Question 19.3: Predict what redox reaction will take place, if any, when mo...

Predict what redox reaction will take place, if any, when molecular bromine (\text{Br}_2) is added to (a) a 1-M solution of \text{NaI} and (b) a 1-M solution of \text{NaCl}. (Assume a temperature of 25°\text{C}.)

Strategy In each case, write the equation for the redox reaction that might take place and use values to determine whether or not the proposed reaction will actually occur.

Setup From Table 19.1:

\text{Br}_2(l) + 2e^– → 2\text{Br}^–(aq)           E° = 1.07  \text{V}

\text{I}_2(s) + 2e^– → 2\text{I}^–(aq)           E° = 0.53 V

\text{Cl}_2(\text{g}) + 2e^– → 2\text{Cl}^–(aq)           E° = 1.36 V

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.
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(a) If a redox reaction is to occur, it will be the oxidation of \text{I}^– ions by \text{Br}_2:

\text{Br}_2(l) + 2\text{I}^–(aq) → 2\text{Br}^–(aq) + \text{I}_2(s)

Because the reduction potential of \text{Br}_2 is greater than that of \text{I}_2, \text{Br}_2 will be reduced to \text{Br}^– and \text{I}^– will be oxidized to \text{I}_2. Thus, the preceding reaction will occur.

(b) In this case, the proposed reaction is the reduction of \text{Br}_2 by \text{Cl}^– ions:

\text{Br}_2(l) + 2\text{Cl}^–(aq) → 2\text{Br}^–(aq) + \text{Cl}_2(\text{g})

However, because the reduction potential of \text{Br}_2 is smaller than that of \text{Cl}_2, this reaction will not occur. \text{Cl}_2 is more readily reduced than \text{Br}_2, so \text{Br}_2 is not reduced by \text{Cl}^–.

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