Question 19.6: Predict whether the following reaction will occur spontaneou...

Predict whether the following reaction will occur spontaneously as written at 298   K:

\text{Co}(s) + \text{Fe}^{2+}(aq) → \text{Co}^{2+}(aq) + \text{Fe}(s)

assuming [\text{Co}^{2+}] = 0.15  M and [\text{Fe}^{2+}] = 0.68   M.

Strategy Use E^\circ values from Table 19.1 to determine E^\circ for the reaction, and use Equation 19.7 to calculate E. If E is positive, the reaction will occur spontaneously.

Setup From Table 19.1,

Cathode  (reduction):          \text{Fe}^{2+}(aq) + 2e^– → \text{Fe}(s)

Anode  (oxidation):          \text{Co}(s) → \text{Co}^{2+}(aq) + 2e^–

E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}

= E^\circ_{\text{Fe}^{2+}/\text{Fe}} − E^\circ_{\text{Co}^{2+}/\text{Co}}

= –0.44  \text{V} − (–0.28  \text{V})

= –0.16  \text{V}

The reaction quotient, Q, for the reaction is [\text{Co}^{2+}]/[\text{Fe}^{2+}]. Therefore, Q = (0.15/0.68) = 0.22

E = E^\circ − \frac{0.0592  \text{V}}{n}  \text{log}  Q           Equation 19.7

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.
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.

From Equation 19.7,

E = E^\circ − \frac{0.0592  \text{V}}{n}  \text{log}  Q

= –0.16 V − \frac{0.0592  \text{V}}{2} \text{log}  0.22

= –0.14  \text{V}

The negative E value indicates that the reaction is not spontaneous as written under the conditions described.

Related Answered Questions