NACE CP1 EXAM 2026/2027 Actual Exam –
Complete Questions & Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Corrosion Theory & Electrochemistry | Q1 – Q10
Section 2 | Cathodic Protection Principles | Q11 – Q20
Section 3 | Field Measurement & Instrumentation | Q21 – Q30
Section 4 | Sacrificial Anode & Impressed Current Systems | Q31 – Q40
Section 5 | Safety, Standards & Report Writing | Q41 – Q50
Instructions: Choose the single best answer. Pass: 40 in 90 minutes.
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SECTION 1: CORROSION THEORY & ELECTROCHEMISTRY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 45-year-old inspector is examining a stainless steel process vessel and notices severe
pitting concentrated in the heat-affected zone adjacent to a recent carbon steel repair
weld. The base metal away from the weld shows no attack. The inspector asks the
junior technician to explain the localized damage.
A. The carbon steel weld metal has become cathodic and protected, causing the
stainless to corrode.
B. The weld heat-affected zone has lost its chromium oxide film, making it anodic to the
base metal. ✓ CORRECT
C. The stainless steel is generating stray current that attacks the weld metal.
D. The tank contents have become acidic and are dissolving only the weld filler material.
Correct Answer: B
Rationale: The heat-affected zone adjacent to welds can lose chromium carbide
precipitation or oxide film integrity, creating an anodic area relative to the unaffected
,stainless base metal. Some inspectors blame the filler metal alone, but the HAZ is
typically the weakest electrochemical link. Welding procedures and post-weld
passivation are critical on process equipment.
Question 2 of 50
A 32-year-old technician is performing a corrosion investigation on an underground
pipeline and measures a voltage drop of 0.1 volts across a 0.5-ohm shunt resistor
installed in a test station. He needs to calculate the current flowing through the circuit
to document the cathodic protection output.
A. He divides 0.1 volts by 0.5 ohms to find 0.2 amperes flowing through the test station
shunt. ✓ CORRECT
B. He multiplies 0.1 volts by 0.5 ohms to find 0.05 amperes.
C. He adds 0.1 volts and 0.5 ohms to estimate 0.6 amperes.
D. He divides 0.5 ohms by 0.1 volts to find 5 amperes.
Correct Answer: A
Rationale: Ohm's law gives current as voltage divided by resistance, so 0.1 V across 0.5
ohms equals 0.2 A. Multiplying or adding the values are common arithmetic errors that
produce meaningless units. Accurate shunt calculations are essential for verifying
cathodic protection current output without breaking circuits.
Question 3 of 50
A 50-year-old engineer is inspecting a reinforced concrete parking garage and notes that
the carbon steel rebar embedded in the high-pH concrete shows no visible rust despite
decades of exposure. He asks a 26-year-old technician why the steel remains
uncorroded.
A. The concrete absorbs all oxygen and prevents any electrochemical reactions.
B. The steel becomes anodic to the concrete and corrodes very slowly.
C. The calcium in concrete acts as a sacrificial anode for the steel reinforcement.
,D. The high pH forms a stable passive oxide film on the steel that suppresses corrosion.
✓ CORRECT
Correct Answer: D
Rationale: Highly alkaline environments like concrete with pH above 12 maintain a
stable passive oxide layer on steel, which drastically reduces corrosion rates. Concrete
does not absorb all oxygen, and steel does not become anodic to concrete; rather, the
chemistry stabilizes the surface. This passivation is why properly maintained reinforced
concrete can last decades.
Question 4 of 50
A 29-year-old technician finds a steel storage tank bottom under a leaky gasket where
the area beneath the leak is heavily corroded while the dry surrounding plate is nearly
pristine. The lead technician asks the assistant to identify the electrochemical cell
responsible.
A. The water leaking from the gasket contains chlorides that uniformly attack the entire
plate.
B. The wet area under the leak is cathodic because it has more oxygen, while the dry
area is anodic.
C. The area under the leak is anodic because oxygen concentration is lower, creating a
concentration cell. ✓ CORRECT
D. The gasket material is more noble than steel and is galvanically protecting the dry
area.
Correct Answer: C
Rationale: Differential aeration cells form where oxygen-starved areas become anodic
and corrode preferentially compared to oxygen-rich surfaces, explaining the localized
attack under the leak. Assuming the wet area is cathodic reverses the actual
mechanism, and gasket materials are rarely noble enough to drive galvanic corrosion of
, steel. Recognizing oxygen concentration cells helps technicians locate hidden corrosion
under seals and deposits.
Question 5 of 50
A 38-year-old technician measures the open-circuit potential of a copper water service
line connected to a steel main and finds the copper at -0.15 V CSE and the steel at -0.65
V CSE. The technician is asked to predict which metal will corrode in the galvanic
couple.
A. The copper will corrode sacrificially because it is more negative.
B. The steel will corrode because it is more active, and the copper will act as the
cathode. ✓ CORRECT
C. Both metals will corrode at equal rates because they are in electrical contact.
D. The potential difference is too small to cause any galvanic corrosion.
Correct Answer: B
Rationale: Steel is more active than copper in the galvanic series, so when coupled,
steel becomes the anode and corrodes while copper becomes the cathode. Reversing
the roles based on potential sign alone is a common mistake; the more negative
potential indicates the more active metal. Dielectric unions are standard practice
wherever copper and steel piping must join.
Question 6 of 50
A 27-year-old trainee is examining a potentiodynamic polarization curve for a steel
specimen in a test cell. Near the corrosion potential, the curve shows a steep slope
where small changes in applied current produce large potential shifts. The mentor asks
what this region indicates.
