3.4: Standard Reduction Potentials

In redox reactions the thermodynamic driving force can be measured as the cell potential. The cell potential results from the difference in the electrical potentials for each electrode. Standard cell potential determined using the standard reduction cell potential and the standard oxidation cell potential. The standard reduction potential is a measure of the likelihood that a species will be reduced. Likewise, the standard oxidation potential is a measure of the likelihood that a species will be oxidized. The equations that relate these three potentials are shown below:

\[ E^o_{cell} = E^o_{reduction} \text{ of reaction at cathode} + E^o_{oxidation} \text{ of reaction at anode}\]

or alternatively

\[ E^o_{cell} = E^o_{reduction} \text{ of reaction at cathode} - E^o_{reduction} \text{ of reaction at anode}\]

The standard oxidation potential and the standard reduction potential are opposite in sign to each other for the same chemical species.

• Relation Between Standard Reduction Potential (SRP) and the Standard Oxidation Potential (SOP)

\[ E_0^o (SRP) = -E_0^o (SOP)\]

How the Standard Reduction Cell Potential is Experimentally Determined

It is impossible to determine the electrical potential of a single electrode; therefore, we use a reference electrode. This means, we assign an electrode an electrical potential value of zero and then compare the electrical potential of other electrodes to it. The standard hydrogen electrode (SHE) is commonly used for assessing the potential of half-cells. The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution at 298 K. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is

\[\ce{2H+}(aq,\: 1\:M)+\ce{2e-}⇌\ce{H2}(g,\:1\: \ce{atm}) \hspace{20px} E°=\mathrm{0\: V} \nonumber \]

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). The voltage is defined as zero for all temperatures.

Figure \(\PageIndex{1}\): 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⁺ (Figure \(\PageIndex{2}\)). Electrons flow from the anode to the cathode. The reactions, which are reversible, are

\[\begin{align*}
&\textrm{Anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq) + \ce{2e-}\\
&\textrm{Cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\\
&\overline{\textrm{Overall: }\ce{Cu^2+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{Cu}(s)}
\end{align*} \nonumber \]

Figure \(\PageIndex{2}\): A galvanic cell can be used to determine the standard reduction potential of Cu²⁺.

The standard reduction potential 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.

\[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber \]

\[\mathrm{+0.34\: V}=E^\circ_{\ce{Cu^2+/Cu}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Cu^2+/Cu}}−0=E^\circ_{\ce{Cu^2+/Cu}} \nonumber \]

Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure \(\PageIndex{3}\), where the electrons flow from left to right, and the reactions are

\[\begin{align*}
&\textrm{anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2e-}\\
&\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\\
&\overline{\textrm{overall: }\ce{2Ag+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2Ag}(s)}
\end{align*} \nonumber \]

Figure \(\PageIndex{3}\): 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.

The standard reduction potential 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.

\[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode} \nonumber \]

\[\mathrm{+0.80\: V}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Ag+/Ag}}−0=E^\circ_{\ce{Ag+/Ag}} \nonumber \]

It is important to note that the potential is not doubled for the cathode reaction.  The reason we don't need to multiply the Ag potential by 2 is that E^o is a measure of the free energy change per electron. Dividing the free energy change by the number of electrons makes E^o an intensive property (like pressure, temperature, etc.).

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, \(E^\circ_\ce{cell}\), for any cell. For example, for the following cell:

\[\begin{align*}
&\textrm{anode (oxidation): }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\\
&\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\\
&\overline{\textrm{overall: }\ce{Cu}(s)+\ce{2Ag+}(aq)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)}
\end{align*} \nonumber \]

\[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Cu^2+/Cu}}=\mathrm{0.80\: V−0.34\: V=0.46\: V} \nonumber \]

Again, note that when calculating \(E^\circ_\ce{cell}\), standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor.

The Activity Series

When solving for the standard cell potential, the species oxidized and the species reduced must be identified. This can be done using an activity series. The Table \(\PageIndex{1}\) shown below is an activity series; which is simply a table of standard reduction potentials in decreasing order. The species at the top have a greater likelihood of being reduced while the ones at the bottom have a greater likelihood of being oxidized. Therefore, when a species at the top is coupled with a species at the bottom, the one at the top will become reduced while the one at the bottom will become oxidized. A more complete list of standard reduction potentials is is provided in Tables P1 or P2.

