Standard potentials (vs SHE)
| Electrode | Abbr. | Potential (Vs SHE) | Temp Coeff (mV/°C) |
|---|---|---|---|
| Copper / Copper Sulfate | CSE | +0.316 V | 0.9 |
| Ag / AgCl (Sea Water) | SSC (SJ) | +0.256 V | -0.33 |
| Saturated Calomel | SCE | +0.244 V | -0.70 |
| Ag / AgCl (Sat KCl) | SSC (LJ) | +0.222 V | -0.70 |
| Zinc (Soil) | ZRE | -0.800 V | N/A |
| Standard Hydrogen | SHE | 0.000 V | 0.0 |
Conversion formulas
General conversion
\[
V_{Target}
=
V_{Measured}
+
(E_{Source} - E_{Target})
\]
\(V_{Target}\) = converted reading on the target electrode scale
\(V_{Measured}\) = measured reading on the source scale
\(E_{Source}\) = standard potential of the source electrode (vs SHE)
\(E_{Target}\) = standard potential of the target electrode (vs SHE)
Temperature correction (to 25°C)
\[
V_{25}
=
V_{Measured}
+
k_t \,\times\, (T - 25)
\]
\(V_{25}\) = corrected potential at 25°C
\(V_{Measured}\) = measured value at temperature \(T\)
\(k_t\) = electrode temperature coefficient
\(T\) = electrode temperature (°C)
Worked examples
Example 1 — SCE to CSE
You measure −0.880 V vs SCE. Convert to the CSE scale.
\[
\begin{aligned}
V_{CSE}
&=
V_{SCE}
+
(E_{SCE} - E_{CSE}) \\[10pt]
&=
-0.880
+
(0.244 - 0.316) \\[10pt]
&=
-0.952\ \text{V}_{CSE}
\end{aligned}
\]
Answer: −0.952 V CSE
Example 2 — CSE to Ag/AgCl (Seawater)
You measure −0.850 V vs CSE. Convert to Ag/AgCl (Seawater).
\[
\begin{aligned}
V_{SSC}
&=
V_{CSE}
+
(E_{CSE} - E_{SSC}) \\[10pt]
&=
-0.850
+
(0.316 - 0.256) \\[10pt]
&=
-0.790\ \text{V}_{SSC}
\end{aligned}
\]
Answer: −0.790 V SSC
Example 3 — ZRE to CSE
You measure +0.250 V vs ZRE. Convert to CSE.
\[
\begin{aligned}
V_{CSE}
&=
V_{ZRE}
+
(E_{ZRE} - E_{CSE}) \\[10pt]
&=
+0.250
+
(-0.800 - 0.316) \\[10pt]
&=
-0.866\ \text{V}_{CSE}
\end{aligned}
\]
Answer: −0.866 V CSE
Example 4 — Ag/AgCl (Sat KCl) to Ag/AgCl (Seawater)
You measure −0.800 V vs Ag/AgCl (Sat KCl). Convert to Ag/AgCl (Seawater).
\[
\begin{aligned}
V_{SSC(sea)}
&=
V_{SSC(KCl)}
+
(E_{SSC(KCl)} - E_{SSC(sea)}) \\[10pt]
&=
-0.800
+
(0.222 - 0.256) \\[10pt]
&=
-0.834\ \text{V}_{SSC(sea)}
\end{aligned}
\]
Answer: −0.834 V SSC(sea)
Practice problems
Problem 1
Convert −0.850 V vs CSE to the SHE scale.
Reveal solution
\[
\begin{aligned}
V_{SHE}
&=
V_{CSE}
+
(E_{CSE} - E_{SHE}) \\[10pt]
&=
-0.850
+
(0.316 - 0.000) \\[10pt]
&=
-0.534\ \text{V}_{SHE}
\end{aligned}
\]
Answer: −0.534 V SHE
Problem 2
Correct −0.850 V CSE measured at 5°C to the 25°C reference temperature.
Reveal solution
\[
\begin{aligned}
V_{25}
&=
V_{Measured}
+
k_t \,\times\, (T - 25) \\[10pt]
&=
-0.850
+
0.0009 \,\times\, (5 - 25) \\[10pt]
&=
-0.868\ \text{V}
\end{aligned}
\]
Answer: −0.868 V
Problem 3
Correct −0.850 V CSE measured at 45°C to the 25°C reference temperature.
Reveal solution
\[
\begin{aligned}
V_{25}
&=
-0.850
+
0.0009 \,\times\, (45 - 25) \\[10pt]
&=
-0.832\ \text{V}
\end{aligned}
\]
Answer: −0.832 V
Problem 4
Convert the −0.850 V CSE protection criterion to the SCE scale.
Reveal solution
\[
\begin{aligned}
V_{SCE}
&=
-0.850
+
(0.316 - 0.244) \\[10pt]
&=
-0.778\ \text{V}_{SCE}
\end{aligned}
\]
Answer: −0.778 V SCE
Problem 5
What voltage difference would a meter read between CSE and ZRE?
Reveal solution
\[
\Delta V
=
(+0.316)
-
(-0.800)
=
1.116\ \text{V}
\]
Answer: 1.116 V
Problem 6
Convert −0.800 V vs Ag/AgCl (Seawater) to the SCE scale.
Reveal solution
\[
\begin{aligned}
V_{SCE}
&=
-0.800
+
(0.256 - 0.244) \\[10pt]
&=
-0.788\ \text{V}
\end{aligned}
\]
Answer: −0.788 V
Problem 7
Convert −0.900 V vs SCE to the SHE scale.
Reveal solution
\[
\begin{aligned}
V_{SHE}
&=
V_{SCE}
+
(E_{SCE} - E_{SHE}) \\[10pt]
&=
-0.900
+
(0.244 - 0.000) \\[10pt]
&=
-0.656\ \text{V}_{SHE}
\end{aligned}
\]
Answer: −0.656 V SHE
Problem 8
Convert −0.750 V vs Ag/AgCl (Sat KCl) to the CSE scale.
Reveal solution
\[
\begin{aligned}
V_{CSE}
&=
V_{SSC}
+
(E_{SSC} - E_{CSE}) \\[10pt]
&=
-0.750
+
(0.222 - 0.316) \\[10pt]
&=
-0.844\ \text{V}_{CSE}
\end{aligned}
\]
Answer: −0.844 V CSE
Problem 9
Convert −0.500 V vs SHE to the CSE scale.
Reveal solution
\[
\begin{aligned}
V_{CSE}
&=
V_{SHE}
+
(E_{SHE} - E_{CSE}) \\[10pt]
&=
-0.500
+
(0.000 - 0.316) \\[10pt]
&=
-0.816\ \text{V}_{CSE}
\end{aligned}
\]
Answer: −0.816 V CSE
Problem 10
Convert −0.780 V vs Ag/AgCl (Seawater) to the CSE scale.
Reveal solution
\[
\begin{aligned}
V_{CSE}
&=
V_{SSC(sea)}
+
(E_{SSC(sea)} - E_{CSE}) \\[10pt]
&=
-0.780
+
(0.256 - 0.316) \\[10pt]
&=
-0.840\ \text{V}_{CSE}
\end{aligned}
\]
Answer: −0.840 V CSE