NACE CP2 Cathodic Protection Technician Exam 2026:
Official Practice Test Bank | 125+ Verified MCQs with Step-
by-Step Math & Engineering Rationales (100% Pass Study
Guide Bundle)
Question 1
Using the Wenner Four-Pin Method to determine soil resistivity, you set up your pins
with an equal spacing of 5.0 feet (1.52 meters). Your resistance meter displays a
reading of 2.15 ohms. What is the calculated soil resistivity at that specific layer depth?
A. 10.75 ohm-cm
B. 67.54 ohm-cm
C. 2,058 ohm-cm
D. 205,800 ohm-cm
Verified Answer: C
Engineering Rationale: The formula for soil resistivity using the Wenner method is ρ =
2π a R, where ρ is resistivity in ohm-cm, a is the pin spacing in centimeters, and R is
the measured resistance in ohms. First, convert the pin spacing from feet to
, centimeters:
\(5.0\text{\ ft}\times 30.48\text{\ cm/ft}=152.4\text{\ cm}\)
Next, plug the values into the Wenner equation:
\(\rho =2\times 3.14159\times 152.4\text{\ cm}\times 2.15\ \Omega \approx 2,058.4\text{\
ohm-cm}\)
Options A, B, and D represent critical conversion errors where feet were not converted
to centimeters or the mathematical constant π was omitted.
Question 2
You are calculating the total circuit resistance of an Impressed Current Cathodic
Protection (ICCP) system. The rectifier output is currently configured at 24 Volts and 8
Amperes. What is the total circuit resistance of this CP system?
A. 0.33 ohms
B. 3.00 ohms
C. 12.00 ohms
D. 192.00 ohms
Verified Answer: B
Engineering Rationale: According to Ohm’s Law, resistance (R) is calculated by
dividing voltage (V) by current (I):
\(R=\frac{V}{I}\)
Plugging in the operational values from the rectifier:
\(R=\frac{24\text{\ V}}{8\text{\ A}}=3.0\ \Omega \)
Option A is the inverse of the correct calculation (). Option C is an arbitrary value,
and Option D is the total power consumption in Watts (V × I = 192 W) rather than
resistance.
Question 3
A pipeline structure-to-soil potential measurement is taken using a saturated Saturated
Calomel Reference Electrode (SCE). The multimeter displays a value of -0.78 V. What
is the equivalent structure potential converted to the standard Copper/Copper Sulfate
(Cu/CuSO₄) reference scale?
A. -0.70 V
B. -0.78 V
C. -0.85 V
D. -0.92 V
Verified Answer: C
, Engineering Rationale: To convert a potential measurement from the Saturated
Calomel Electrode (SCE) scale to the Saturated Copper/Copper Sulfate (CSE) scale,
you must add approximately -0.07 V (or -72 mV) to the SCE reading because the
Cu/CuSO₄ electrode has a higher half-cell potential than the Calomel electrode.
\(\text{Potential\ vs.\ CSE}=-0.78\text{\ V}+(-0.07\text{\ V})=-0.85\text{\ V}\)
This shows that a reading of -0.78 V vs. SCE exactly satisfies the classic NACE -0.85 V
polarized criterion on the CSE scale.
Question 4
While performing troubleshooting on an ICCP system, you discover that the rectifier
fuse keeps blowing immediately upon turning the system on. A visual inspection reveals
no damage to the rectifier cabinet. What is the most probable field fault?
A. The anode bed has reached the end of its operational lifespan and gone high-
resistance.
B. There is a complete break or open-circuit fault in the positive header cable.
C. There is a direct dead-short circuit between the positive anode cable and the
negative structure cable.
D. The soil surrounding the groundbed has dried out completely.
Verified Answer: C
Engineering Rationale: An immediate, repeating blown fuse indicates an overcurrent
condition (amperage spike) caused by extremely low circuit resistance. A direct short
circuit between the positive (anode) and negative (structure) DC cables creates a zero-
resistance path, bypassing the soil electrolyte completely and causing current to spike
beyond the capacity of the fuse. Options A, B, and D would all increase circuit
resistance, which drops the current output down toward zero Amperes, preventing a
fuse from blowing.
Question 5
When assessing a pipeline system, which of the following scenarios represents the
biggest threat regarding the creation of localized shielding, preventing CP current from
reaching the steel structure?
