L1 - Reduction Potentials

objectives

Understandings

A voltaic cell generates an electromotive force (EMF) resulting in the movement of electrons from the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The EMF is termed the cell potential (Eº).

The standard hydrogen electrode (SHE) consists of an inert platinum electrode in contact with 1 mol dm-3 hydrogen ion and hydrogen gas at 100 kPa and 298 K. The standard electrode potential (Eº) is the potential (voltage) of the reduction half-equation under standard conditions measured relative to the SHE. Solute concentration is 1 mol dm-3 or 100 kPa for gases. Eº of the SHE is 0 V.

Applications

Calculation of cell potentials using standard electrode potentials.

Prediction of whether a reaction is spontaneous or not using Eº values.

Standard Hydrogen Electrode

The cell potential in Galvanic Cells (+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 Figure 1 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:

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 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.

Determining Standard Reduction Potentials

A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure 2).
In cell notation, the reaction is:

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

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.

Figure 2. 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.

Example

Figure 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.

Consider the cell shown in Figure 3, where:

Electrons flow from left to right, and the reactions are:

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.

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_∘cell∘, for any cell.
For example, for the cell shown above:

Again, note that when calculating Ecell∘, standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor.

Using the Electrochemical Series - Alternative Method

Predicting Feasibility using Reduction Potentials

E° values

Can be used to predict the feasibility of redox and cell reactions
In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK
In practice, it proceeds if the E° value of the reaction is greater than + 0.40V
An equation with a more positive E° value reverse a less positive one

Worked Example

What happens if an Sn(s) / Sn²⁺(aq) and a Cu(s) / Cu²⁺(aq) cell are connected?

Write out the equations

Cu²⁺(aq) + 2e¯ Cu(s) ; E° = +0.34V

Sn²⁺(aq) + 2e¯ Sn(s) ; E° = -0.14V

the half reaction with the more positive E° value is more likely to work

it gets the electrons by reversing the half reaction with the lower E° value

2. therefore Cu²⁺(aq) ——> Cu(s) and Sn(s) ——> Sn²⁺(aq)

3. the overall reaction is Cu²⁺(aq) + Sn(s) ——> Sn²⁺(aq) + Cu(s)

4. the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V

Will this reaction be spontaneous?

Sn(s) + Cu²⁺(aq) ——> Sn²⁺(aq) + Cu(s)

Write out the appropriate equations Cu²⁺(aq) + 2e¯ Cu(s) ; E° = +0.34V

as reductions with their E° values Sn²⁺(aq) + 2e¯ Sn(s) ; E° = - 0.14V

The reaction which takes place will involve the more positive one reversing the other

i.e. Cu²⁺(aq) ——> Cu(s) and Sn(s) ——> Sn²⁺(aq)

The cell voltage will be the difference

in E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V

If this is the equation you want then it will be spontaneous

If it is the opposite equation (going the other way) it will not be spontaneous

ReDuCtion Potential Activity 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.

Using the electrochemical series, the reactions involved in the galvanic cell, both written as reductions, are

Au³⁺(aq)+3e⁻⟶Au(s)E_∘Au³⁺_/Au=+1.498V

Ni²⁺(aq)+2e⁻⟶Ni(s)E_∘Ni²⁺_/Ni=−0.257V

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:

Anode\;(oxidation):Ni(s)Ni2+(aq)+2e− Eanode∘=ENi2+/Ni∘=−0.257V

Cathode\;(reduction):Au3+(aq)+3e−Au(s) Ecathode∘=EAu3+/Au∘=+1.498V

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

3Ni(s)+2Au³⁺(aq)⟶3Ni²⁺(aq)+2Au(s)3Ni(s)+2Au3+(aq)⟶3Ni2+(aq)+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_∘cell=E_∘cathode−E_∘anode=1.498V−(−0.257V)=1.755VEcell∘=Ecathode∘−Eanode∘=1.498V−(−0.257V)=+1.755V

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

Reduction potentials activity

Copy of 11 - Exercise 2 - electrochemical cells.docx