n F V$ and WHY $n = 4$ ). Electrical energy = charge × voltage = $n F \times V_{app}$ , because charge is $n F$ and energy is charge times the volts pushed across it. We use $n = 4$ because splitting $2$ mol of water ( $2 H_{2} O \to 2 H_{2} + O_{2}$ ) moves exactly four electrons per turnover — the same electron count that fixed $1.23 V$ in L3·Q1: $W_{in} = 4 \times 96485 \times 1.85 = 714000 J = 714 kJ .$ (Check: at the ideal $1.23 V$ this would be $474 kJ$ — matches $Δ G^{\circ}$ , good.) (b) Output (WHY multiply by $0.83$ ). The fuel cell returns only its ideal fraction of the stored energy, because $83%$ is how much of $Δ G^{\circ}$ it converts to electricity: $W_{out} = 0.83 \times 474 kJ = 393 kJ .$ (c) Round-trip (WHY output ÷ input). Efficiency = what you got back ÷ what you put in: $η_{round} = \frac{393 kJ}{714 kJ} = 0.551 \approx 55% .$ Meaning: roughly half the input survives the store-and-release loop. Losses split between electrolysis overpotential (input inflated to $714$ from $474 kJ$ ) and fuel-cell irreversibility (output cut to $83%$ ). Cleaner electrodes shrink both.

The energy-flow (Sankey-style) bar below shows exactly where the $714 kJ$ input goes — dark chunk lost to overpotential, then a further chunk lost in the cell, leaving $393 kJ$ .

Level 5 — Mastery

L5·Q1 — Full-loop carbon accounting (grey vs green)

A factory needs $10.0 kg$ of $H_{2}$ . (a) If made grey by steam reforming, $CH_{4} + 2 H_{2} O \to CO_{2} + 4 H_{2}$ , how many kg of $CO_{2}$ are emitted? (Assume the reaction is the only $CO_{2}$ source.) (b) If made green by electrolysis, and then all this $H_{2}$ is later burned in fuel cells, what is the only product? (c) A "blue" retrofit captures $90%$ of the grey $CO_{2}$ using $Ca(OH)_{2} + CO_{2} \to CaCO_{3} + H_{2} O$ . How many kg of $CaCO_{3}$ solid form? ( $M_{CaCO_{3}} = 100 g/mol$ .)

Recall Solution

(a) Grey $CO_{2}$ . Moles $H_{2} = 10 000/2.0 = 5000 mol$ (mass ÷ molar mass). Stoichiometry: the balanced equation makes $4 H_{2}$ per $1 CO_{2}$ , so divide by 4 to convert $H_{2}$ moles into $CO_{2}$ moles: $n_{CO_{2}} = 5000/4 = 1250 mol$ . Mass = moles × molar mass = $1250 \times 44.0 = 55000 g = 55.0 kg C O_{2}$ . (b) Green. Combustion / fuel-cell product of $H_{2}$ is only water, $H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ . (Provided the electricity was renewable — otherwise the $CO_{2}$ just moved upstream to the power plant.) (c) Blue capture. Captured $CO_{2} = 0.90 \times 1250 = 1125 mol$ . The reaction locks $1$ mol $CO_{2}$ into $1$ mol $CaCO_{3}$ (a $1 : 1$ ratio), so $n_{CaCO_{3}} = 1125 mol$ . Mass = moles × molar mass = $1125 \times 100 = 112500 g = 112.5 kg CaC O_{3}$ . $CO_{2}$ is a mild acid anhydride; the strong base $Ca(OH)_{2}$ neutralises it into stable limestone — permanent solid storage.

L5·Q2 — The one-line reversal

Write the overall electrolysis reaction and the overall fuel-cell reaction side by side, state the sign of $Δ G^{\circ}$ and of $E_{cell}^{\circ}$ for each, and explain in one sentence why the magnitude $1.23 V$ is identical.

Recall Solution

	Overall reaction	$Δ G^{\circ}$	$E_{cell}^{\circ}$
Electrolysis	$2 H_{2} O \to 2 H_{2} + O_{2}$	$+ (non-spontaneous)$	$- 1.23 V$ (must apply $\geq 1.23$ )
Fuel cell	$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$	$- (spontaneous)$	$+ 1.23 V$ (delivers)

They are the same redox reaction reversed. Reversing a reaction flips the sign of both $Δ G^{\circ}$ and $E^{\circ}$ but not their magnitude — because $Δ G^{\circ} = - n F E^{\circ}$ ties them to the same water-formation energy either way.

L5·Q3 — Design a low-carbon loop end to end

A remote solar farm has surplus daytime electricity and wants to deliver reliable night-time power. Sketch the round-trip chain in words, name the reaction and the sign of $Δ G^{\circ}$ at each conversion, and state the one condition that makes the whole loop genuinely low-carbon.

Recall Solution

Daytime (store surplus). Run electrolysis: $2 H_{2} O \to 2 H_{2} + O_{2}$ , with $Δ G^{\circ} > 0$ (non-spontaneous), so the surplus electricity is pushed in and banked as $H_{2}$ — needing $\geq 1.23 V$ (more in practice, from overpotential). Store the $H_{2}$ . Night-time (release). Run the fuel cell: $H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ , with $Δ G^{\circ} < 0$ (spontaneous), delivering $+ 1.23 V$ of electricity back — dodging the Carnot limit that would cap a turbine. The one condition. The loop is low-carbon only if the electrolysis electricity is renewable (here, solar) — i.e. green hydrogen. If the electrons came from a fossil grid, the $CO_{2}$ merely moved upstream, exactly the L5·Q1 lesson and a core green-chemistry principle. Round-trip cost. From L4·Q2, only ~ $55%$ of the stored energy returns — the price of storing electricity as chemistry rather than in a wire.

Recall One-line self-test (do this last)

Cover every answer. Which single quantity is the shared currency linking Faraday mass, the $2 : 1$ gas ratio, the $1.23 V$ , and the electron count $n$ ? ::: The mole of electrons — charge $Q = n_{e^{-}} F$ threads through all of them.