Chapter 6 · Irreversible Oxidation · Ten playable problems

Ch 6 · P1 — Transaction or destruction

PROBLEM 51 · IRREVERSIBLE OXIDATION — TWO KINDS OF ELECTRON TRANSFER

Every oxidation moves electrons. Sort these three by what the transfer costs — ask what, if anything, stopped working.

Live model

the question is always: what stopped working?Drag each electron transfer into its column.

Ch 6 · P2 — No loyalty

PROBLEM 52 · IRREVERSIBLE OXIDATION — THE TREATMENT DILEMMA

The chlorine is aimed at the biofilm. The chapter says oxidants are not loyal to their intended target — why does that create a dilemma?

Live model

enough power to kill · not enough to spare the metalThe chlorine is aimed at the biofilm. Why is the pipe nervous?

Ch 6 · P3 — Four kill switches

PROBLEM 53 · IRREVERSIBLE OXIDATION — THE FOUR REQUIREMENTS

Electrochemical corrosion needs an anode, a cathode, an electrolyte, and a metal path — all four at once. Prove it: shut the cell down four different ways.

Live model

CURRENT FLOWING0 / 4 PROVEN

remove any one and the reaction haltsThe cell is running. Shut it down four different ways — 0 of 4 proven.

Ch 6 · P4 — The temperature paradox

PROBLEM 54 · IRREVERSIBLE OXIDATION — LESS OXYGEN, MORE DAMAGE

Warm the water from 70°F to 120°F. Oxygen solubility falls — and corrosion gets worse anyway. Watch the three bars, then explain it in one line.

Live model

less fuel, faster fire — the fire wins to a pointWarm the water from 70 to 120°F, then explain the paradox.

Ch 6 · P5 — Warm the water

PROBLEM 55 · IRREVERSIBLE OXIDATION — OXYGEN SOLUBILITY

Move from 77°F to 100°F and read the dissolved oxygen that remains on the line.

Live model

TARGET 100°F

linear estimate between two referencesSlide the temperature to 100°F.

Ch 6 · P6 — Three ten degree steps

PROBLEM 56 · IRREVERSIBLE OXIDATION — THE DOUBLING STAIRCASE

Corrosion rate roughly doubles every 10°C. Predict where 50°C lands, then run the staircase and see.

Live model

rate = 2^(ΔT / 10)Every 10°C doubles the rate. Predict where 50°C lands, then run the staircase.

Ch 6 · P7 — The price of rust

PROBLEM 57 · IRREVERSIBLE OXIDATION — 4Fe + 3O₂ → 2Fe₂O₃

Ten pounds of rust are scraped off a pipe. Pull iron out of the wall until the rust pile hits 10 lb, and read what the wall gave up.

Live model

TARGET 10 lb RUST

iron = rust × 0.699Drag iron out of the pipe wall until the rust pile reaches 10 pounds.

Ch 6 · P8 — Two probes, one honest answer

PROBLEM 58 · IRREVERSIBLE OXIDATION — READING ORP

Same pH, same conductivity. Tower A reads +650 mV, Tower B reads +200 mV. What can you reasonably conclude — and no more?

Live model

ORP is a response, not a concentrationTwo towers, same pH and conductivity. What do the probes license you to say?

Ch 6 · P9 — The cold deaerator

PROBLEM 59 · IRREVERSIBLE OXIDATION — THE FAILED STEAM VALVE

The DA's steam valve fails and feedwater drops to 140°F, air-saturated, at 50,000 lb/hr. Pick the concentration this broken machine actually delivers and read the oxygen marching into the boiler.

Live model

50,000 lb/hr × 4.4 ppm ÷ 10⁶ = 0.22 lb O₂/hrPick the concentration this broken deaerator actually delivers, and drop it on the line.

Ch 6 · P10 — Oxygen's unfair advantages

PROBLEM 60 · IRREVERSIBLE OXIDATION — THE PERSISTENT THREAT

Oxygen is not the strongest oxidant, yet it is the most persistent corrosion threat in industrial water. Pick its two advantages no applied oxidant can match.

Live model

0 / 2 PICKED

present by default · replenished for freePick oxygen’s two unmatched advantages. 0 of 2 picked.

