Electrochemistry Review

Half-Reaction and Standard Reduction Potentials

E{cell}^\circ = E{red}^\circ - E_{ox}^\circ
Table 11.1 lists standard reduction potentials at 25°C (298 K) for many common half-reactions.

Calculating E^\circ and \Delta G^\circ

Example 1 involves calculating E^\circ and \Delta G^\circ for a given reaction.
Values from the table are used to calculate cell potential.

Standard Reduction Potentials Examples

F_2 + 2e^- \rightarrow 2F^- E^\circ = 2.87 V
Ag^{2+} + e^- \rightarrow Ag^+ E^\circ = 1.99 V
Co^{3+} + e^- \rightarrow Co^{2+} E^\circ = 1.82 V
H2O2 + 2H^+ + 2e^- \rightarrow 2H_2O E^\circ = 1.78 V
Ce^{4+} + e^- \rightarrow Ce^{3+} E^\circ = 1.70 V
PbO2 + 4H^+ + SO4^{2-} + 2e^- \rightarrow PbSO4 + 2H2O E^\circ = 1.69 V
MnO4^- + 4H^+ + 3e^- \rightarrow MnO2 + 2H_2O E^\circ = 1.68 V
IO4^- + 2H^+ + 2e^- \rightarrow IO3^- + H_2O E^\circ = 1.60 V
MnO4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H2O E^\circ = 1.51 V
Au^{3+} + 3e^- \rightarrow Au E^\circ = 1.50 V
PbO2 + 4H^+ + 2e^- \rightarrow Pb^{2+} + 2H2O E^\circ = 1.46 V
Cl_2 + 2e^- \rightarrow 2Cl^- E^\circ = 1.36 V
Cr2O7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O E^\circ = 1.33 V
O2 + 4H^+ + 4e^- \rightarrow 2H2O E^\circ = 1.23 V
MnO2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H2O E^\circ = 1.21 V
IO3^- + 6H^+ + 5e^- \rightarrow I2 + 3H_2O E^\circ = 1.20 V
Br_2 + 2e^- \rightarrow 2Br^- E^\circ = 1.09 V
VO2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H2O E^\circ = 1.00 V
AuCl_4^- + 3e^- \rightarrow Au + 4Cl^- E^\circ = 0.99 V
NO3^- + 4H^+ + 3e^- \rightarrow NO + 2H2O E^\circ = 0.96 V
ClO2 + e^- \rightarrow ClO2^- E^\circ = 0.954 V
2Hg^{2+} + 2e^- \rightarrow Hg_2^{2+} E^\circ = 0.91 V
Ag^+ + e^- \rightarrow Ag E^\circ = 0.80 V
Hg_2^{2+} + 2e^- \rightarrow 2Hg E^\circ = 0.80 V
Fe^{3+} + e^- \rightarrow Fe^{2+} E^\circ = 0.77 V
O2 + 2H^+ + 2e^- \rightarrow H2O_2 E^\circ = 0.68 V
MnO4^- + e^- \rightarrow MnO4^{2-} E^\circ = 0.56 V
I_2 + 2e^- \rightarrow 2I^- E^\circ = 0.54 V
Cu^+ + e^- \rightarrow Cu E^\circ = 0.52 V
O2 + 2H2O + 4e^- \rightarrow 4OH^- E^\circ = 0.40 V
Cu^{2+} + 2e^- \rightarrow Cu E^\circ = 0.34 V
Hg2Cl2 + 2e^- \rightarrow 2Hg + 2Cl^- E^\circ = 0.27 V
AgCl + e^- \rightarrow Ag + Cl^- E^\circ = 0.22 V
SO2 + 4H^+ + 2e^- \rightarrow H2SO3 + H2O E^\circ = 0.20 V
Cu^{2+} + e^- \rightarrow Cu^+ E^\circ = 0.16 V
2H^+ + 2e^- \rightarrow H_2 E^\circ = 0.00 V
Fe^{3+} + 3e^- \rightarrow Fe E^\circ = -0.036 V
Pb^{2+} + 2e^- \rightarrow Pb E^\circ = -0.13 V
Sn^{2+} + 2e^- \rightarrow Sn E^\circ = -0.14 V
Ni^{2+} + 2e^- \rightarrow Ni E^\circ = -0.23 V
PbSO4 + 2e^- \rightarrow Pb + SO4^{2-} E^\circ = -0.35 V
Cd^{2+} + 2e^- \rightarrow Cd E^\circ = -0.40 V
Fe^{2+} + 2e^- \rightarrow Fe E^\circ = -0.44 V
Cr^{3+} + e^- \rightarrow Cr^{2+} E^\circ = -0.50 V
Cr^{3+} + 3e^- \rightarrow Cr E^\circ = -0.73 V
Zn^{2+} + 2e^- \rightarrow Zn E^\circ = -0.76 V
2H2O + 2e^- \rightarrow H2 + 2OH^- E^\circ = -0.83 V
Mn^{2+} + 2e^- \rightarrow Mn E^\circ = -1.18 V
Al^{3+} + 3e^- \rightarrow Al E^\circ = -1.66 V
H_2 + 2e^- \rightarrow 2H^- E^\circ = -2.23 V
Mg^{2+} + 2e^- \rightarrow Mg E^\circ = -2.37 V
La^{3+} + 3e^- \rightarrow La E^\circ = -2.37 V
Na^+ + e^- \rightarrow Na E^\circ = -2.71 V
Ca^{2+} + 2e^- \rightarrow Ca E^\circ = -2.76 V
Ba^{2+} + 2e^- \rightarrow Ba E^\circ = -2.90 V
K^+ + e^- \rightarrow K E^\circ = -2.92 V
Li^+ + e^- \rightarrow Li E^\circ = -3.05 V

