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EC-012 · Faraday's Law and Corrosion Rates July 26, 2026
EC TRACK · ELECTROCHEMISTRY & THE GALVANIC SERIES

Faraday's Law and Corrosion Rates

Current into pounds, pounds into years - the arithmetic behind anode life, corrosion rates, and the expected-life line on every galvanic quote.

Foundation ~10 minutes PDH/CEC eligible

Apply · three problems

Run the chain

Three problems, one chain — run forward, run from the middle, then run in both directions at once. In the first, the creek-bottom tank from the Read lesson finally settles up. In the second, a records review ends in the exact sentence this set once promised you’d be able to say. In the third, one meter reading writes two bills, and you get to read both.

Work each one on paper (or in your head on the tailgate) before you open the reveal. The reveals walk the reasoning step by step — the point isn’t the answer, it’s the path.

Ground rules. Everything you need is from the Read lesson: actual capacities — magnesium ≈ 500 A-hr/lb, zinc ≈ 335 A-hr/lb · steel’s exchange rate ≈ 20.1 lb per amp-year · 8,760 hours in a year · and Ohm’s Law off a shunt, I = V ÷ R. Round sensibly — field answers carry a “give or take,” not four decimals.

Problem 1

The creek bottom settles up

The job. The creek-bottom water tank finally gets its answer. Its zinc string went in with the tank; each bar connects through the test station with a 0.1 Ω shunt in the lead. Kneel at the test head, put the meter across the first bar’s shunt: 12.0 mV.
SITE B — CREEK BOTTOM
Soilwet gray clay
Resistivity~400 Ω·cm
Moisturesaturated year-round
THE STRING
Metalhigh-purity zinc
Barsstandard 30-lb
Capacity, actual≈ 335 A-hr/lb
Shunt0.1 Ω

A — What’s this bar’s output, and which way is the current headed?

Show solution — part A

Ohm’s Law, straight off the shunt:

I = V ÷ R = 0.0120 V ÷ 0.1 Ω = 0.120 A = 120 mA

And the polarity of that drop is information, not decoration: it tells you the current is flowing from the bar to the tank — the anode is feeding the structure, not the other way around. If that sign ever flips on you, stop trusting the hookup and find out why before you write anything down.

B — Run the chain: how many years does this bar have?

Show solution — part B

Pounds to the bank, bank to hours, hours to years:

The bank: 30 lb × 335 A-hr/lb = 10,050 A-hr
The hours: 10,050 ÷ 0.120 A = 83,750 hr
The years: 83,750 ÷ 8,760 ≈ 9.6 years

Tailgate answer: “Call it nine and a half — ten if the summers run dry.”

Put it next to the hilltop and the pattern shows: the magnesium up there promised about nineteen years on 50 mA. This bar carries nearly two and a half times the current, because wet clay barely resists. Wet ground works its anodes harder — more metal per year is the price of easy soil.

C — The operator remembers the hilltop’s nineteen years: “Should we have put magnesium down here instead?”

Show solution — part C

Run it before you answer. A hilltop-size 17-lb magnesium bar asked to carry this same 120 mA:

17 lb × 500 A-hr/lb = 8,500 A-hr ÷ 0.120 A = 70,833 hr ≈ 8.1 years

Shorter, not longer — the smaller bank empties faster at the bigger draw. And in practice it’s worse than that: magnesium’s stronger push in soil this easy wouldn’t hold 120 mA — it would drive more, spend itself faster still, and feed the tank current it doesn’t need.

Zinc was the right call twice over: the gentle pusher matched to easy dirt when the string was chosen, and now the arithmetic that proves it. Same verdict, two kinds of evidence.

Wet ground works its anodes harder. The right metal isn’t the strongest pusher — it’s the one whose push matches the dirt, and the chain turns that judgment into a number you can put in a report.


Problem 2

Eleven more years

The job. Different line, same standard bars. A short bare tie-in has been protected by a small string of 17-lb standard-alloy magnesium anodes since a project five years ago. The annual reads have held steady around 60 mA per bar. The operator’s question this time isn’t about new anodes — it’s about the ones already in the ground: “How much life is left?”

A — What’s already been spent, and what’s left in the bank?

