AP TOPIC 9.7 ELECTROCHEMISTRY: GALVANIC (VOLTAIC) CELLS

ANTHONY MCCALL.

“BETWEEN YOU AND I” (2006)

ELECTROCHEMICAL REACTIONS

In electrochemical reactions, electrons are transferred from one species to another. In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

REDUCTION AND OXIDATION: “OIL RIG”

A species is oxidized when it loses electrons.

(OIL – Oxidation is Loss of Electrons)

• Here, zinc loses two electrons to go from neutral zinc metal to the Zn²⁺ ion.

REDUCTION AND OXIDATION: “OIL RIG”

A species is reduced when it gains electrons.

(RIG – Reduction is Gain of Electrons)

• Here, each of the H⁺ gains an electron and they combine to form H₂.

REDUCTION AND OXIDATION: “OIL RIG”

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

Zn 🡪 Zn²⁺ + 2 e^-

Cu²⁺ + 2 e^- 🡪 Cu

Zn + Cu²⁺ 🡪 Zn²⁺ + Cu

GALVANIC (VOLTAIC) CELLS

• We can use that energy to do work if we make the electrons flow through an external device.
• We call such a setup a galvanic (or voltaic) cell.
• A galvanic cell separates the two half reactions and allows for electrons to move from one to the other through a wire connecting them.

GALVANIC CELLS

• A typical cell looks like this.
• One side of the cell has both atoms (a piece of the metal) and ions (a solution of a salt containing ions of the metal) of the element that will be oxidized.
• The other side of the cell has both atoms and ions of the element that will be reduced.

GALVANIC CELLS

• The pieces of metal where the reaction occurs are called electrodes.
• The oxidation half reaction electrode is called the anode.
• The reduction half reaction electrode is called the cathode.
• “Red Cat / An Ox” (reduction cathode/anode oxidation)

GALVANIC CELLS

• The pieces of metal where the reaction occurs are called electrodes.
• The oxidation half reaction electrode is called the anode.
• The reduction half reaction electrode is called the cathode.
• “Red Cat / An Ox” (reduction cathode/anode oxidation)

GALVANIC CELLS

The electrons produced in the oxidation half-reaction travel through a wire that connects the two electrodes. Those electrons are used in the reduction half-reaction. Viola! A battery!

A lightbulb or other electrical device could be powered by being hooked up along the wire.

GALVANIC CELLS

• As positive ions leave the anode, the solution it’s in becomes more positive because it is gaining the positive ions produced by the oxidation half-reaction.
• The same is true at the cathode...negative charge is building up.
• Unless charge is balanced the reaction will not continue.

GALVANIC CELLS

To balance the charge in each solution, we use a salt bridge.

A salt bridge is usually a U-shaped tube that contains a salt solution.

Ions from the salt bridge flow into the solutions to balance the charge.

GALVANIC CELLS

• Electrons only spontaneously flow one way in a redox reaction—from higher electrical potential energy to lower electrical potential energy.

GALVANIC CELLS

• Electrons only spontaneously flow one way in a redox reaction—from higher electrical potential energy to lower electrical potential energy.
• As an analogy, water only spontaneously flows one way in a waterfall.

ELECTROMOTIVE FORCE (EMF)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
• It is also called the cell potential and is designated E_cell.
• Cell potential is measured in volts (V).
• Think of cell potential as the push of electrons. More voltage, more push on the electrons.

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STANDARD REDUCTION POTENTIALS

Reduction potentials for many electrodes have been measured and tabulated.

STANDARD REDUCTION POTENTIALS

Reversing the reduction half-reaction gives the oxidation half-reaction.

The potential will have the opposite sign (First Law of Thermo!)

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STANDARD HYDROGEN ELECTRODE (SHE)

• EMF values are referenced to a standard hydrogen electrode (SHE).
• By definition the reduction potential for hydrogen is 0 V:

2 H⁺ (aq, 1M) + 2 e⁻ ⎯⎯→ H₂ (g, 1 atm)

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STANDARD CELL POTENTIALS

The cell potential at standard conditions can be found through this equation:

E^o_cell = E^o_red + E^o_ox

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+E^o_cell indicates a thermodynamically favorable reaction

-E^o_cell indicates a thermodynamically unfavorable reaction

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EXAMPLE

What is the cell potential of the following reaction?

