Corrosion Authority
Part of
Certification Path (CP1–CP4)

CP 3 Practice: Reference Electrode Temperature Correction

Practice normalizing structure-to-electrolyte potentials to 25 °C using reference electrode temperature coefficients. Use consistent units (V or mV), label the reference, and show each step.

Standard layout: define values → line → formula → substitute → solve.

Formula used

\[ V_{25}=V_{Measured}+k_t(T-25) \]

Use k in mV/°C and convert to V/°C when working in volts.

Problems

Problem 1: CSE — Cold

Convert -0.850 VCSE measured at 5°C to 25°C. Use k = +0.9 mV/°C.

Show solution
\[ \begin{array}{c} V_{CSE}=-0.850\quad | \quad k_t=0.9\ \text{mV/°C}\quad | \quad 5°C \\[10pt] \color{white}{\rule{22em}{0.6pt}} \\[16pt] V_{25}=V_{Measured}+k_t(T-25) \\[16pt] V_{25}=-0.850\ \text{V}+0.0009\ \text{V}(5-25) \\[16pt] V_{25}=-0.868\ \text{V} \end{array} \]
Answer: -0.868 V25 CSE
Problem 2: CSE — Cold (mV)

Convert -850 mVCSE measured at 5°C to 25°C. Use k = +0.9 mV/°C.

Show solution
\[ \begin{array}{c} mV_{CSE}=-850\quad | \quad k_t=0.9\ \text{mV/°C}\quad | \quad 5°C \\[10pt] \color{white}{\rule{22em}{0.6pt}} \\[16pt] V_{25}=V_{Measured}+k_t(T-25) \\[16pt] V_{25}=-850\ \text{mV}+0.9\ \text{mV}(5-25) \\[16pt] V_{25}=-868\ \text{mV} \end{array} \]
Answer: -868 mV25 CSE
Problem 3: Ag / AgCl (Seawater) — Hot

Convert -0.800 VAg/AgCl(SW) measured at 40°C to 25°C. Use k = −0.33 mV/°C.

Show solution
\[ \begin{array}{c} V_{SW}=-0.800\quad | \quad k_t=-0.33\ \text{mV/°C}\quad | \quad 40°C \\[10pt] \color{white}{\rule{22em}{0.6pt}} \\[16pt] V_{25}=V_{Measured}+k_t(T-25) \\[16pt] V_{25}=-0.800\ \text{V}+(-0.00033\ \text{V})(40-25) \\[16pt] V_{25}=-0.805\ \text{V} \end{array} \]
Answer: -0.805 V25 Ag/AgCl(SW)
Problem 4: Ag / AgCl (Sat. KCl) — Cold

Convert -0.820 VAg/AgCl(KCl) measured at 13°C to 25°C. Use k = −0.70 mV/°C.

Show solution
\[ \begin{array}{c} V_{KCl}=-0.820\quad | \quad k_t=-0.70\ \text{mV/°C}\quad | \quad 13°C \\[10pt] \color{white}{\rule{22em}{0.6pt}} \\[16pt] V_{25}=V_{Measured}+k_t(T-25) \\[16pt] V_{25}=-0.820\ \text{V}+(-0.00070\ \text{V})(13-25) \\[16pt] V_{25}=-0.812\ \text{V} \end{array} \]
Answer: -0.812 V25 Ag/AgCl(KCl)