A. Applying a premium, factory-applied fusion-bonded epoxy (FBE) coating.
B. A disbonded, high-electrical-resistance solid polyethylene tape wrap.
C. Utilizing a highly conductive calcined petroleum coke breeze backfill around anodes.
D. Installing magnesium anodes in low-resistivity native clay soil.
Verified Answer: B
, Engineering Rationale: Cathodic protection shielding occurs when a non-conductive
barrier physically detaches from the pipe steel (disbondment) but prevents moisture and
electrical current from flowing underneath it. Solid tape wraps are notorious shields
because they block CP current from reaching the steel surface while allowing moisture
to seep behind the wrap, creating an unmitigated environment for accelerated localized
corrosion. FBE (Option A) fails by cracking or blistering without forming wide-scale
dielectric shields.
Question 6
An instant-off potential survey is conducted on an underground storage tank (UST). The
"on" reading is -1420 mV vs. CSE. When the current interruption cycles, the "instant-off"
reading drops rapidly to -810 mV vs. CSE, and then over the next 24 hours drifts down
to -690 mV vs. CSE. Does this system satisfy the 100 mV polarization decay criterion?
A. Yes, because the total polarization decay equals 120 mV, which exceeds the
requirement.
B. No, because the instant-off potential failed to achieve the absolute -850 mV baseline.
C. Yes, because the difference between the "on" and "off" readings is 610 mV.
D. No, because the potential drifted to a less negative state over time.
Verified Answer: A
Engineering Rationale: The 100 mV polarization criterion states that the structure
must exhibit at least 100 mV of electrochemical polarization decay between the true
polarized potential (instant-off) and the fully depolarized static potential (measured after
the system has been off for an extended duration).
\(\text{Polarization\ Decay}=|\text{Instant-Off\ Potential}|-|\text{Fully\ Depolarized\
Potential}|\)
\(\text{Polarization\ Decay}=810\text{\ mV}-690\text{\ mV}=120\text{\ mV}\)
Because 120 mV is greater than the 100 mV NACE requirement, the structure is
adequately protected. Option C incorrectly uses the "on" potential, which contains
invalid soil IR drop errors.
Question 7
A standard galvanic anode system is being designed for a buried pipeline. You are
evaluating the driving voltage of different sacrificial anodes. If a magnesium anode has
an open-circuit potential of -1.75 V vs. CSE and the steel pipeline's polarized potential is
-0.90 V vs. CSE, what is the net driving voltage of the system?
A. -2.65 V
B. -0.85 V
Official Practice Test Bank | 125+ Verified MCQs with Step-
by-Step Math & Engineering Rationales (100% Pass Study
Guide Bundle)
Question 1
Using the Wenner Four-Pin Method to determine soil resistivity, you set up your pins
with an equal spacing of 5.0 feet (1.52 meters). Your resistance meter displays a
reading of 2.15 ohms. What is the calculated soil resistivity at that specific layer depth?
A. 10.75 ohm-cm
B. 67.54 ohm-cm
C. 2,058 ohm-cm
D. 205,800 ohm-cm
Verified Answer: C
Engineering Rationale: The formula for soil resistivity using the Wenner method is ρ =
2π a R, where ρ is resistivity in ohm-cm, a is the pin spacing in centimeters, and R is
the measured resistance in ohms. First, convert the pin spacing from feet to
, centimeters:
\(5.0\text{\ ft}\times 30.48\text{\ cm/ft}=152.4\text{\ cm}\)
Next, plug the values into the Wenner equation:
\(\rho =2\times 3.14159\times 152.4\text{\ cm}\times 2.15\ \Omega \approx 2,058.4\text{\
ohm-cm}\)
Options A, B, and D represent critical conversion errors where feet were not converted
to centimeters or the mathematical constant π was omitted.
Question 2
You are calculating the total circuit resistance of an Impressed Current Cathodic
Protection (ICCP) system. The rectifier output is currently configured at 24 Volts and 8
Amperes. What is the total circuit resistance of this CP system?
A. 0.33 ohms
B. 3.00 ohms
C. 12.00 ohms
D. 192.00 ohms
Verified Answer: B
Engineering Rationale: According to Ohm’s Law, resistance (R) is calculated by
dividing voltage (V) by current (I):
\(R=\frac{V}{I}\)
Plugging in the operational values from the rectifier:
\(R=\frac{24\text{\ V}}{8\text{\ A}}=3.0\ \Omega \)
Option A is the inverse of the correct calculation (). Option C is an arbitrary value,
and Option D is the total power consumption in Watts (V × I = 192 W) rather than
resistance.
