Chemistry » Electrochemistry » Standard Reduction Potentials

# Standard Reduction Potentials

## Standard Reduction Potentials

The cell potential in this figure (+0.46 V) results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in the figure below and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is

$${\text{2H}}^{\text{+}}(aq\text{, 1}\phantom{\rule{0.2em}{0ex}}M)+2{\text{e}}^{\text{−}}⇌{\text{H}}_{2}(g,\phantom{\rule{0.2em}{0ex}}\text{1 atm})\phantom{\rule{5em}{0ex}}E\text{°}=\text{0 V}$$

E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). This voltage is defined as zero for all temperatures.

Hydrogen gas at 1 atm is bubbled through 1 M HCl solution. Platinum, which is inert to the action of the 1 M HCl, is used as the electrode. Electrons on the surface of the electrode combine with H+ in solution to produce hydrogen gas.

A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (see the figure below). In cell notation, the reaction is

$$\text{Pt}(s)│{\text{H}}_{2}(g,\phantom{\rule{0.2em}{0ex}}\text{1 atm})│{\text{H}}^{\text{+}}(aq,\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}M)║{\text{Cu}}^{2+}(aq,\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}M)│\phantom{\rule{0.2em}{0ex}}\text{Cu}(s)$$

Electrons flow from the anode to the cathode. The reactions, which are reversible, are

$$\begin{array}{l}\underset{¯}{\begin{array}{l}\text{Anode (oxidation):}\phantom{\rule{5.2em}{0ex}}{\text{H}}_{2}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}^{\text{+}}(aq{\text{) + 2e}}^{\text{−}}\\ \text{Cathode (reduction):}\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cu}(s)\end{array}}\\ \text{Overall:}\phantom{\rule{4.6em}{0ex}}{\text{Cu}}^{2+}(aq)+{\text{H}}_{2}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}^{\text{+}}(aq)+\text{Cu}(s)\end{array}$$

The standard cell potential, E°cell, can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction. The minus sign is necessary because oxidation is the reverse of reduction.

$${E}_{\text{cell}}^{°}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{\text{cathode}}^{°}-{E}_{\text{anode}}^{°}$$

$$\text{+0.34 V}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Cu}}^{2+}\text{/Cu}}^{°}-{E}_{{\text{H}}^{\text{+}}{\text{/H}}_{2}}^{°}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Cu}}^{2+}\text{/Cu}}^{°}-0\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Cu}}^{2+}\text{/Cu}}^{°}$$

A galvanic cell can be used to determine the standard reduction potential of Cu2+.

Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in the figure below, where

$$\text{Pt}(s)│{\text{H}}_{2}(g,\phantom{\rule{0.2em}{0ex}}\text{1 atm})│{\text{H}}^{\text{+}}(aq\text{, 1}\phantom{\rule{0.2em}{0ex}}M)║{\text{Ag}}^{\text{+}}(aq\text{, 1}\phantom{\rule{0.2em}{0ex}}M)│\text{Ag}(s)$$

Electrons flow from left to right, and the reactions are

$$\begin{array}{l}\underset{¯}{\begin{array}{l}\text{anode (oxidation):}\phantom{\rule{5.4em}{0ex}}{\text{H}}_{2}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}^{\text{+}}(aq)+{\text{2e}}^{\text{−}}\\ \text{cathode (reduction):}\phantom{\rule{0.2em}{0ex}}2{\text{Ag}}^{\text{+}}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Ag}(s)\end{array}}\\ \text{overall:}\phantom{\rule{4.6em}{0ex}}2{\text{Ag}}^{\text{+}}(aq)+{\text{H}}_{2}(g)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}^{\text{+}}(aq)+\text{2Ag}(s)\end{array}$$

The standard cell potential, E°cell, can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.

$${E}_{\text{cell}}^{°}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{\text{cathode}}^{°}-{E}_{\text{anode}}^{°}$$

$$\text{+0.80 V}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{°}-{E}_{{\text{H}}^{\text{+}}{\text{/H}}_{2}}^{°}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{°}-0\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}{E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{°}$$

It is important to note that the potential is not doubled for the cathode reaction.

The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, $${E}_{\text{cell}}^{°},$$ for any cell. For example, for the cell shown in this lesson,

$$\text{Cu}(s)│{\text{Cu}}^{2+}(aq,\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}M)║{\text{Ag}}^{\text{+}}(aq,\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}M)│\text{Ag}(s)$$

$$\begin{array}{}\\ \underset{¯}{\begin{array}{l}\text{anode (oxidation):}\phantom{\rule{5.5em}{0ex}}\text{Cu}(s)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}(aq)+{\text{2e}}^{\text{−}}\\ \text{cathode (reduction):}\phantom{\rule{0.2em}{0ex}}2{\text{Ag}}^{\text{+}}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Ag}(s)\end{array}}\\ \text{overall:}\phantom{\rule{4.7em}{0ex}}\text{Cu}(s)+{\text{2Ag}}^{\text{+}}(aq)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}(aq)+\text{2Ag}(s)\end{array}$$

$${E}_{\text{cell}}^{°}={E}_{\text{cathode}}^{°}-{E}_{\text{anode}}^{°}={E}_{{\text{Ag}}^{\text{+}}\text{/Ag}}^{°}-{E}_{{\text{Cu}}^{2+}\text{/Cu}}^{°}=\text{0.80 V}-\text{0.34 V}=0.4\text{6 V}$$

Again, note that when calculating $${E}_{\text{cell}}^{°},$$ standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in the table below. A more complete list is provided in this appendix.

A galvanic cell can be used to determine the standard reduction potential of Ag+. The SHE on the left is the anode and assigned a standard reduction potential of zero.

