Experiment 7 (electrochemistry)

Question 1

galvanic/voltaic cell

Answer 1

-any electrochemical cell in which spontaneous oxidation-reduction (chemical) reactions produce electrical energy
-daniell cells are a type of galvanic cells that specifically use zinc and copper electordes

Question 2

Daniell cell

Answer 2

-type of electrochemical cell that converts chemical energy into electrical energy
-consists of a zinc anode and a copper cathode, each immersed in a solution of their respective salts
-Zn / Zn2+ (1.00 M) // Cu2+ (1.00 M) / Cu
–Zn / Zn2+ = Anode/Anode soln (oxidized)
–Cu2+ /Cu = Cathode soln/Cathode (reduced)
-has salt bridge
-Δε° = +1.10 V
—Cell potential changes with the concentrations of solutions

Question 3

Salt bridge

Answer 3

-maintains electrical neutrality and completes the circuit
-allows ions to flow between the two half-cells
-preventing the build-up of charge and maintaining a steady flow of electrons through the external circuit

Question 4

Half cell reactions of Daniell cell

Answer 4

-OIL (oxidation half-cell): Zn → Zn2+ + 2e-
—Oxidation occurs at the anode
-Rig (reduction half-cell): Cu2+ + 2e- → Cu
—Reduction occurs in cathode

Question 5

Direction in which electrons flow in electrochemical/daniell cells

Answer 5

-Electrons always flow from anode to cathode
-Zn + Cu2+ → Zn2+ + Cu

Question 6

Nernst equation

Answer 6

-formula that allows calculation of reduction potential under non-standard conditions
-Δε = Δε° - RT/nF(ln(Q))
-for daniell cells: Δε = Δε° - RT/nF (ln( [Zn2+] / [Cu2+] )
-n is number of electrons transferred
-ε(re) = -ε(ox)
-RT/F is often replaced with 0.0257

Question 7

Concentration cell

Answer 7

-an electrolytic cell that is comprised of two half-cells with the same electrodes, but differing in concentrations
-acts to dilute the more concentrated solution and concentrate the more dilute solution, creating a voltage as the cell reaches an equilibrium

Question 8

Example of a concentration cell

Answer 8

-concentration cell of Fe(CN)6^3- and Fe(CN)6^4-
-Half-cell reaction:
-Fe(CN)6^3- + e → Fe(CN)6^4- ε° = 0.36 V
-ε = ε° - RT/nF (ln( [Fe(CN)6^4-] / [Fe(CN)6^3-] )
-Δε = ε(right) - ε(left) = ε° - RT/nF(ln(Q(right)) - ε° - RT/nF(ln(Q(left))
–ε° cancels out

Question 9

Graph of Δε (V)

Answer 9

-Cell potential vs ln(Q)
-Slope of line = RT/nF
–R= Gas constant (8.314 J/K mol)
–T= temperature in K
–N = # of electrons transferred
–F = Farraday constant (96,485 A/mol)