Question 3
A pipeline structure-to-soil potential measurement is taken using a saturated Saturated
Calomel Reference Electrode (SCE). The multimeter displays a value of -0.78 V. What
is the equivalent structure potential converted to the standard Copper/Copper Sulfate
(Cu/CuSO₄) reference scale?
A. -0.70 V
B. -0.78 V
C. -0.85 V
D. -0.92 V
Verified Answer: C
, Engineering Rationale: To convert a potential measurement from the Saturated
Calomel Electrode (SCE) scale to the Saturated Copper/Copper Sulfate (CSE) scale,
you must add approximately -0.07 V (or -72 mV) to the SCE reading because the
Cu/CuSO₄ electrode has a higher half-cell potential than the Calomel electrode.
\(\text{Potential\ vs.\ CSE}=-0.78\text{\ V}+(-0.07\text{\ V})=-0.85\text{\ V}\)
This shows that a reading of -0.78 V vs. SCE exactly satisfies the classic NACE -0.85 V
polarized criterion on the CSE scale.
Question 4
While performing troubleshooting on an ICCP system, you discover that the rectifier
fuse keeps blowing immediately upon turning the system on. A visual inspection reveals
no damage to the rectifier cabinet. What is the most probable field fault?
A. The anode bed has reached the end of its operational lifespan and gone high-
resistance.
B. There is a complete break or open-circuit fault in the positive header cable.
C. There is a direct dead-short circuit between the positive anode cable and the
negative structure cable.
D. The soil surrounding the groundbed has dried out completely.
Verified Answer: C
Engineering Rationale: An immediate, repeating blown fuse indicates an overcurrent
condition (amperage spike) caused by extremely low circuit resistance. A direct short
circuit between the positive (anode) and negative (structure) DC cables creates a zero-
resistance path, bypassing the soil electrolyte completely and causing current to spike
beyond the capacity of the fuse. Options A, B, and D would all increase circuit
resistance, which drops the current output down toward zero Amperes, preventing a
fuse from blowing.
Question 5
When assessing a pipeline system, which of the following scenarios represents the
biggest threat regarding the creation of localized shielding, preventing CP current from
reaching the steel structure?
A. Applying a premium, factory-applied fusion-bonded epoxy (FBE) coating.
B. A disbonded, high-electrical-resistance solid polyethylene tape wrap.
C. Utilizing a highly conductive calcined petroleum coke breeze backfill around anodes.
D. Installing magnesium anodes in low-resistivity native clay soil.
Verified Answer: B
, Engineering Rationale: Cathodic protection shielding occurs when a non-conductive
barrier physically detaches from the pipe steel (disbondment) but prevents moisture and
electrical current from flowing underneath it. Solid tape wraps are notorious shields
because they block CP current from reaching the steel surface while allowing moisture
to seep behind the wrap, creating an unmitigated environment for accelerated localized
corrosion. FBE (Option A) fails by cracking or blistering without forming wide-scale
dielectric shields.
Question 6
An instant-off potential survey is conducted on an underground storage tank (UST). The
"on" reading is -1420 mV vs. CSE. When the current interruption cycles, the "instant-off"
reading drops rapidly to -810 mV vs. CSE, and then over the next 24 hours drifts down
to -690 mV vs. CSE. Does this system satisfy the 100 mV polarization decay criterion?
A. Yes, because the total polarization decay equals 120 mV, which exceeds the
requirement.
B. No, because the instant-off potential failed to achieve the absolute -850 mV baseline.
C. Yes, because the difference between the "on" and "off" readings is 610 mV.
D. No, because the potential drifted to a less negative state over time.
Verified Answer: A
Engineering Rationale: The 100 mV polarization criterion states that the structure
must exhibit at least 100 mV of electrochemical polarization decay between the true
polarized potential (instant-off) and the fully depolarized static potential (measured after
the system has been off for an extended duration).
\(\text{Polarization\ Decay}=|\text{Instant-Off\ Potential}|-|\text{Fully\ Depolarized\
Potential}|\)
\(\text{Polarization\ Decay}=810\text{\ mV}-690\text{\ mV}=120\text{\ mV}\)
Because 120 mV is greater than the 100 mV NACE requirement, the structure is
adequately protected. Option C incorrectly uses the "on" potential, which contains
invalid soil IR drop errors.
Question 7
A standard galvanic anode system is being designed for a buried pipeline. You are
evaluating the driving voltage of different sacrificial anodes. If a magnesium anode has
an open-circuit potential of -1.75 V vs. CSE and the steel pipeline's polarized potential is
-0.90 V vs. CSE, what is the net driving voltage of the system?
A. -2.65 V
B. -0.85 V
