Standard Reduction Potentials

Overview

Standard reduction potentials ($E°$) measure the tendency of a half-reaction to occur as a reduction. They are used to predict spontaneity of redox reactions and calculate cell potentials.

Standard Reduction Potential Table

Strong Oxidizing Agents (High E°)

Half-Reaction	E° (V)
$\text{F}_2 + 2e^- \rightarrow 2\text{F}^-$	+2.87
$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O}$	+1.78
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$	+1.51
$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	+1.50
$\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$	+1.36
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$	+1.33
$\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$	+1.23
$\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$	+1.07
$\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}$	+0.96
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	+0.80
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$	+0.77

Moderate Potentials

Half-Reaction	E° (V)
$\text{I}_2 + 2e^- \rightarrow 2\text{I}^-$	+0.54
$\text{Cu}^+ + e^- \rightarrow \text{Cu}$	+0.52
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	+0.34
$\text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+}$	+0.15
$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$	0.00
$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$	-0.13
$\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$	-0.14
$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$	-0.26
$\text{Co}^{2+} + 2e^- \rightarrow \text{Co}$	-0.28
$\text{Cd}^{2+} + 2e^- \rightarrow \text{Cd}$	-0.40
$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$	-0.44

Strong Reducing Agents (Low E°)

Half-Reaction	E° (V)
$\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$	-0.74
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$	-0.76
$\text{Mn}^{2+} + 2e^- \rightarrow \text{Mn}$	-1.18
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$	-1.66
$\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}$	-2.37
$\text{Na}^+ + e^- \rightarrow \text{Na}$	-2.71
$\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$	-2.87
$\text{K}^+ + e^- \rightarrow \text{K}$	-2.93
$\text{Li}^+ + e^- \rightarrow \text{Li}$	-3.04

Using the Table

Rule 1: Higher E° = Stronger Oxidizing Agent

Species with high E° readily accept electrons.

Rule 2: Lower E° = Stronger Reducing Agent

Species with low E° (more negative) readily donate electrons.

Rule 3: Spontaneous Reactions

Oxidizing agent with higher E° will react with reducing agent from a lower E° pair.

Predicting Spontaneity

A reaction is spontaneous if:

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} > 0$$

Example 1: Will Cu reduce Fe³⁺?

$$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \quad E° = +0.77 \text{ V}$$ $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E° = +0.34 \text{ V}$$

Cu is lower, so Cu can be oxidized by Fe³⁺.

$$\text{Cu} + 2\text{Fe}^{3+} \rightarrow \text{Cu}^{2+} + 2\text{Fe}^{2+}$$ $$E°_{\text{cell}} = 0.77 - 0.34 = +0.43 \text{ V} > 0 \checkmark \text{ Spontaneous}$$

Example 2: Will Zn reduce H⁺?

$$2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E° = 0.00 \text{ V}$$ $$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E° = -0.76 \text{ V}$$

Zn is lower, so Zn can reduce H⁺.

$$\text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2$$ $$E°_{\text{cell}} = 0.00 - (-0.76) = +0.76 \text{ V} > 0 \checkmark \text{ Spontaneous}$$

Calculating Cell Potential

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$$

Important Notes

1. Don't multiply E° by stoichiometric coefficients
2. E° is an intensive property (doesn't depend on amount)
3. Reverse sign when reversing reaction

Example: Galvanic Cell

For $\text{Al} \mid \text{Al}^{3+} \| \text{Cu}^{2+} \mid \text{Cu}$

$$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E° = +0.34 \text{ V (cathode)}$$ $$\text{Al}^{3+} + 3e^- \rightarrow \text{Al} \quad E° = -1.66 \text{ V (anode)}$$ $$E°_{\text{cell}} = 0.34 - (-1.66) = +2.00 \text{ V}$$

Disproportionation

A species can disproportionate if it appears in two different oxidation states with appropriate E° values.

Example: Cu⁺ in solution

$$\text{Cu}^+ + e^- \rightarrow \text{Cu} \quad E° = +0.52 \text{ V}$$ $$\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \quad E° = +0.15 \text{ V}$$

Cu⁺ can act as both oxidizing and reducing agent:

$$2\text{Cu}^+ \rightarrow \text{Cu} + \text{Cu}^{2+}$$ $$E°_{\text{cell}} = 0.52 - 0.15 = +0.37 \text{ V} > 0 \checkmark \text{ Spontaneous}$$

Relationship to Other Quantities

Gibbs Free Energy

$$\Delta G° = -nFE°$$

Equilibrium Constant

$$E° = \frac{0.0592}{n} \log K \quad \text{(at 25°C)}$$

Summary Table

E°	$\Delta G°$	K	Reaction
> 0	< 0	> 1	Spontaneous, products favored
= 0	= 0	= 1	At equilibrium
< 0	> 0	< 1	Non-spontaneous, reactants favored

Example Calculations

Example: Finding K from E°

For the reaction:

$$2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+} \quad E° = 0.46 \text{ V}$$

Find K:

$$E° = \frac{0.0592}{n} \log K$$ $$0.46 = \frac{0.0592}{2} \log K$$ $$\log K = \frac{0.46 \times 2}{0.0592} = 15.5$$ $$K = 3.5 \times 10^{15}$$

Example: Finding $\Delta G°$ from E°

For the same reaction:

$$\Delta G° = -nFE°$$ $$\Delta G° = -(2)(96,485 \text{ C/mol})(0.46 \text{ V})$$ $$\Delta G° = -88,800 \text{ J/mol} = -88.8 \text{ kJ/mol}$$