Cell Potential Calculation Example

E_{cell} = 1.23V - 1.09V = 0.14V

Gibbs Free Energy Calculation

\Delta G^\circ = -nFE^\circ
For the reaction: 4Ag + O2 + 4H^+ \rightarrow 4Ag^+ + 2H2O
\Delta G = -(4 \text{moles}) \times (96485 \frac{C}{\text{mol} \cdot e^-}) \times (0.14V) = -54 \text{kJ}
The reaction is spontaneous.

Nernst Equation

Describes the dependence of cell potential on concentration.
Directly related to the dependence of free energy on concentration.
For a reaction not at equilibrium: aA \rightleftharpoons bB
Q = \frac{[B]^b}{[A]^a}
\Delta G = \Delta G^\circ + RT \ln(Q)
-nFE = -nFE^\circ + RT \ln(Q)
E = E^\circ - \frac{RT}{nF} \ln(Q)
Nernst Equation: E = E^\circ - \frac{RT}{nF} \ln(Q)
Calculate E^\circ before any current flows to find the maximum potential.

Nernst Equation Example

Calculate the potential for the given reaction at 10.0 °C:
Given:
- Na^+ (aq) + e^- \rightarrow Na (s) \quad E^\circ = -2.71 V
- Cu^{2+} (aq) + 2e^- \rightarrow Cu (s) \quad E^\circ = 0.34 V
Overall Reaction: 2Na(s) + Cu^{2+}(aq) \rightarrow Cu(s) + 2Na^+(aq)
Given concentrations: [Na^+] = 0.010 M, [Cu^{2+}] = 0.80 M
Oxidation: 2Na \rightarrow 2Na^+ + 2e^-
Reduction: Cu^{2+} + 2e^- \rightarrow Cu
n = 2 \text{ moles of } e^-
E^\circ = E{red}^\circ - E{ox}^\circ = 0.34 V - (-2.71 V) = 3.05 V
Q = \frac{[Na^+]^2}{[Cu^{2+}]} = \frac{(0.010)^2}{0.80} = 1.25 \times 10^{-4}
E = 3.05V - \frac{(8.314 \frac{J}{\text{mol} \cdot K})(283.15 K)}{(2 \text{ mol}) (96485 \frac{C}{\text{mol}})} \ln(1.25 \times 10^{-4}) = 3.16 V

