Question 18.PS.9: Equilibrium Constant for a Redox Reaction Calculate the equi......
Equilibrium Constant for a Redox Reaction
Calculate the equilibrium constant K_c for the reaction
Fe(s) + Cd^{2+}(aq) \rightleftharpoons Fe^{2+}(aq) + Cd(s)
using the standard reduction potentials listed in Table 18.1.
| Table 18.1 Standard Reduction Potentials in Aqueous Solution at 25 °C* | ||
| Reduction Half-Reaction | E° (V) | |
| F_2(g) + 2 e^- | →2 F^-(aq) | + 2.87 |
| H_2O_2(aq) + 2 H_3O^+(aq) + 2 e^- | →4 H2O(\ell) | + 1.77 |
| PbO_2(s) + SO_4^{2-} (aq) + 4 H_3O^+(aq) +2 e^- | →PbSO_4 (s) +6 H_2O(\ell) | + 1.685 |
| MnO_4^-(aq) + 8 H_3O^+(aq) + 5 e^- | →Mn^{2+}(aq) +12 H_2O(\ell) | + 1.51 |
| Au^{3+}(aq) + 3 e^- | →Au(s) | + 1.50 |
| Cl_2(g) + 2 e^- | →2 Cl^-(aq) | + 1.358 |
| Cr_2O_7^{2-}(aq) + 14 H_3O^+(aq) + 6 e^- | →2 Cr^{3+}(aq) + 21 H_2O(\ell) | + 1.33 |
| O_2(g) + 4 H_3O^+(aq) + 4 e^- | →6 H_2O(\ell) | + 1.229 |
| Br_2 (\ell) + 2 e^- | →2 Br^-(aq) | + 1.066 |
| NO_3^-(aq) + 4 H_3O^+(aq) + 3 e^- | →NO(g) + 6 H_2O(\ell) | + 0.96 |
| OCl^-(aq) + H_2O(\ell) + 2 e^- | →Cl^-(aq) + 2 OH^-(aq) | + 0.89 |
| Hg^{2+}(aq) + 2 e^- | →Hg(\ell) | + 0.855 |
| Ag^+(aq) + e^- | →Ag(s) | + 0.7994 |
| Hg_2^{2+}(aq) + 2 e^- | →2 Hg(\ell) | + 0.789 |
| Fe^{3+}(aq) + e^- | →Fe^{2+}(aq) | + 0.771 |
| I_2 (s) + 2 e^- | →2 I^-(aq) | + 0.535 |
| O_2(g) + 2 H_2O(\ell) + 4 e^- | →4 OH^-(aq) | + 0.403 |
| Cu^{2+}(aq) + 2 e^- | →Cu(s) | + 0.337 |
| Sn^{4+}(aq) + 2 e^- | →Sn^{2+}(aq) | + 0.15 |
| 2 H_3O^+(aq) + 2 e^- | →H_2(g) + 2 H_2O(\ell) | 0.00 |
| Sn^{2+}(aq) + 2 e^- | →Sn(s) | – 0.14 |
| Ni^{2+}(aq) + 2 e^- | →Ni(s) | – 0.25 |
| PbSO_4 (s) + 2 e^- | →Pb(s) + SO_4^{2-}(aq) | – 0.356 |
| Cd^{2+}(aq) + 2 e^- | →Cd(s) | – 0.403 |
| Fe^{2+}(aq) + 2 e^- | →Fe(s) | – 0.44 |
| Zn^{2+}(aq) + 2 e^- | →Zn(s) | – 0.763 |
| 2 H_2O(\ell) + 2 e^- | →H_2(g) + 2 OH^-(aq) | – 0.8277 |
| Al^{3+}(aq) + 3 e^- | →Al(s) | – 1.66 |
| Mg^{2+}(aq) + 2 e^- | →Mg(s) | – 2.37 |
| Na^+(aq) + e^- | →Na(s) | – 2.714 |
| K^+(aq) + e^- | →K(s) | – 2.925 |
| Li^+(aq) + e^- | →Li (s) | – 3.045 |
| *In volts (V) versus the standard hydrogen electrode. | ||
Strategy and Explanation We first need to calculate E^\circ_{cell} . To do so we separate the reaction into its two half-reactions.
Fe( s ) → Fe^{2+}(aq) + 2 e^- E^\circ_{anode} = – 0.44 V
\underline{Cd^{2+}(aq) + 2 e^- → Cd(s) E^\circ_{cathode} = – 0.40 V}
Fe( s ) + Cd^{2+}(aq) → Fe^{2+}(aq) + Cd(s) E^\circ_{cell} = + 0.04 V
Two moles of electrons are transferred.
log K° = \frac{nE^\circ_{cell}}{0.0592 V} = \frac{(2)(0.04 V)}{0.0592 V} = 1.35 and K = 10^{1.35} = 22
The K° value is larger than 1, which shows that the reaction is product-favored as written. Since the reaction occurs in aqueous solution, K_c = K° = 22.
Reasonable Answer Check The value for E^\circ_{cell} is positive, which indicates a product-favored reaction, as does K° > 1.