Question 19.4: Calculate the standard free-energy change for the following ...
Calculate the standard free-energy change for the following reaction at 25°\text{C}:
2\text{Au}(s) + 3\text{Ca}^{2+}(1.0 M) \rightleftarrows 2\text{Au}^{3+}(1.0 M) + 3\text{Ca}(s)
Strategy Use E^\circ values from Table 19.1 to calculate E^\circ for the reaction, and then use Equation 19.3 to calculate the standard free-energy change.
Setup The half-cell reactions are
Cathode (reduction) : 3\text{Ca}^{2+}(aq) + 6e^– → 3\text{Ca}(s)
Anode (oxidation) : 2\text{Au}(s) → 2\text{Au}^{3+}(aq) + 6e^–
From Table 19.1, E^\circ_{\text{Ca}^{2+}/\text{Ca}} = –2.87 \text{V} and E^\circ_{\text{Au}^{3+}/\text{Au}} = 1.50 \text{V}
ΔG^\circ= –nFE^\circ_{cell} Equation 19.3
| TA B L E 1 9 . 1 Standard Reduction Potentials at 25°\text{C} ^* | |||
| \text{Increasing strength as oxidizing agent}
\uparrow |
Half-Reaction | E°(\text{V}) |
\text{ncreasing strength as reducing agent} \downarrow |
| \text{F}_2(\text{g}) + 2e^– → 2\text{F}^–(aq) | +2.87 | ||
| \text{O}_3(\text{}g) + 2\text{H}^+(aq) + 2e^– → \text{O}_2(\text{g}) + \text{H}_2\text{O}(l) | +2.07 | ||
| \text{Co}^{3+}(aq) + e^– → \text{Co}^{2+}(aq) | +1.82 | ||
| \text{H}_2\text{O}_2(aq) + 2\text{H}^+(aq) + 2e^– → 2\text{H}_2\text{O}(l) | +1.77 | ||
| \text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^2–(aq) + 2e^– → \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) | +1.70 | ||
| \text{Ce}^{4+}(aq) + e^– → \text{Ce}^{3+}(aq) | +1.61 | ||
| \text{MnO}_4^– (aq) + 8\text{H}^+(aq) + 5e^– → \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) | +1.51 | ||
| \text{Au}^{3+}(aq) + 3e^– → \text{Au}(s) | +1.50 | ||
| \text{Cl}_2(\text{g}) + 2e^– → 2\text{Cl}^–(aq) | +1.36 | ||
| \text{Cr}_2\text{O}_7^{2–}(aq) + 14\text{H}^+(aq) + 6e^– → 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) | +1.33 | ||
| \text{MnO}_2(s) + 4\text{H}^+(aq) + 2e^– → \text{Mn}^{2+}(aq) + 2\text{H}_2\text{O}(l) | +1.23 | ||
| \text{O}_2(\text{g}) + 4\text{H}^+(aq) + 4e^– → 2\text{H}_2\text{O}(l) | +1.23 | ||
| \text{Br}_2(l) + 2e^– → 2\text{Br}^–(aq) | +1.07 | ||
| \text{NO}_3^– (aq) + 4\text{H}^+(aq) + 3e^– → \text{NO(g)} + 2\text{H}_2\text{O}(l) | +0.96 | ||
| 2\text{Hg}^{2+}(aq) + 2e^– → \text{Hg}_2^{2+}(aq) | +0.92 | ||
| \text{Hg}_2^{2+}(aq) + 2e^– → 2\text{Hg}(l) | +0.85 | ||
| \text{Ag}^+(aq) + e^– → \text{Ag}(s) | +0.80 | ||
| \text{Fe}^{3+}(aq) + e^– → \text{Fe}^{2+}(aq) | +0.77 | ||
