NACE CP2 CERTIFICATION SCRIPT
2026 QUESTIONS WITH SOLUTIONS
GRADED A+
◍ Nernst Equation.
Answer: Demonstrates how the metal ion concentration in the electrolyte
affects the potential of the electrode
◍ EMF series.
Answer: The arrangement of elements according to their electrode potential,
or their tendency to corrode vs SHE standard hydrogen electrode
◍ Corrosion of the metal will cease once the applied cathodic protection
current _______ the corrosion current.A) equalsB) equals or exceeds C)
falls below.
Answer: b
◍ An environment where cathodic protection is NOT effective is:A) liquidB)
soilC) atmospherically exposed surfaceD) concrete.
Answer: c
◍ The cathodic protection criteria for steel includes which of the following?
(select all that apply)A) a potential of -850 mVcse or more negative with
current applied after IR drop is considered.B) a polarized potential of -850
mVcse or more negativeC) polarization of 100 mVcse or greaterD) a
polarized potential of -790 mVssc or more negative.
Answer: a, b, c, d-790 mVssc = -850 mVcse
◍ Given:E1 = 20 VR1 = 10 ΩR2 = 10 Ω R3 = 10 ΩR4 = 10 ΩWhat is the
current R1?.
Answer: Answer on next slide
, ◍ Galvanic Series.
Answer: List of metals and alloys arranged according to their corrosion
potentials in a common electrolyte
◍ EMF (electromotive force).
Answer: Driving force for corrosion (voltage difference)
◍ Volt.
Answer: joule per coulomb where a joule is a unit of energy
◍ Amps (Amperes).
Answer: Rate of flow of charge, 1 coulomb per second
◍ Metal ion concentration HIGH.
Answer: less active the metal will be in that particular electrolyte
◍ Given:E1 = 20 VR1 = 10 ΩR2 = 10 Ω R3 = 10 ΩR4 = 10 ΩWhat is the
current R1?A) 0.29 AB) 0.67 AC) 1.0 AD) 2.0 A.
Answer: CR2,3,4 = 1/30 + 1/30 + 1/30R2,3,4 = 0.033 + 0.033 + 0.033R2,3,4
= .099R2,34 = 10.1 ΩRt = R1 + R2,3,4Rt = 10 Ω + 10.1 ΩRt = 20.1 ΩI
= 20 V / 20.1 ΩI = 0.995 A or 1.0 A
◍ Temperature.
Answer: Temp Corrosion Temp Corrosion
◍ Dezincification.
Answer: Zinc Corrosion/leaching in brass alloys
◍ Graphitic Corrosion.
Answer: Iron Corrosion/leaching in cast and ductile iron
◍ Oxygen Concentration.
Answer: Oxygen Corrosion Oxygen Corrosion
◍ Salt Concentration.
Answer: Salt Corrosion Salt Corrosion
◍ When cathodic protection is applied to acorroding steel surface, it shifts the
corrosion cell potential at the _____________A) no change in potentialB)
2026 QUESTIONS WITH SOLUTIONS
GRADED A+
◍ Nernst Equation.
Answer: Demonstrates how the metal ion concentration in the electrolyte
affects the potential of the electrode
◍ EMF series.
Answer: The arrangement of elements according to their electrode potential,
or their tendency to corrode vs SHE standard hydrogen electrode
◍ Corrosion of the metal will cease once the applied cathodic protection
current _______ the corrosion current.A) equalsB) equals or exceeds C)
falls below.
Answer: b
◍ An environment where cathodic protection is NOT effective is:A) liquidB)
soilC) atmospherically exposed surfaceD) concrete.
Answer: c
◍ The cathodic protection criteria for steel includes which of the following?
(select all that apply)A) a potential of -850 mVcse or more negative with
current applied after IR drop is considered.B) a polarized potential of -850
mVcse or more negativeC) polarization of 100 mVcse or greaterD) a
polarized potential of -790 mVssc or more negative.
Answer: a, b, c, d-790 mVssc = -850 mVcse
◍ Given:E1 = 20 VR1 = 10 ΩR2 = 10 Ω R3 = 10 ΩR4 = 10 ΩWhat is the
current R1?.
Answer: Answer on next slide
, ◍ Galvanic Series.
Answer: List of metals and alloys arranged according to their corrosion
potentials in a common electrolyte
◍ EMF (electromotive force).
Answer: Driving force for corrosion (voltage difference)
◍ Volt.
Answer: joule per coulomb where a joule is a unit of energy
◍ Amps (Amperes).
Answer: Rate of flow of charge, 1 coulomb per second
◍ Metal ion concentration HIGH.
Answer: less active the metal will be in that particular electrolyte
◍ Given:E1 = 20 VR1 = 10 ΩR2 = 10 Ω R3 = 10 ΩR4 = 10 ΩWhat is the
current R1?A) 0.29 AB) 0.67 AC) 1.0 AD) 2.0 A.
Answer: CR2,3,4 = 1/30 + 1/30 + 1/30R2,3,4 = 0.033 + 0.033 + 0.033R2,3,4
= .099R2,34 = 10.1 ΩRt = R1 + R2,3,4Rt = 10 Ω + 10.1 ΩRt = 20.1 ΩI
= 20 V / 20.1 ΩI = 0.995 A or 1.0 A
◍ Temperature.
Answer: Temp Corrosion Temp Corrosion
◍ Dezincification.
Answer: Zinc Corrosion/leaching in brass alloys
◍ Graphitic Corrosion.
Answer: Iron Corrosion/leaching in cast and ductile iron
◍ Oxygen Concentration.
Answer: Oxygen Corrosion Oxygen Corrosion
◍ Salt Concentration.
Answer: Salt Corrosion Salt Corrosion
◍ When cathodic protection is applied to acorroding steel surface, it shifts the
corrosion cell potential at the _____________A) no change in potentialB)