Problem set

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Problem 1
What is the difference between transactional oxidation and destructive oxidation? Why is the formation of NaCl considered transactional while the rusting of iron is destructive?
Authored answer
Transactional oxidation transfers electrons without destroying a useful structure. Sodium and chlorine exchange electrons to form a stable salt, but no engineered material or living structure has lost its function. Destructive oxidation strips electrons from a working material, forcing it into a new and less useful form. Rusting converts iron metal into iron oxide; the pipe wall does not spontaneously rebuild itself under normal system conditions.
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Problem 2
The chapter states that oxidants are not loyal to their intended target. Explain what this means and why it creates a dilemma for water treatment.
Authored answer
Oxidizers are added to attack vulnerable biological structures, but the same electron-accepting chemistry can also drive metal corrosion when system conditions permit. Their reactions are selective, but that selectivity is controlled by chemistry rather than by the treatment objective. This creates a fundamental tension: enough oxidizing power must be maintained for biological control without creating unacceptable damage to system metallurgy.
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Problem 3
Four requirements must be present for electrochemical corrosion to occur in water. Name all four and explain why removing any single one stops the reaction.
Authored answer
Anode (a material capable of being oxidized), cathode (the electrically connected surface where electrons are consumed), electrolyte (the water, which allows dissolved ions to move), and metal path (allows electrons to move). Remove the anode: no material to oxidize. Remove the cathode: no electron acceptor. Remove the electrolyte: no ionic pathway to complete circuit. Remove the metal path: electrons cannot flow between anode and cathode. Without all four, electrochemical corrosion in water comes to a halt.
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Problem 4
In one sentence, explain the temperature paradox: why can corrosion get worse as water warms from 70°F to 120°F even though oxygen solubility decreases?
Authored answer
Warming decreases oxygen solubility but increases reaction kinetics. In warm water, there may still be enough dissolved oxygen present for the faster kinetics to dominate, producing higher net oxidation rates.
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Problem 5
Oxygen solubility is approximately 8.3 mg/L at 77°F and 4.4 mg/L at 140°F. Assuming linear behavior, estimate the solubility at 100°F. (Note: actual solubility follows a curve. This question uses linear approximation for simplicity.)
Authored answer
Linear interpolation: 100°F is 23/63 of the way from 77°F to 140°F. Δ = 8.3 − 4.4 = 3.9 mg/L. Decrease: 3.9 × (23/63) = 1.4 mg/L. Estimated DO at 100°F ≈ 8.3 − 1.4 = 6.9 mg/L.
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Problem 6
Using the Arrhenius rule of thumb (rate doubles per 10°C rise), calculate the approximate kinetic rate multiplier from 20°C to 50°C.
Authored answer
20°C → 50°C = +30°C = three 10°C steps. Rate multiplier = 2³ = 8×.
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Problem 7
The simplified rust reaction is: 4Fe + 3O₂ → 2Fe₂O₃. To produce 10 pounds of rust (Fe₂O₃), how many pounds of iron must be destroyed? (Fe = 55.8, O = 16, Fe₂O₃ = 159.6 g/mol)
Authored answer
Mass fraction of Fe in Fe₂O₃: (2 × 55.8) / 159.6 = 111.6 / 159.6 = 0.699 (≈70%). 10 lb rust × 0.699 = 6.99 lb of iron destroyed.
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Problem 8
Two cooling towers operate at the same pH and conductivity. Tower A reads +650 mV and Tower B reads +200 mV. What can you reasonably conclude from these readings? What additional information would you need before comparing biological control or corrosion risk?
Authored answer
Tower A is exhibiting a more oxidizing ORP response at the probe. That supports the conclusion that its bulk redox environment is more oxidizing at the time of measurement. However, ORP alone cannot establish oxidizer concentration, biological kill, or corrosion rate. The oxidizer used, residual concentration, temperature, demand, contact time, metallurgy, inhibitor program, flow conditions, and corrosion measurements would also be needed.
Problem 9
A deaerator normally heats boiler feedwater to 220°F, removing nearly all oxygen. The steam valve fails and feedwater drops to 140°F. If the system flows 50,000 lb/hr, how many pounds of dissolved oxygen per hour are now entering the boiler? Assume the feedwater is exposed to air and reaches approximately air-saturated dissolved oxygen at 140°F.
Authored answer
At 140°F, dissolved O₂ ≈ 4.4 mg/L = 4.4 ppm. 1 ppm = 1 lb O₂ per million lb water. 50,000 lb/hr ÷ 1,000,000 × 4.4 = 0.22 lb O₂/hr entering the boiler.
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Problem 10
The chapter says oxygen is not the strongest oxidant but has two advantages no applied oxidant can match. Name those advantages, and explain why they make oxygen the most persistent corrosion threat in industrial water systems.
Authored answer
Oxygen’s two unmatched advantages: (1) it is dissolved in almost every water system on Earth, and (2) it is constantly replenished by contact with air. Applied oxidants like chlorine or ozone must be dosed and maintained. Oxygen requires no deliberate addition. It is present by default in most waters and can be continuously replenished whenever the system contacts air.
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