NACE CP1 TESTER EXAM 2026/2027 Actual Exam
– Complete Questions & Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Corrosion Principles & Electrochemistry | Q1 – Q10
Section 2 | Cathodic Protection Fundamentals | Q11 – Q20
Section 3 | Field Testing & Instrumentation | Q21 – Q30
Section 4 | Safety & Regulatory Compliance | Q31 – Q40
Section 5 | Data Interpretation & Reporting | Q41 – Q50
Instructions: Choose the single best answer. Pass: 40 in 90 minutes.
══════════════════════════════════════
SECTION 1: CORROSION PRINCIPLES & ELECTROCHEMISTRY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 42-year-old inspector is examining an above-ground firewater storage tank and
notices severe rusting on the carbon steel shell directly below a brass overflow fitting
that was added during a recent modification. The steel elsewhere shows only light
surface oxidation. The inspector asks the junior technician to explain why the localized
attack is concentrated at the brass-to-steel interface.
A. The brass is acting as the anode and forcing the steel to become cathodic, which
accelerates oxidation of the brass fitting itself.
B. The steel and brass have formed a concentration cell because the oxygen level is
higher at the bottom of the tank.
C. The steel has become the anode in a galvanic couple with the more noble brass,
causing accelerated corrosion of the steel shell. ✓ CORRECT
D. The brass fitting is generating stray direct current that is being discharged onto the
steel surface.
,Correct Answer: C
Rationale: In a galvanic couple, the less noble metal becomes the anode and corrodes
sacrificially while the more noble metal serves as the cathode. Brass is more noble than
carbon steel in most environments, so the steel sacrifices itself rather than the brass.
This is a textbook example of galvanic corrosion that field technicians encounter
whenever dissimilar metals are joined without isolation.
Question 2 of 50
A 29-year-old technician is performing a corrosion survey on an underground pipeline
and measures a current of 0.5 amperes flowing through a 2-ohm resistance path in the
soil between a holiday on the pipe and a neighboring structure. He needs to calculate
the voltage driving this corrosion cell to document the severity in his report.
A. He multiplies 0.5 amperes by 2 ohms to obtain 1 volt, applying Ohm’s law directly to
the electrolyte path. ✓ CORRECT
B. He divides 0.5 amperes by 2 ohms to obtain 0.25 volts, which represents the
pipe-to-soil potential.
C. He adds 0.5 amperes and 2 ohms to obtain 2.5 volts, which is the standard driving
voltage for steel in soil.
D. He multiplies 0.5 amperes by the pipe length in feet to determine the total corrosion
voltage.
Correct Answer: A
Rationale: Ohm’s law states that voltage equals current multiplied by resistance, so 0.5
A times 2 ohms yields 1 V across that specific soil path. Dividing current by resistance
is a common arithmetic error that inverts the relationship and produces meaningless
values. Understanding Ohm’s law in soil circuits is essential for estimating stray current
effects and voltage drops during field surveys.
Question 3 of 50
,A 38-year-old corrosion technician is examining a section of buried steel water main
that has developed a large tubercle with reddish-brown rust. Using a half-cell and a
high-impedance voltmeter, he measures the potential at the rusted area and compares it
to a nearby shiny metallic area on the same pipe. The rusted area reads -0.72 V CSE
while the shiny area reads -0.55 V CSE.
A. The shiny area is the anode because it is more negative and electrons are flowing
toward the rusted tubercle.
B. The rusted area is the anode because it is more negative and corroding, while the
shiny area acts as the cathode. ✓ CORRECT
C. Both areas are cathodes because steel always acts as a cathode when it is buried in
moist soil.
D. The potential difference proves the pipe is under cathodic protection and both areas
are fully protected.
Correct Answer: B
Rationale: The more negative potential indicates the anodic site where oxidation and
metal loss are actively occurring, while the less negative shiny region is the cathode
where reduction reactions take place. Reversing the anode and cathode is a frequent
mistake made by technicians who associate shiny metal with active corrosion rather
than cathodic protection. Recognizing polarity in corrosion cells is fundamental to
locating active corrosion before it causes leaks.
