Experiment #10 Electrochemical Cell

What are the goals of this experiment? To build a Cu-Zn galvanic cell. To study the effect of changing concentration of the electrolytes on the voltage output of the galvanic cell

Electrochemistry A study of chemical changes produced by electric current and with the production of electricity by chemical reactions. All electrochemical reactions involve the transfer of electrons and are therefore oxidation-reduction reactions. The reacting system is contained in a cell, and an electric current enters or exits by electrodes.

Electrochemistry The sites of oxidation and reduction are separated physically so that oxidation occurs at one location while reduction occurs at the other. Electrons flow from site of oxidation to the site of reduction. Flow of electrons implies flow of electric current. Current is usually denoted by symbol “I” and unit amperes (amps). Current ‘I’ is related to voltage ‘V’ through, ohms Law. R is called the resistance.

Electrochemistry Cu-Zn galvanic cell Metallic Cu (s) ----- Electrode Metallic Zinc (s) ----- Electrode CuSO4 (aq) --------- Electrolyte ZnSO4 (aq) --------- Electrolyte

What is oxidation? Current definition: Loss of Electrons is Oxidation (LEO) Na Na+ + e- Positive charge represents electron deficiency ONE POSITIVE CHARGE MEANS DEFICIENT BY ONE ELECTRON Oxidation occurs at the Anode

What is reduction? Current definition: Gain of Electrons is Reduction (GER) Cl + e- Cl - Negative charge represents electron richness ONE NEGATIVE CHARGE MEANS RICH BY ONE ELECTRON Reduction occurs at the cathode

Sign conventions in a galvanic cell A battery has a positive terminal and a negative terminal Cathode is assigned a positive (+) sign Anode is assigned a negative (-) sign Anodic Oxidation (AO) and Cathodic Reduction

Will this setup sustain flow of electrons? Reducing half, will have negative Charges. Will have excess anions. SO42- Oxidizing half, will have positive Charges. Will have excess cations (Zn2+) but deficient in anions. No, this set up does not sustain flow of electrons due to charge build up.

Will this setup sustain flow of electrons? We therefore need to provide a pathway for the anions to flow from the area where they are no longer needed (where Cu2+ is being converted to neutral Cu) to the area where they are needed (where neutral Zn is being converted to Zn2+). In other words, the solutions must be connected so that ions can flow to keep the net charge in each compartment zero

How do we connect the two halves?

How do we connect the two halves?

Why do electrons migrate spontaneously from one electrode to the other? The electrons move through the wire because by doing so they will move from a higher-energy to a lower-energy.

How to identify the oxidizing half-cell and reducing half-cell? When two half-cells are connected, the one with the larger reduction potential (the one with greater tendency to undergo reduction) acquires electrons from the half-cell with lower reduction potential, which is therefore forced to undergo oxidation. The half-cell with the higher standard reduction potential acts as the cathode (Reduction) and the half-cell with the lower standard reduction potential acts as the anode (Oxidation).

What is reduction potential? It is the voltage that is generated when a half-cell undergoes reduction. Where can I obtain these values from? From the table of standard reduction potentials

standard reduction potential? What is a standard electrode? Why is it called standard reduction potential? Because these voltages were measured, when the half-cells underwent reduction under a set of standard conditions in combination with a standard electrode. What is a standard electrode? It is an electrode, whose reduction/ oxidation potential is known.

standard reduction potential A reference electrode has been arbitrarily chosen and its standard reduction potential has been assigned a value of exactly 0 V. This reference electrode is called the standard hydrogen electrode. 2 H+(aq, 1.00 M) + 2 e- H2(g, 1atm) E= 0.00 V Standard conditions: Concentration = 1.0 M Pressure = 1.0 atm

standard reduction potential Units Volts (V)

standard reduction potential

An Example: What if I were to build a Cu-Ag galvanic cell? If we were to build a galvanic cell, we should keep in mind that, there are two half-cells. One of them undergoes oxidation and the other undergoes reduction. For that we will need to know the values of standard Reduction potential of each half cell. Cu2+(aq) +2e- Cu(s) E = + 0.34 V Ag +(aq) + e- Ag (S) E= + 0.80 V

An Example: What if I were to build a Cu-Ag galvanic cell? Cu2+(aq) +2e- Cu(s) E = + 0.34 V (Oxidizing, Anode) Ag +(aq) + e- Ag (S) E= + 0.80 V (Reducing, Cathode) So, Ag electrode is the positive(+) electrode and Cu electrode is the negative (-) electrode.

An Example: How can I find the Standard reduction potential of the Cu-Ag galvanic cell? Cu2+(aq) +2e- Cu2+(aq) E = + 0.34 V (Oxidizing, Anode) Ag +(aq) + e- Ag (S) E= + 0.80 V (Reducing, Cathode)

An Example: How can I find the potential of the Cu-Ag galvanic cell under any concentration condition? Cu2+(aq) +2e- Cu (s) E = + 0.34 V (Oxidizing, Anode) Ag +(aq) + e- Ag (S) E= + 0.80 V (Reducing, Cathode) {Cu(s) Cu2+(aq) + 2e-} ×1 {Ag +(aq) + e- Ag (S) } × 2 2 e- involved in the overall process. Cu(s) + 2Ag+ Cu2+(aq) + 2Ag

Nernst Equation Concentration of pure solids are to be omitted.!! Where Ecell= Voltage measured for the reactant and product concentrations summarized by the reaction quotient Q; Ecell= Voltage measured when reactant and product concentrations are 1 M (or 1 atm for gases) R = gas constant in units appropriate for the system, 8.314 JK-1mol-1; T = absolute temperature in K; n = number of moles of electrons transferred in the oxidation-reduction reaction F = 96,485 coul/mol e-; ln Q = natural logarithm of the reaction quotient for the reaction. Concentration of pure solids are to be omitted.!!

An Example: How can I find the potential of the Cu-Ag galvanic cell under any concentration condition? 2 e- involved in the overall process. Cu(s) + 2Ag+ Cu2+(aq) + 2Ag(s)

An Example: How can I find the potential of the Cu-Ag galvanic cell under any concentration condition? 2 e- involved in the overall process. Cu(s) + 2Ag+ Cu2+(aq) + 2Ag(s)

An Example: How can I find the potential of the Cu-Ag galvanic cell under any concentration condition?

An Example: How can I find the potential of the Cu-Zn galvanic cell under any concentration condition? 1 Cu2+(aq) + 1 Zn(s)  1 Cu(s) + 1 Zn2+(aq)

An Example: How can I find the potential of the Cu-Zn galvanic cell under any concentration condition?

An Example: How does the voltage output of the Cu-Zn galvanic cell change with change in concentration of Cu2+(aq) and Zn2+(aq)? [Zn2+(aq)] (M) [Cu2+(aq)] (M) Ecell (V) 2.00 1.00 Ecell – 8.89  10-3 Ecell Ecell + 8.89  10-3

Standard free energy change (G) and Free energy change G) Free energy in this context refers to the energy that is generated by the movement of electrons for doing work. Stands for change G stands for Gibbs free energy

Units of free energy change