Aim # 22: How can a redox reaction be used to produce an electric current? H.W. # 22 Study pp. 839-849 (sec. 18.2 – 18.3) Ans. ques. p. 879 # 41, 43(for exercise 41 only), 45, 47

I Galvanic Cell - a device in which electron transfer takes place through an external pathway rather than directly between reactants. A galvanic cell spontaneously converts chemical energy to electrical energy as a result of a redox reaction.

A galvanic cell makes use of two chemical half-reactions to produce an electric current. 1. At the Zn electrode (the anode) – oxidation occurs ox. Half-reaction: Zn(s) → Zn2+(aq) + 2e- 2. At the Cu electrode (the cathode) – reduction occurs red. Half-reaction: Cu2+(aq) + 2e- → Cu(s) 3. Ions formed at the anode (Zn2+ cations) enter the solution of the anode’s half-cell. 4. Ions attracted to the cathode (Cu2+ anions) “plate out” as atoms on the surface of the cathode. 5. Salt Bridge (or porous plug) – connects the half-cells and allows the migration of ions to neutralize the charge buildup occurring in the half-cells. anions (SO42-, NO3-) migrate to the anode. cations (Zn2+, K+, Cu2+) migrate to the cathode.

6. anode (ox. ) Zn(s) → Zn2+(aq) + 2e- cathode (red 6. anode (ox.) Zn(s) → Zn2+(aq) + 2e- cathode (red.) Cu2+(aq) + 2e- → Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) II Cell emf (electromotive force) A. Potential Difference – the difference in potential energy per unit charge between the two electrodes. It provides the driving force that pushes electrons through the external wire. It is also called the electromotive force (emf), and is measured in volts (V). 1 V = 1 Joule 1V = 1 J/C Coulomb Note: the charge on one electron is 1.60 x 10-19 C

B. Cell Potential – the emf of the cell (εcell) Its value depends upon the specific reactions that occur at the cathode and anode, the concentration of reactants and products, an the temperature. C. Standard Cell Potential (Standard emf, ε0cell) – the emf of a cell under standard conditions: 250C, 1M concentrations, an 1atm pressure for gases. e.g. For the Zn/Cu galvanic cell at 250C, where Zn(s) + Cu2+(aq, 1 M) → Zn2+(aq,1M) + Cu(s) ε0cell = + 1.10 V ε0cell = ε0red(cathode) – ε0red(anode) where ε0red is the potential for reduction to occur at a particular electrode.

STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 250C

III Calculating Standard Cell Potentials D. Standard Reduction Potentials (ε0red) – standard electrode potentials calculated and tabulated for reduction half-reactions (in a standard cell). III Calculating Standard Cell Potentials A. The Standard Hydrogen Electrode 2H+(aq, 1M) + 2e- → H2(g, 1 atm) by convention, ε0red = 0.0 V

B. Calculation of Standard Reduction Potentials for various half-cells (Half-Cell Potentials) A specific half-cell is connected to the hydrogen half-cell with the hydrogen half-cell electrode as the cathode. Problem: For the Zn/H2 standard cell, ε0 = +0.76 V. Calculate the standard reduction potential for the Zn/Zn2+ half-cell. Ans: ε0cell = +0.76 V = ε0red(cathode) - ε0red(anode) = 0 - ε0red(anode) ε0red(anode) = -0.76 V WHICH WAY WILL THE CURRENT FLOW?

Problem: For the zn/Cu galvanic cell, ε0cell = +1. 10 V Problem: For the zn/Cu galvanic cell, ε0cell = +1.10 V. Calculate the standard half-cell potential for the Cu half-cell. Ans: ε0cell = ε0red(cathode) - ε0red(anode) = +1.10 V ε0red – (-0.76 V) = + 1.10 V For the half-cell, ε0red(cathode) = + 0.34 V C. Calculating cell emf using half-cell potentials Problem: Using data from the table of standard reduction potentials, calculate the standard emf for a cell that employs the following overall reaction: 2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq)

Ans: Identify the half-reactions (red.) Cathode: I2(s) + 2e- → 2I-(aq) ε0 = +0.54 V For 3I2(s) + 6e- → 6I-(aq) Note: Potentials are intensive properties (ox.) Anode: Al3+(aq) + 3e- → Al(s) For 2Al3+(aq) + 6e- → 2Al(s) ε0 = -1.66 V Note: This reaction will run in reverse! 2Al(s) → 2Al3+(aq) + 6e- Overall: 3I2(s) + 6e- → 6I-(aq) 2Al(s) +3I2(s) → 2Al3+(aq) +6I-(aq)

ε0cell = ε0(cathode) - ε0(anode) ε0cell = +0. 54 V - (-1 ε0cell = ε0(cathode) - ε0(anode) ε0cell = +0.54 V - (-1.66 V) ε0cell = +2.20 V Note: the sign of the voltage indicates that electricity will flow spontaneously from the anode to the cathode. Note: The more positive the value of ε0cell , the greater the driving force of the reaction. There must be a positive value of ε0cell for the cell to operate spontaneously. WHAT WOULD BE THE SIGNIFICANCE OF AN ε0cell < 0?

Problem: A voltaic cell is composed of a Cu+/Cu half-cell and a Cu2+/Cu+ half-cell. a) What reaction occurs at the anode? b) What is the standard cell potential? Ans: From the Ref. Table Cu+(aq) + e- → Cu(s) ε0 = +0.521 V Cu2+(aq) + e- → Cu+(aq) ε0 = + .153 V a) The “more positive” (less negative) half-reacion will be higher in the table and will be the cathode reaction (reduction) ••• Anode: Cu2+(aq) + e- → Cu+ (aq) b) ε0cell = ε0red(cathode) - ε0red(anode) = +0.521 v – (+.153 V) ε0cell = +0.368 V

Its time to practice. Zumdahl (8th ed.) p. 863 # 36, 46 line notation- # 42 (for problem 36 only)