Question 4 of 50
A 47-year-old field engineer is reviewing polarization data from a newly commissioned
impressed-current cathodic protection system on a tank farm. He notices that as the
applied current density increases, the pipe-to-soil potential initially shifts rapidly
negative and then levels off even though the current continues to rise. He asks the
technician to identify the dominant polarization mechanism controlling the later plateau.
, A. The plateau indicates the steel has become immune due to complete passivation of
the surface.
B. The plateau is caused by ohmic polarization where the soil resistance has dropped to
zero.
C. The plateau represents activation polarization where every available anodic site has
been fully suppressed.
D. The plateau is primarily concentration polarization as the rate of oxygen diffusion to
the cathode becomes the limiting factor. ✓ CORRECT
Correct Answer: D
Rationale: At higher current densities, the reduction reaction at the cathode often
becomes limited by how fast oxygen can diffuse through the soil electrolyte to the pipe
surface, causing concentration polarization. Activation polarization dominates at lower
current densities when the energy barrier of the reaction itself is the limiting step. Field
technicians who recognize concentration polarization understand why simply increasing
rectifier output sometimes yields diminishing returns.
Question 5 of 50
A 33-year-old assistant technician is collecting soil samples alongside a natural gas
pipeline route and finds that the pH at one location is 4.5 due to acidic runoff from a
nearby mine. The senior technician explains that this acidic environment will influence
the corrosion rate of the buried steel.
A. The acidic soil will form a passive oxide film that completely stops corrosion at pH
values below 5.0.
B. The low pH will increase soil resistivity and therefore reduce the corrosion rate of the
pipeline.
C. The acidic environment accelerates the cathodic hydrogen evolution reaction and
increases corrosion of the steel. ✓ CORRECT
D. The low pH converts the steel into a cathode and protects it by making the potential
more positive.
Correct Answer: C
– Complete Questions & Detailed Rationales – Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Corrosion Principles & Electrochemistry | Q1 – Q10
Section 2 | Cathodic Protection Fundamentals | Q11 – Q20
Section 3 | Field Testing & Instrumentation | Q21 – Q30
Section 4 | Safety & Regulatory Compliance | Q31 – Q40
Section 5 | Data Interpretation & Reporting | Q41 – Q50
Instructions: Choose the single best answer. Pass: 40 in 90 minutes.
══════════════════════════════════════
SECTION 1: CORROSION PRINCIPLES & ELECTROCHEMISTRY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 42-year-old inspector is examining an above-ground firewater storage tank and
notices severe rusting on the carbon steel shell directly below a brass overflow fitting
that was added during a recent modification. The steel elsewhere shows only light
surface oxidation. The inspector asks the junior technician to explain why the localized
attack is concentrated at the brass-to-steel interface.
A. The brass is acting as the anode and forcing the steel to become cathodic, which
accelerates oxidation of the brass fitting itself.
B. The steel and brass have formed a concentration cell because the oxygen level is
higher at the bottom of the tank.
C. The steel has become the anode in a galvanic couple with the more noble brass,
causing accelerated corrosion of the steel shell. ✓ CORRECT
D. The brass fitting is generating stray direct current that is being discharged onto the
steel surface.
,Correct Answer: C
Rationale: In a galvanic couple, the less noble metal becomes the anode and corrodes
sacrificially while the more noble metal serves as the cathode. Brass is more noble than
carbon steel in most environments, so the steel sacrifices itself rather than the brass.
This is a textbook example of galvanic corrosion that field technicians encounter
whenever dissimilar metals are joined without isolation.
Question 2 of 50
A 29-year-old technician is performing a corrosion survey on an underground pipeline
and measures a current of 0.5 amperes flowing through a 2-ohm resistance path in the
soil between a holiday on the pipe and a neighboring structure. He needs to calculate
the voltage driving this corrosion cell to document the severity in his report.