Obtaining Equilibrium Constants from Electrochemical Potentials

At equilibrium, no net reaction occurs, so E = 0 V
0 = E^\circ - \frac{RT}{nF} \ln(Q)
Since at equilibrium, Q = K (equilibrium constant), then:
E^\circ = \frac{RT}{nF} \ln(K)
\ln(K) = \frac{E^\circ (nF)}{RT}
K = e^{\frac{E^\circ (nF)}{RT}}
If E^\circ is given, the temperature T is whatever temperature was used to obtain E^\circ.

Equilibrium Constant Calculation Example

Reaction at 25 °C:
- Reduction: 2Cr^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 2e^-
- Oxidation: Co^{2+}(aq) + 2e^- \rightarrow Co(s)
- Overall: 2Cr^{2+}(aq) + Co^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + Co(s) \quad E^\circ = 0.22 V
Calculate the equilibrium constant K:
K = e^{\frac{(0.22 \frac{J}{C}) (2 \text{ moles } e^-) (96485 \frac{C}{\text{mol}})}{(8.314 \frac{J}{\text{mol} \cdot K})(298.15 K)}}
K = e^{17.144} = 2.8 \times 10^7

Concentration Cell

E{cell}^\circ = E{red}^\circ - E{ox}^\circ = 0 V, but E{cell} \neq 0 V
Example: Nickel concentration cell with 0.05 M and 2.0 M NiCl_2 solutions.
Electrons flow from low to high concentration.
Ni is produced to increase concentration on the left cell to reach equilibrium (equal concentrations).
Ni(s) + Ni^{2+}(aq, 2.0M) \rightarrow Ni^{2+}(aq, 0.05M) + Ni(s)
E = E^\circ - \frac{RT}{nF} \ln(Q) \quad E^\circ = 0V
E = 0V - \frac{(8.314 \frac{J}{\text{mol} \cdot K})(298.15 K)}{(2 \text{ moles } e^-)(96485 \frac{C}{\text{mol}})} \ln(\frac{0.05}{2.0}) = 0.047 V

Electrolysis

Electrolytic Cell: Uses electrical energy to produce chemical change.
Example: 2Al2O3(s) \rightarrow 4Al(s) + 3O_2(g) \quad E^\circ = -2.06 V
Reverse of the spontaneous reaction. Not useful for Al(s) production by itself.
1 A = 1 \frac{C}{s}
F = 96485 \frac{C}{\text{mol } e^-}
Need to pass high energy electrons through a mixture of Al2O3 to drive the forward reaction.
General idea: Stoichiometry of electrolytic processes:
\text{Current } + \text{ Time } \rightarrow \text{ Moles } e^- \rightarrow \text{ Moles of substance produced } \rightarrow \text{ Mass of substance produced}
Use balanced redox reaction and Faraday's constant (F).

Electrolytic Cell Example

Generating Al(s) from Al^{3+} using an electrolytic cell.
Passing 5.00 A of current through a molten Al^{3+} salt for 20.0 min.
What mass of Al(s) can be produced under these conditions?
3e^- + Al^{3+}(aq) \rightarrow Al(s)
5.00 A = 5.00 \frac{C}{s}
Moles of electrons passed through the salt:
5.00 \frac{C}{s} \times 20.0 \text{ min} \times 60 \frac{s}{\text{min}} \times \frac{1 \text{ mol } e^-}{96485 C} = 0.06219 \text{ moles } e^-
Moles of Al(s) produced:
0.06219 \text{ moles } e^- \times \frac{1 \text{ mol } Al(s)}{3 \text{ moles } e^-} = 0.02073 \text{ mol } Al(s)
Mass of Al produced:
0.02073 \text{ mol } Al \times \frac{26.98 g}{1 \text{ mol } Al} = 0.559 g

Note

0.0(0)

Take a practice test

Chat with Kai

undefined Flashcards

0 Cards0.0(0)