Electrochemical Potential

Outline

• Measuring Energy Changes in Voltaic cells
• Practice with Concepts
• Discussion Questions
• Optional Website Reading

Electrochemical energy in Voltaic Cells

First, a disclaimer. Electromotive force is not a force at all. It's energy divided by charge, where the energy is the potential energy difference between two points which happen to be the terminals of a battery cell. Just as gravitational potential energy is the difference between potential energy at two points in a gravitational field that slopes "down" as we move away from the center of mass, electric potential is the difference between two points in an electric field that slopes in some direction. If we place a charge in the field, it will respond to the slope by moving down it or up it, depending on the signs and the amount of the charges involved. So voltage is the potential divided by the charge: the amount of energy necessary the charge gains as it moves through the field.

For electrons moving on wires strung between two terminals of a battery (or through wet cells if they happen to be part of the circuit), the field slopes "down" from the negative terminal toward the positive terminal. The potential energy lost as the electrons "fall" down this gradient equals the electromotive force. So "emf" isn't really a force, but an energy change relative to a given charge.

Electric potential changes turn out to be critical for some life functions. Neurons fire if the normal "resting" potential of 75 microVolts between the inside and outside of the neuron is disturbed. Again, the difference is the critical measurement here.

Measuring Energy Changes in Voltaic Cells

With gravitational potential, we chose some base point, such as the floor or the surface of the Earth, and measure "up" away from the center of mass to determine the difference between two gravitational potential states.

With electrical circuits, we chose one part of the circuit to be the base for our measurement. If we have two half-cells as our battery, then the half-cell where hydrogen gas forms from hydrogen ions is the base "zero" point. We measure all differences against this state.

As we've already seen, a circuit with two half-cells supports two reactions:

Metal solid + cations in solution → metal ions in solutions + cation solid or gas

An example would be $$\mathrm{Zn}(s)+2H^{+}(\mathrm{aq})\to {\mathrm{Zn}}^{2+}(\mathrm{aq})+H_2(g)$$

But these two reactions occur in isolation from one another in the half-cells. We actually have a two-step reaction: $$\begin{array}{l}\mathrm{Zn}(s)\to {\mathrm{Zn}}^{2+}(\mathrm{aq})+2e^{-}\\ 2H^{+}(\mathrm{aq})+2e^{-}\to H_2(g)\end{array}$$

The electrons produced from the zinc travel the wire as measurable current to the second beaker, where they allow the hydrogen ions to form as water. The voltage difference measured from this current is the standard potential energy required to pull the electrons off the zinc atom. The zinc is oxidized (it loses electrons) and the hydrogen reduced (it gains electrons).

We can set up a series of half-cell reactions with different metals reacting against a hydrogen half-cell, and measure the potentials. This gives us a sequence of reduction half-reactions which we can rank in order of standard reduction potentials, high at the top the list, decreasing as we go down to hydrogen (0.00 on the chart) and becoming increasingly negative below hydrogen. Those with a higher (positive) standard reduction potential are better oxidizing agents than hydrogen or the other metals below them. Those with a lower (negative) standard reduction potential are better reducing agents than hydrogen or the other metals above them.

The Standard Electrode potential table here lists most well-established potentials in order, high to low.

Now we can use the potentials to help us determine what happens if we use something other than hydrogen for the second half-cell. Assume that we chose two half-cells, one with a zinc electrode and one with an iron electrode. From our standard reduction potentials chart, $$\begin{array}{l}{\mathrm{Fe}}^{2+}(\mathrm{aq})+2e^{-}\to \mathrm{Fe}(s)E^0=-0.44\\ {\mathrm{Zn}}^{2+}(\mathrm{aq})+2e^{-}\to \mathrm{Zn}(s)E^0=-0.763\end{array}$$

Iron has a more positive potential than zinc, so zinc is the better reducing agent. It will be oxidized and iron reduced in any reaction involving these electrodes in a half-cell arrangement. $$\begin{array}{l}{\mathrm{REDUCTION}\mathrm{at}\mathrm{Cathode}:\mathrm{Fe}}^{2+}(\mathrm{aq})+2e^{-}\to \mathrm{Fe}(s)E^0=-0.44\\ {\mathrm{OXIDATION}\mathrm{at}\mathrm{Anode}:\mathrm{Zn}(s)\to \mathrm{Zn}}^{2+}(\mathrm{aq})+2e^{-}{E}^0=+0.763\end{array}$$

The net potential change through the voltmeter will be: $$\begin{array}{l}{E^0}_{\mathrm{cell}}={E^0}_{\mathrm{cathode}}-{E^0}_{\mathrm{anode}}\\ {E^0}_{\mathrm{cell}}=-0.44-(-0.763)=-0.44+0.763=+0.323V\end{array}$$

Practice with the Concepts

Determining the potential

Two half-cells are set up with copper and silver electrods in water, connected by a voltmeter and a salt bridge. Which electrode is the anode?

What is the net potenital for the reaction?

Is the reaction $$2\mathrm{Ag}(s)+{\mathrm{Cu}}^{2+}(\mathrm{aq})\to 2{\mathrm{Ag}}^{+}(\mathrm{aq})+\mathrm{Cu}(s)$$

reactant favored or product favored?

Discussion Questions

• In cellular respiration, glucose is oxidized by combustion with oxygen to form carbon dioxide and water. How are electrons transfered in this reaction? How does this produce energy?

Optional Readings

• This Iron and copper sulphate reaction video shows a typical iron and copper sulfate reaction, where the copper is displaced from its solution and deposited on the iron, copper-plating the iron.

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