A. He multiplies 0.5 amperes by 2 ohms to obtain 1 volt, applying Ohm’s law directly to
the electrolyte path. ✓ CORRECT
B. He divides 0.5 amperes by 2 ohms to obtain 0.25 volts, which represents the
pipe-to-soil potential.
C. He adds 0.5 amperes and 2 ohms to obtain 2.5 volts, which is the standard driving
voltage for steel in soil.
D. He multiplies 0.5 amperes by the pipe length in feet to determine the total corrosion
voltage.
Correct Answer: A
Rationale: Ohm’s law states that voltage equals current multiplied by resistance, so 0.5
A times 2 ohms yields 1 V across that specific soil path. Dividing current by resistance
is a common arithmetic error that inverts the relationship and produces meaningless
values. Understanding Ohm’s law in soil circuits is essential for estimating stray current
effects and voltage drops during field surveys.
Question 3 of 50
,A 38-year-old corrosion technician is examining a section of buried steel water main
that has developed a large tubercle with reddish-brown rust. Using a half-cell and a
high-impedance voltmeter, he measures the potential at the rusted area and compares it
to a nearby shiny metallic area on the same pipe. The rusted area reads -0.72 V CSE
while the shiny area reads -0.55 V CSE.
A. The shiny area is the anode because it is more negative and electrons are flowing
toward the rusted tubercle.
B. The rusted area is the anode because it is more negative and corroding, while the
shiny area acts as the cathode. ✓ CORRECT
C. Both areas are cathodes because steel always acts as a cathode when it is buried in
moist soil.
D. The potential difference proves the pipe is under cathodic protection and both areas
are fully protected.
Correct Answer: B
Rationale: The more negative potential indicates the anodic site where oxidation and
metal loss are actively occurring, while the less negative shiny region is the cathode
where reduction reactions take place. Reversing the anode and cathode is a frequent
mistake made by technicians who associate shiny metal with active corrosion rather
than cathodic protection. Recognizing polarity in corrosion cells is fundamental to
locating active corrosion before it causes leaks.
Question 4 of 50
A 47-year-old field engineer is reviewing polarization data from a newly commissioned
impressed-current cathodic protection system on a tank farm. He notices that as the
applied current density increases, the pipe-to-soil potential initially shifts rapidly
negative and then levels off even though the current continues to rise. He asks the
technician to identify the dominant polarization mechanism controlling the later plateau.
, A. The plateau indicates the steel has become immune due to complete passivation of
the surface.
B. The plateau is caused by ohmic polarization where the soil resistance has dropped to
zero.
C. The plateau represents activation polarization where every available anodic site has
been fully suppressed.
D. The plateau is primarily concentration polarization as the rate of oxygen diffusion to
the cathode becomes the limiting factor. ✓ CORRECT
Correct Answer: D
Rationale: At higher current densities, the reduction reaction at the cathode often
becomes limited by how fast oxygen can diffuse through the soil electrolyte to the pipe
surface, causing concentration polarization. Activation polarization dominates at lower
current densities when the energy barrier of the reaction itself is the limiting step. Field
technicians who recognize concentration polarization understand why simply increasing
rectifier output sometimes yields diminishing returns.
Question 5 of 50
A 33-year-old assistant technician is collecting soil samples alongside a natural gas
pipeline route and finds that the pH at one location is 4.5 due to acidic runoff from a
nearby mine. The senior technician explains that this acidic environment will influence
the corrosion rate of the buried steel.
A. The acidic soil will form a passive oxide film that completely stops corrosion at pH
values below 5.0.
B. The low pH will increase soil resistivity and therefore reduce the corrosion rate of the
pipeline.
C. The acidic environment accelerates the cathodic hydrogen evolution reaction and
increases corrosion of the steel. ✓ CORRECT
D. The low pH converts the steel into a cathode and protects it by making the potential
more positive.
Correct Answer: C