A. The metal is in the passive region and will not corrode.
B. The corrosion potential is dominated by concentration polarization.
Complete Questions & Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Corrosion Theory & Electrochemistry | Q1 – Q10
Section 2 | Cathodic Protection Principles | Q11 – Q20
Section 3 | Field Measurement & Instrumentation | Q21 – Q30
Section 4 | Sacrificial Anode & Impressed Current Systems | Q31 – Q40
Section 5 | Safety, Standards & Report Writing | Q41 – Q50
Instructions: Choose the single best answer. Pass: 40 in 90 minutes.
══════════════════════════════════════
SECTION 1: CORROSION THEORY & ELECTROCHEMISTRY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 45-year-old inspector is examining a stainless steel process vessel and notices severe
pitting concentrated in the heat-affected zone adjacent to a recent carbon steel repair
weld. The base metal away from the weld shows no attack. The inspector asks the
junior technician to explain the localized damage.
A. The carbon steel weld metal has become cathodic and protected, causing the
stainless to corrode.
B. The weld heat-affected zone has lost its chromium oxide film, making it anodic to the
base metal. ✓ CORRECT
C. The stainless steel is generating stray current that attacks the weld metal.
D. The tank contents have become acidic and are dissolving only the weld filler material.
Correct Answer: B
Rationale: The heat-affected zone adjacent to welds can lose chromium carbide
precipitation or oxide film integrity, creating an anodic area relative to the unaffected
,stainless base metal. Some inspectors blame the filler metal alone, but the HAZ is
typically the weakest electrochemical link. Welding procedures and post-weld
passivation are critical on process equipment.
Question 2 of 50
A 32-year-old technician is performing a corrosion investigation on an underground
pipeline and measures a voltage drop of 0.1 volts across a 0.5-ohm shunt resistor
installed in a test station. He needs to calculate the current flowing through the circuit
to document the cathodic protection output.
A. He divides 0.1 volts by 0.5 ohms to find 0.2 amperes flowing through the test station
shunt. ✓ CORRECT
B. He multiplies 0.1 volts by 0.5 ohms to find 0.05 amperes.
C. He adds 0.1 volts and 0.5 ohms to estimate 0.6 amperes.
D. He divides 0.5 ohms by 0.1 volts to find 5 amperes.
Correct Answer: A
Rationale: Ohm's law gives current as voltage divided by resistance, so 0.1 V across 0.5
ohms equals 0.2 A. Multiplying or adding the values are common arithmetic errors that
produce meaningless units. Accurate shunt calculations are essential for verifying
cathodic protection current output without breaking circuits.
Question 3 of 50
A 50-year-old engineer is inspecting a reinforced concrete parking garage and notes that
the carbon steel rebar embedded in the high-pH concrete shows no visible rust despite
decades of exposure. He asks a 26-year-old technician why the steel remains
uncorroded.
A. The concrete absorbs all oxygen and prevents any electrochemical reactions.
B. The steel becomes anodic to the concrete and corrodes very slowly.
C. The calcium in concrete acts as a sacrificial anode for the steel reinforcement.
,D. The high pH forms a stable passive oxide film on the steel that suppresses corrosion.
✓ CORRECT
Correct Answer: D
Rationale: Highly alkaline environments like concrete with pH above 12 maintain a
stable passive oxide layer on steel, which drastically reduces corrosion rates. Concrete
does not absorb all oxygen, and steel does not become anodic to concrete; rather, the
chemistry stabilizes the surface. This passivation is why properly maintained reinforced
concrete can last decades.
Question 4 of 50
A 29-year-old technician finds a steel storage tank bottom under a leaky gasket where
the area beneath the leak is heavily corroded while the dry surrounding plate is nearly
pristine. The lead technician asks the assistant to identify the electrochemical cell
responsible.
A. The water leaking from the gasket contains chlorides that uniformly attack the entire
plate.
B. The wet area under the leak is cathodic because it has more oxygen, while the dry
area is anodic.
C. The area under the leak is anodic because oxygen concentration is lower, creating a
concentration cell. ✓ CORRECT
D. The gasket material is more noble than steel and is galvanically protecting the dry
area.
Correct Answer: C
Rationale: Differential aeration cells form where oxygen-starved areas become anodic
and corrode preferentially compared to oxygen-rich surfaces, explaining the localized
attack under the leak. Assuming the wet area is cathodic reverses the actual
mechanism, and gasket materials are rarely noble enough to drive galvanic corrosion of
, steel. Recognizing oxygen concentration cells helps technicians locate hidden corrosion
under seals and deposits.
Question 5 of 50
A 38-year-old technician measures the open-circuit potential of a copper water service
line connected to a steel main and finds the copper at -0.15 V CSE and the steel at -0.65
V CSE. The technician is asked to predict which metal will corrode in the galvanic
couple.
A. The copper will corrode sacrificially because it is more negative.
B. The steel will corrode because it is more active, and the copper will act as the
cathode. ✓ CORRECT
C. Both metals will corrode at equal rates because they are in electrical contact.
D. The potential difference is too small to cause any galvanic corrosion.
Correct Answer: B
Rationale: Steel is more active than copper in the galvanic series, so when coupled,
steel becomes the anode and corrodes while copper becomes the cathode. Reversing
the roles based on potential sign alone is a common mistake; the more negative
potential indicates the more active metal. Dielectric unions are standard practice
wherever copper and steel piping must join.
Question 6 of 50
A 27-year-old trainee is examining a potentiodynamic polarization curve for a steel
specimen in a test cell. Near the corrosion potential, the curve shows a steep slope
where small changes in applied current produce large potential shifts. The mentor asks
what this region indicates.
A. The metal is in the passive region and will not corrode.
B. The corrosion potential is dominated by concentration polarization.