Show solution — part A

Five years of steady draw, converted to amp-hours:

Spent: 0.060 A × 5 yr × 8,760 hr/yr = 2,628 A-hr
The bank: 17 lb × 500 A-hr/lb = 8,500 A-hr
Remaining: 8,500 − 2,628 ≈ 5,870 A-hr

One honest footnote: this assumes the output really did hold near 60 the whole time — which is exactly what the annual reads are for. Records turn an assumption into an estimate.

B — At today’s draw, how many more years?

Show solution — part B
5,870 A-hr ÷ 0.060 A = 97,833 hr ÷ 8,760 ≈ 11.2 years

So the records review closes with one sentence, said out loud: “That anode lasts eleven more years.”

Current into pounds, pounds into years — the whole module in one line on a tailgate. That sentence is the skill.

C — So the replacement dig goes on the eleven-year plan?

Show solution — part C

No — eleven is when the math says empty, not when the plan says dig. Anodes come out around the 85% line, because the last stretch of a bar can’t hold dependable output:

Usable bank: 0.85 × 8,500 = 7,225 A-hr
Room left before the line: 7,225 − 2,628 = 4,597 A-hr ÷ 0.060 A ≈ 76,600 hr ≈ 8.7 years

Put the dig in the eight-to-nine-year window. You’d rather replace on schedule than on failure — and the schedule is the same chain with one planning factor on top.

Remaining life is the same chain entered from the middle: subtract what’s spent, divide by the draw. The 85% line is what turns the estimate into a schedule.


Problem 3

Same current, two bills

The find. Mid-survey, you turn up an old abandoned service line still metallically tied to the protected main — no isolation, no notes, just continuity where nobody expected it. A shunt at the accessible bond point reads a steady 250 mA flowing off your system into the freeloader, and the paper trail says the tie was probably missed on a project three years ago. The anode string on this segment is zinc.

A — What has three years of freeloading cost the anode side?

Show solution — part A

Convert the draw to amp-hours per year, then to zinc:

Per year: 0.250 A × 8,760 hr = 2,190 A-hr
In zinc: 2,190 ÷ 335 A-hr/lb ≈ 6.5 lb per year
Three years: ≈ 19.6 lb of zinc

Call it twenty pounds of zinc — bought, buried, and spent on a line nobody uses.

B — Translate the same 250 mA to the steel side: what corrosion bill has the system been paying off?

Show solution — part B
20.1 lb/A-yr × 0.250 A × 3 yr ≈ 15.1 lb of steel

Notice the trade is not pound-for-pound: the system spent about twenty pounds of zinc covering roughly fifteen pounds’ worth of steel demand. That’s normal, and it’s the whole idea — zinc pounds are budgeted and replaceable; steel pounds are pipe.

C — Why would those fifteen pounds matter far more if the current had been leaving the coated main at a few holidays, instead of feeding a bare abandoned line?

Show solution — part C

No arithmetic in this one — just the principle that decides whether pounds become paperwork or become a leak. On a bare line, demand spreads across the whole surface: fifteen pounds over hundreds of square feet is a haze nobody will ever measure. On a coated main, everything funnels to the few square inches where the coating’s gone — the same pounds delivered to three quarter-size addresses is pitting, and pitting on a schedule is a leak.

The law counts the pounds; the circuit picks the address. That’s why the freeloader hunt is worth the survey day — the bill always lands somewhere, and the somewhere is what leaks.

One shunt reading, two invoices — anode pounds and steel pounds, priced by the same law. Cathodic protection is a transfer of the bill; the survey’s job is knowing whose account it lands on.

What this Apply lesson was after. Three runs of one chain. Forward at the creek bottom — shunt to years, with the numbers backing last module’s metal call. From the middle on a records review — what’s spent, what’s left, and the difference between when the math says empty and when the plan says dig. And both directions at once on a freeloader — the same 250 milliamps written as a zinc bill and a steel bill, which is the entire idea of cathodic protection in one meter reading.

Every problem started at a shunt, because that’s where the field hands you the number. The law does the rest — and now, so can you. The quiz is ten questions, eighty percent to pass, and you’ve already done everything it asks. After that, the reference electrodes are waiting.