Selected Standard Reduction Potentials at 25 °C
Half-ReactionE° (V)
$${\text{F}}_{2}(g)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2F}}^{\text{−}}(aq)$$+2.866
$${\text{PbO}}_{2}(s)+{\text{SO}}_{4}{}^{2-}(aq)+{\text{4H}}^{\text{+}}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{PbSO}}_{4}(s)+{\text{2H}}_{2}\text{O}(l)$$+1.69
$${\text{MnO}}_{4}{}^{\text{−}}(aq)+{\text{8H}}^{\text{+}}(aq)+{\text{5e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mn}}^{2+}(aq)+{\text{4H}}_{2}\text{O}(l)$$+1.507
$${\text{Au}}^{3+}(aq)+{\text{3e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Au}(s)$$+1.498
$${\text{Cl}}_{2}(g)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Cl}}^{\text{−}}(aq)$$+1.35827
$${\text{O}}_{2}(g)+{\text{4H}}^{\text{+}}(aq)+{\text{4e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}(l)$$+1.229
$${\text{Pt}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Pt}(s)$$+1.20
$${\text{Br}}_{2}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Br}}^{\text{−}}(aq)$$+1.0873
$${\text{Ag}}^{\text{+}}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ag}(s)$$+0.7996
$${\text{Hg}}_{2}{}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Hg}(l)$$+0.7973
$${\text{Fe}}^{3+}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}^{2+}(aq)$$+0.771
$${\text{MnO}}_{4}{}^{\text{−}}(aq)+{\text{2H}}_{2}\text{O}(l)+{\text{3e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{MnO}}_{2}(s)+{\text{4OH}}^{\text{−}}(aq)$$+0.558
$${\text{I}}_{2}(s)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2I}}^{\text{−}}(aq)$$+0.5355
$${\text{NiO}}_{2}(s)+{\text{2H}}_{2}\text{O}(l)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ni(OH)}}_{2}(s)+{\text{2OH}}^{\text{−}}(aq)$$+0.49
$${\text{Cu}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cu}(s)$$+0.34
$${\text{Hg}}_{2}{\text{Cl}}_{2}(s)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Hg}(l)+{\text{2Cl}}^{\text{−}}(aq)$$+0.26808
$$\text{AgCl}(s)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ag}(s)+{\text{Cl}}^{\text{−}}(aq)$$+0.22233
$${\text{Sn}}^{4+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Sn}}^{2+}(aq)$$+0.151
$${\text{2H}}^{\text{+}}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)$$0.00
$${\text{Pb}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Pb}(s)$$−0.1262
$${\text{Sn}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Sn}(s)$$−0.1375
$${\text{Ni}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ni}(s)$$−0.257
$${\text{Co}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Co}(s)$$−0.28
$${\text{PbSO}}_{4}(s)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Pb}(s)+{\text{SO}}_{4}{}^{2-}(aq)$$−0.3505
$${\text{Cd}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cd}(s)$$−0.4030
$${\text{Fe}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Fe}(s)$$−0.447
$${\text{Cr}}^{3+}(aq)+{\text{3e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cr}(s)$$−0.744
$${\text{Mn}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Mn}(s)$$−1.185
$${\text{Zn(OH)}}_{2}(s)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Zn}(s)+{\text{2OH}}^{\text{−}}(aq)$$−1.245
$${\text{Zn}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Zn}(s)$$−0.7618
$${\text{Al}}^{3+}(aq)+{\text{3e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Al}(s)$$−1.662
$${\text{Mg}}^{2}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Mg}(s)$$−2.372
$${\text{Na}}^{\text{+}}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Na}(s)$$−2.71
$${\text{Ca}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ca}(s)$$−2.868
$${\text{Ba}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ba}(s)$$−2.912
$${\text{K}}^{\text{+}}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{K}(s)$$−2.931
$${\text{Li}}^{\text{+}}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Li}(s)$$−3.04

Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.

## Example

### Cell Potentials from Standard Reduction Potentials

What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.

### Solution

Using the table above, the reactions involved in the galvanic cell, both written as reductions, are

$${\text{Au}}^{3+}(aq)+3{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Au}(s)\phantom{\rule{4em}{0ex}}{E}_{{\text{Au}}^{3+}\text{/Au}}^{°}=\text{+1.498 V}$$

$${\text{Ni}}^{2+}(aq)+2{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ni}(s)\phantom{\rule{4em}{0ex}}{E}_{{\text{Ni}}^{2+}\text{/Ni}}^{°}=\text{−0.257 V}$$

Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:

$$\begin{array}{}\\ \\ \text{Anode (oxidation):}\phantom{\rule{5.7em}{0ex}}\text{Ni}(s)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ni}}^{2+}(aq)+{\text{2e}}^{\text{−}}\phantom{\rule{2em}{0ex}}{E}_{\text{anode}}^{°}={E}_{{\text{Ni}}^{2+}\text{/Ni}}^{°}=\text{−0.257 V}\\ {\text{Cathode (reduction): Au}}^{3+}(aq)+{\text{3e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Au}(s)\phantom{\rule{4em}{0ex}}{E}_{\text{cathode}}^{°}={E}_{{\text{Au}}^{3+}\text{/Au}}^{°}=+1.498 V\end{array}$$

The least common factor is six, so the overall reaction is

$$\text{3Ni}(s)+{\text{2Au}}^{3+}(aq)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3Ni}}^{2+}(aq)+\text{2Au}(s)$$

The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.

$${E}_{\text{cell}}^{°}={E}_{\text{cathode}}^{°}-{E}_{\text{anode}}^{°}=\text{1.498 V}-(-0.2\text{57 V})=1.7\text{55 V}$$

From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.