Table \(\PageIndex{1}\): Activity Series - Standard Reduction Potential for Selected Species

Reduction Half-Reaction	Standard Reduction Potential (V)
F2(g)+2e^- → 2F^-(aq)	+2.87
S2O8^2^-(aq)+2e^- → 2SO4^2^-(aq)	+2.01
O2(g)+4H^+(aq)+4e^- → 2H2O(l)	+1.23
Br2(l)+2e^- → 2Br^-(aq)	+1.09
Ag^+(aq)+e^- → Ag(s)	+0.80
Fe^3^+(aq)+e^- → Fe^2^+(aq)	+0.77
I2(l) + 2e^- → 2I^-(aq)	+0.54
Cu^2^+(aq)+2e^- → Cu(s)	+0.34
Sn^4^+(aq)+2e^- → Sn^2^+(aq)	+0.15
S(s)+2H^+(aq)+2e^- → H2S(g)	+0.14
2H^+(aq)+2e^- → H2(g)	0.00
Sn^2^+(aq)+2e^- → Sn(g)	-0.14
V^3^+(aq)+e^- → V^2^+(aq)	-0.26
Fe^2^+(aq)+2e^- → Fe(s)	-0.44
Cr^3^+(aq)+3e^- → Cr(s)	-0.74
Zn^2^+(aq)+2e^- → Zn(s)	-0.76
Mn^2^+(aq)+2e^- → Mn(s)	-1.18
Na^+(aq)+e^- → Na(s)	-2.71
Li^+(aq)+e^- → Li(s)	-3.04

Problems

Exercise \(\PageIndex{1}\): 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.

Answer

Using Table \(\PageIndex{1}\), the reactions involved in the galvanic cell, both written as reductions, are

\[\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \nonumber \]

\[\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s) \hspace{20px} E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V} \nonumber \]

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:

\[\begin{align*}
&\textrm{Anode (oxidation): }\ce{Ni}(s)⟶\ce{Ni^2+}(aq)+\ce{2e-} \hspace{20px} E^\circ_\ce{anode}=E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V}\\
&\textrm{Cathode (reduction): }\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_\ce{cathode}=E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V}
\end{align*} \nonumber \]

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

\(\ce{3Ni}(s)+\ce{2Au^3+}(aq)⟶\ce{3Ni^2+}(aq)+\ce{2Au}(s)\)

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

\[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{1.498\: V−(−0.257\: V)=1.755\: V} \nonumber \]

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

Exercise \(\PageIndex{1}\)

A galvanic cell consists of a Mg electrode in 1 M Mg(NO₃)₂ solution and a Ag electrode in 1 M AgNO₃ solution. Calculate the standard cell potential at 25 °C.

Answer

\[\ce{Mg}(s)+\ce{2Ag+}(aq)⟶\ce{Mg^2+}(aq)+\ce{2Ag}(s) \hspace{20px} E^\circ_\ce{cell}=\mathrm{0.7996\: V−(−2.372\: V)=3.172\: V} \nonumber \]

Exercise \(\PageIndex{3}\)

True or False?

1. Hydrogen has oxidation potentials of 0.
2. The standard oxidation potential is not much like the standard reduction potential.
3. The standard reduction cell potential and the standard oxidation cell potential can never be combined.

Answer

1. True
2. False: the standard oxidation potential is much like the standard reduction potential
3. False: The standard reduction cell potential and the standard oxidation cell potential can be combined to determine the overall cell potential

Exercise \(\PageIndex{4}\)

1. What does the standard reduction potential measure?
2. What are the differences between the standard reduction potential and standard oxidation potential, and how are the two related?
3. What conditions must be met for a potential to be standard?
4. When standard reduction potentials are measured, what are the potentials relative to?
5. How is a standard reduction potential measured?
6. Explain how the activity series is used.
7. Explain how standard reduction potentials or standard oxidation potentials are applied.
8. Draw and label a SHE.

Answer

1. Standard reduction potential measures the tendency for a given chemical species to be reduced.
2. The standard oxidation potential measures the tendency for a given chemical species to be oxidized as opposed to be reduced. For the same chemical species the standard reduction potential and standard oxidation potential are opposite in sign.
3. The cell must be at 298K, 1atm, and all solutions must be at 1M. Standard reduction potentials are measured with relativity to hydrogen which has be universally set to have a potential of zero.
4. A standard reduction potential is measured using a galvanic cell which contains a SHE on one side and an unknown chemical half cell on the other side.
5. The amount of charge that passes between the cells is measured using a voltmeter.
6. The activity series is a list of standard reduction potentials in descending order of the tendency for chemical species to be reduced. Species at the top are more likely to be reduced while species at the bottom are more likely to be oxidized.
7. Standard reduction and oxidation potentials can be applied to solve for the standard cell potential of two different non hydrogen species. Examples can be seen in Cell Potentials.
8. See Figure (2).

Exercise \(\PageIndex{5}\)

Based on the activity series, which species will be oxidized and reduced: Zn²⁺ or H⁺.

Answer

H⁺ is farther up on the activity series then Zn²⁺ so H⁺ is reduced while Zn²⁺ is oxidized.

Exercise \(\PageIndex{6}\)

The standard reduction potential of Fe³⁺ is +0.77V. What is its standard oxidation potential.

Answer

The standard oxidation potential and standard reduction potential are always opposite in sign for the same species. The oxidation potential is -0.77V.