Question 19.7: Relating E°cell and K Use the tabulated electrode potentials...
Relating E_{cell}^\circ and K
Use the tabulated electrode potentials to calculate K for the oxidation of copper by H^+ (at 25 °C):
Cu(s) + 2 H^+(aq) \longrightarrow Cu^{2+}(aq) + H_2( g)
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Learn more on how we answer questions.
SORT You are given a redox reaction and asked to find K. | GIVEN: Cu(s) + 2 H^+(aq) \longrightarrow Cu^{2+}(aq) + H_2( g)
FIND: K |
STRATEGIZE Refer to the values of electrode potentials in Table 19.1 to calculate E_{cell}^\circ . Then use Equation 19.6 to calculate K from E_{cell}^\circ .
E_{cell}^\circ =\frac{0.0592 V}{n} log K [19.6] |
CONCEPTUAL PLAN
E_{an}^\circ ,E_{cat}^\circ \longrightarrow E_{cell}^\circ
E_{cell}^\circ \underset{E_{cell}^\circ=\frac{0.0592 V}{n}log K }{\longrightarrow} K |
SOLVE Separate the reaction into oxidation and reduction half-reactions and find the standard electrode potentials for each. Find E_{cell}^\circ by subtracting E_{an} from E_{cat}. | SOLUTION
Oxidation (Anode): Cu(s) \longrightarrow Cu^{2+}(aq) + \cancel{2 e^-} E° = 0.34 V Reduction (Cathode): 2 H^+(aq) + \cancel{2 e^-} \longrightarrow H_2(g) E° = 0.00 V \\ \overline{ Cu(s) + 2 H^+(aq) \longrightarrow Cu^{2+}(aq) + H_2(g) E_{cell}^\circ=E_{cat}^\circ -E_{an}^\circ = -0.34 V } |
Calculate K from E_{cell}^\circ . The value of n (the number of moles of electrons) corresponds to the number of electrons that are canceled in the half-reactions. | E_{cell}^\circ =\frac{0.0592 V}{n} log K
log K =E_{cell}^\circ \frac{n}{0.0592 V} log K = -0.34 \cancel{V} \frac{2}{0.0592 \cancel{V}} =-11.4\underline{8}6
K = 10^{-11.4\underline{8}6} =3.3 \times 10^{-12} |
CHECK The answer has no units, as expected for an equilibrium constant. The magnitude of the answer is small, indicating that the reaction lies far to the left at equilibrium, as expected for a reaction in which E_{cell}^\circ is negative. |
Table 19.1
Standard Electrode Potentials at 25 °C | ||||
Reduction Half-Reaction | E°(V) | |||
![]() |
F_2(g) + 2 e^- | \longrightarrow 2 F^-(aq) | 2.87 | ![]() |
H_2O_2(aq) + 2 H^+(aq) + 2 e^- | \longrightarrow 2 H_2O(l) | 1.78 | ||
PbO_2(s) + 4 H^+(aq) + SO_4{}^{2-}(aq) + 2 e^- | \longrightarrow PbSO_4(s) + 2 H_2O(l) | 1.69 | ||
MnO_4^-(aq) + 4 H^+(aq) + 3 e^- | \longrightarrow MnO_2(s) + 2 H_2O(l) | 1.68 | ||
MnO_4{}^-(aq) + 8 H^+(aq) + 5 e^- | \longrightarrow Mn^{2 +}(aq) + 4 H_2O(l) | 1.51 | ||
Au^{3 +}(aq) + 3 e^- | \longrightarrow Au(s) | 1.50 | ||
PbO_2(s) + 4 H^+(aq) + 2 e^- | \longrightarrow Pb^{2 +}(aq) + 2 H_2O(l) | 1.46 | ||
Cl_2(g) + 2 e^- | \longrightarrow 2 Cl^-(aq) | 1.36 | ||
