IB DP Chemistry -R3.2.12 Standard electrode potentials - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.2.12 Standard electrode potentials – Study Notes – New Syllabus
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Reactivity 3.2.12 — The Hydrogen Half-Cell and Standard Electrode Potentials
Reactivity 3.2.12 — The Hydrogen Half-Cell and Standard Electrode Potentials
Standard Hydrogen Electrode (SHE):
The standard hydrogen electrode (SHE) is the reference electrode with an assigned potential of 0.00 V.
Half-equation: \( \text{H}^+(aq, 1\ \text{mol dm}^{-3}) + e^- \rightleftharpoons \dfrac{1}{2} \text{H}_2(g, 100\ \text{kPa}) \)
- Platinum electrode is used as an inert surface.
- Standard conditions: 1 mol dm-3, 100 kPa, 298 K.
Standard Electrode Potential (\( E^\circ \)):
It is the voltage measured when a half-cell is connected to the SHE under standard conditions.Measures the tendency of a substance to be reduced.
- More positive \( E^\circ \) → greater tendency to gain electrons (reduction).
- More negative \( E^\circ \) → greater tendency to lose electrons (oxidation).
Interpretation of Standard Electrode Potentials:
- Given as standard reduction potentials.
- Positive \( E^\circ \) → species more readily reduced (oxidizing agent).
- Negative \( E^\circ \) → species more readily oxidized (reducing agent).
Standard Electrode Potentials (IB DP Chemistry Data Booklet Examples):
| Half-Equation (Reduction) | \( E^\circ \) / V |
|---|---|
| \( \text{F}_2(g) + 2e^- \rightarrow 2\text{F}^-(aq) \) | +2.87 |
| \( \text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \) | +1.51 |
| \( \text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \) | +1.36 |
| \( \text{O}_2(g) + 4H^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l) \) | +1.23 |
| \( \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \) | +0.80 |
| \( \text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \) | +0.77 |
| \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \) | +0.34 |
| \( \text{H}^+(aq) + e^- \rightarrow \dfrac{1}{2} \text{H}_2(g) \) | 0.00 (by convention) |
| \( \text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \) | -0.44 |
| \( \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \) | -0.76 |
| \( \text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \) | -1.66 |
| \( \text{Na}^+(aq) + e^- \rightarrow \text{Na}(s) \) | -2.71 |
Example
Will zinc displace copper from copper(II) sulfate solution?
▶️Answer/Explanation
From the data:
\( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76\ \text{V} \)
\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = +0.34\ \text{V} \)
Zinc has a more negative \( E^\circ \) → it is the reducing agent → gets oxidized.
Copper(II) is the oxidizing agent → gets reduced.
Oxidation: \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \)
Reduction: \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \)
Overall: \( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \)
\( E^\circ_{\text{cell}} = +0.34 – (-0.76) = +1.10\ \text{V} \)
Since \( E^\circ_{\text{cell}} > 0 \), the reaction is spontaneous.
Example
The following half-equations are given:
\( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \quad E^\circ = -0.44\ \text{V} \)
\( \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad E^\circ = +0.80\ \text{V} \)
Determine the overall redox reaction and cell potential.
▶️Answer/Explanation
Ag+ has the higher \( E^\circ \) → it gets reduced.
Fe is oxidized.
Oxidation: \( \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^- \)
Reduction: \( 2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s) \)
Overall: \( \text{Fe}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Fe}^{2+}(aq) + 2\text{Ag}(s) \)
\( E^\circ_{\text{cell}} = +0.80 – (-0.44) = +1.24\ \text{V} \)
Example
Using the following half-equations:
\( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad E^\circ = +1.51\ \text{V} \)
\( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad E^\circ = +0.77\ \text{V} \)
Determine if the reaction is feasible and write the balanced overall equation.
▶️Answer/Explanation
MnO4− has the higher \( E^\circ \) → it is reduced.
Fe2+ is oxidized.
Multiply iron equation ×5 to match electrons:
Ox: \( 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^- \)
Red: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
Overall: \( \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)
\( E^\circ_{\text{cell}} = +1.51 – (+0.77) = +0.74\ \text{V} \) → spontaneous.