Zn(s) + Cu²⁺ (aq) 🡪 Zn²⁺(aq) + Cu(s)

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Cu²⁺ is reduced. Look up its EMF on the standard reduction chart.

Cu²⁺ + 2 e^- 🡪 Cu +0.34 V

EXAMPLE

What is the cell potential of the following reaction?

Zn(s) + Cu²⁺ (aq) 🡪 Zn²⁺(aq) + Cu(s)

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Zn is oxidized. Look up its EMF on the standard reduction chart.

Because Zn is oxidized, we must reverse the reaction in the chart. Therefore we must change the sign of the listed potential.

Zn 🡪 Zn²⁺ + 2 e^- +0.76 V

EXAMPLE

Remember, E^o_cell = E^o_red + E^o_ox

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E^o_cell = 0.34 V + 0.76 V = 1.10 V

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This is a thermodynamically favorable reaction due to the +E^o_cell

CELL NOTATION

Another way to write the reaction occurring in a galvanic cell is the cell notation.

Zn(s) + Cu²⁺ (aq) 🡪 Zn²⁺(aq) + Cu(s)

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Zn | Zn²⁺ || Cu²⁺ | Cu

oxidation reduction

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EXAMPLE 2: COEFFICIENTS DO NOT AFFECT E^O_CELL

Write the cell notation and calculate the cell potential for the following reaction.

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3 Fe²⁺(aq) + 2 Al(s) 🡪 3 Fe(s) + 2 Al³⁺(aq)

Cell notation: Al(s) | Al³⁺(aq) || Fe²⁺(aq) | Fe(s)

• Reduced: Fe²⁺ + 2e^- 🡪 Fe E_red = -0.45 V
• Oxidized: Al 🡪 Al³⁺ + 3e^- E_ox = 1.66 V

E^o_cell = E^o_red + E^o_ox

= -0.45 V + 1.66 V

= 1.21 V

EXAMPLE 3

Calculate E^o_cell for the reaction utilizing the following two half reactions and write the net ionic equation for the reaction that occurs.

Al³⁺(aq) + 3e^- 🡪 Al(s)

Mg²⁺(aq) + 2e^- 🡪 Mg(s)

1. Determine which half reaction must be flipped (oxidized instead of reduced to give a positive E^o_cell.

E^o_cell = E_red + E_ox

The only way this can be positive is for Al to be oxidized and Mg²⁺ to be reduced.

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

Al³⁺(aq) + 3e^- 🡪 Al(s) E^o = -1.66 V

Mg (s) 🡪 Mg²⁺ (aq) + 2e^- E^o = 2.37

E^o_cell = E_red + E_ox

= -1.66 V + 2.37

= 0.71 V

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

Al³⁺(aq) + 3e^- 🡪 Al(s)

Mg (s) 🡪 Mg²⁺ (aq) + 2e^-

2) To find the net ionic equation, we must add the two half reactions together and then balance the resulting equation.

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

Al³⁺(aq) + 3e^- 🡪 Al(s)

Mg (s) 🡪 Mg²⁺ (aq) + 2e^-

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

2x[Al³⁺(aq) + 3e^- 🡪 Al(s)]

3x[Mg (s) 🡪 Mg²⁺ (aq) + 2e^-]

(balance electrons)

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

2x[Al³⁺(aq) + 3e^- 🡪 Al(s)]

3x[Mg (s) 🡪 Mg²⁺ (aq) + 2e^-]

2 Al³⁺(s) + 3 Mg (aq) + 6e^- 🡪 2 Al (aq) + 3Mg²⁺ (s) + 6e^-

(add the two half reactions)

EXAMPLE 3

Write the net ionic equation for the reaction that occurs utilizing the following two half reactions and calculate E^o_cell for the reaction.

2x[Al³⁺(aq) + 3e^- 🡪 Al(s)]

3x[Mg (s) 🡪 Mg²⁺ (aq) + 2e^-]

2 Al³⁺(s) + 3 Mg (aq) 🡪 2 Al (aq) + 3Mg²⁺ (s)

(cancel the electrons from each side)

(check for balance atoms and balanced charge)

OXIDIZING VS REDUCING

The more easily an element is reduced, the less easily it is oxidized and vice verse.

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For example, E^o_red for Fe²⁺ + 2e^- 🡪 Fe is -0.45.

But E^o_ox for Fe 🡪 Fe²⁺ + 2e^- is 0.45.

OXIDIZING VS REDUCING

Less easily reduced

More easily oxidized