Cr_2O_7{}^{2 -}(aq) + 14 H^+(aq) + 6 e^- | \longrightarrow 2 Cr^{3 +}(aq) + 7 H_2O(l) | 1.33 | ||
O_2(g) + 4 H^+(aq) + 4 e^- | \longrightarrow 2 H_2O(l) | 1.23 | ||
MnO_2(s) + 4 H^+(aq) + 2 e^- | \longrightarrow Mn^{2 +}(aq) + 2 H_2O(l) | 1.21 | ||
IO_3^-(aq) + 6 H^+(aq) + 5 e^- | \longrightarrow \frac{1}{2} I_2(aq) + 3 H_2O(l) | 1.20 | ||
Br_2(l) + 2 e^- | \longrightarrow 2 Br^-(aq) | 1.09 | ||
VO_2^{+}(aq) + 2 H^+(aq) + e^- | \longrightarrow VO^{2 +}(aq) + H_2O(l) | 1.00 | ||
NO_3{}^{-}(aq) + 4 H^+(aq) + 3 e^- | \longrightarrow NO(g) + 2 H_2O(l) | 0.96 | ||
ClO_2(g) + e^- | \longrightarrow ClO_2{}^-(aq) | 0.95 | ||
Ag^+(aq) + e^- | \longrightarrow Ag(s) | 0.80 | ||
Fe^{3 +}(aq) + e^- | \longrightarrow Fe^{2 +}(aq) | 0.77 | ||
O_2(g) + 2 H^+(aq) + 2 e^- | \longrightarrow H_2O_2(aq) | 0.70 | ||
MnO_4{}^-(aq) + e^- | \longrightarrow MnO_4{}^{2-}(aq) | 0.56 | ||
I_2(s) + 2 e^- | \longrightarrow 2 I^-(aq) | 0.54 | ||
Cu^+(aq) + e^- | \longrightarrow Cu(s) | 0.52 | ||
O_2(g) + 2 H_2O(l) + 4 e^- | \longrightarrow 4 OH^-(aq) | 0.40 | ||
Cu^{2+}(aq) + 2 e^- | \longrightarrow Cu(s) | 0.34 | ||
SO_4{}^{2 -}(aq) + 4 H^+(aq) + 2 e^- | \longrightarrow H_2SO_3(aq) + H_2O(l) | 0.20 | ||
Cu^{2+}(aq) + e^- | \longrightarrow Cu^+(aq) | 0.16 | ||
Sn^{4+}(aq) + 2 e^- | \longrightarrow Sn^{2 +}(aq) | 0.15 | ||
2 H^+(aq) + 2 e^- | \longrightarrow H_2(g) | 0 | ||
Fe^{3+}(aq) + 3 e^- | \longrightarrow Fe(s) | –0.036 | ||
Pb^{2+}(aq) + 2 e^- | \longrightarrow Pb(s) | -0.13 | ||
Sn^{2+}(aq) + 2 e^- | \longrightarrow Sn(s) | -0.14 | ||
Ni^{2+}(aq) + 2 e^- | \longrightarrow Ni(s) | -0.23 | ||
Cd^{2+}(aq) + 2 e^- | \longrightarrow Cd(s) | -0.40 | ||
Fe^{2+}(aq) + 2 e^- | \longrightarrow Fe(s) | -0.45 | ||
Cr^{3+}(aq) + e^- | \longrightarrow Cr^{2+}(aq) | -0.50 | ||
Cr^{3+}(aq) + 3 e^- | \longrightarrow Cr(s) | -0.73 | ||
Zn^{2+}(aq) + 2 e^- | \longrightarrow Zn(s) | -0.76 | ||
2 H_2O(l) + 2 e^- | \longrightarrow H_2(g) + 2 OH^-(aq) | -0.83 | ||
Mn^{2+}(aq) + 2 e^- | \longrightarrow Mn(s) | -1.18 | ||
Al^{3+}(aq) + 3 e^- | \longrightarrow Al(s) | -1.66 | ||
Mg^{2+}(aq) + 2 e^- | \longrightarrow Mg(s) | -2.37 | ||
Na^+(aq) + e^- | \longrightarrow Na(s) | -2.71 | ||
Ca^{2+}(aq) + 2 e^- | \longrightarrow Ca(s) | -2.76 | ||
Ba^{2+}(aq) + 2 e^- | \longrightarrow Ba(s) | -2.90 | ||
K^+(aq) + e^- | \longrightarrow K(s) | -2.92 | ||
Li^+(aq) + e^- | \longrightarrow Li(s) | -3.04 |
Related Answered Questions
Question: 19.5
Verified Answer:
(a) Fe(s) + Mg^{2+}(aq) \longrightarrow Fe^...
Question: 19.8
Verified Answer:
Question: 19.3
Verified Answer:
To balance redox reactions occurring in basic solu...