| \text{O}_2(\text{g}) + 2\text{H}^+(aq) + 2e^– → \text{H}_2\text{O}_2(aq) | +0.68 | ||
| \text{MnO}_4^– (aq) + 2\text{H}_2\text{O}(l) + 3e^– → \text{MnO}_2(s) + 4\text{OH}^–(aq) | +0.59 | ||
| \text{I}_2(s) + 2e^– → 2\text{I}^–(aq) | +0.53 | ||
| \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(l) + 4e^– → 4\text{OH}^–(aq) | +0.40 | ||
| \text{Cu}^{2+}(aq) + 2e^– → \text{Cu}(s) | +0.34 | ||
| \text{AgCl}(s) + e^– → \text{Ag}(s) + \text{Cl}^–(aq) | +0.22 | ||
| \text{SO}_4^{2–}(aq) + 4\text{H}^+(aq) + 2e^– → \text{SO}_2(\text{g}) + 2\text{H}_2\text{O}(l) | +0.20 | ||
| \text{Cu}^{2+}(aq) + e^– → \text{Cu}^+(aq) | +0.15 | ||
| \text{Sn}^{4+}(aq) + 2e^– → \text{Sn}^{2+}(aq) | +0.13 | ||
| 2\text{H}^+(aq) + 2e^– → \text{H}_2(\text{g}) | 0.00 | ||
| \text{Pb}^{2+}(aq) + 2e^– → \text{Pb}(s) | –0.13 | ||
| \text{Sn}^{2+}(aq) + 2e^– → \text{Sn}(s) | –0.14 | ||
| \text{Ni}^{2+}(aq) + 2e^– → \text{Ni}(s) | –0.25 | ||
| \text{Co}^{2+}(aq) + 2e^– → \text{Co}(s) | –0.28 | ||
| \text{PbSO}_4(s) + 2e^– → \text{Pb}(s) + \text{SO}_4^{2–}(aq) | –0.31 | ||
| \text{Cd}^{2+}(aq) + 2e^– → \text{Cd}(s) | –0.40 | ||
| \text{Fe}^{2+}(aq) + 2e^– → \text{Fe}(s) | –0.44 | ||
| \text{Cr}^{3+}(aq) + 3e^– → \text{Cr}(s) | –0.74 | ||
| \text{Zn}^{2+}(aq) + 2e^– → \text{Zn}(s) | –0.76 | ||
| 2\text{H}_2\text{O}(l) + 2e^– → \text{H}_2(\text{g}) + 2\text{OH}^–(aq) | –0.83 | ||
| \text{Mn}^{2+}(aq) + 2e^– → \text{Mn}(s) | –1.18 | ||
| \text{Al3}^+(aq) + 3e^– → \text{Al}(s) | –1.66 | ||
| \text{Be}^{2+}(aq) + 2e^– → \text{Be}(s) | –1.85 | ||
| \text{Mg}^{2+}(aq) + 2e^– → \text{Mg}(s) | –2.37 | ||
| \text{Na}^+(aq) + e^– → \text{Na}(s) | –2.71 | ||
| \text{Ca}^{2+}(aq) + 2e^– → \text{Ca}(s) | –2.87 | ||
| \text{Sr}^{2+}(aq) + 2e^– → \text{Sr}(s) | –2.89 | ||
| \text{Ba}^{2+}(aq) + 2e^– → \text{Ba}(s) | –2.90 | ||
| \text{K}^+(aq) + e^– → \text{K}(s) | –2.93 | ||
| Li+(aq) + e– → Li(s) | –3.05 | ||
| ^* For all half-reactions the concentration is 1 M for dissolved species and the pressure is 1 atm for gases. These are the standard-state values. | |||
E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}
= E^\circ_{\text{Ca}^{2+}/\text{Ca}} − E^\circ_{\text{Au}^{3+}/\text{Au}}
= –2.87 \text{V} − 1.50 \text{V}
= –4.37 \text{V}
Next, substitute this value of E^\circ into Equation 19.3 to obtain ∆G^\circ:
∆G^\circ = –nFE^\circ
The overall reaction shows that n = 6, so
∆G^\circ = –(6 \text{mol} e^–)(96,500 \text{J/V ⋅ mol} e^–)(–4.37 \text{V})
= 2.53 × 10^6 \text{J/mol}
= 2.53 × 10^3 \text{kJ/mol}