Question 19.7: Predicting Whether a Redox Reaction Is Spontaneous Predict f...

Chemistry

Predicting Whether a Redox Reaction Is Spontaneous
Predict from Table 19.1 whether $Pb^{2+}(aq)$ can oxidize Al(s) or Cu(s) under standard state conditions. Calculate E° for each reaction at 25 °C.

STRATEGY
To predict whether a redox reaction is spontaneous, remember that an oxidizing agent can oxidize any reducing agent that lies below it in the table but can’t oxidize one that lies above it. To calculate E° for a redox reaction, sum the E° values for the reduction and oxidation half-reactions. If the value of E° is positive then the reaction is spontaneous.

TABLE 19.1 Standard Reduction Potentials at 25 °C
	Reduction Half-Reaction	E° (V)
	$F_{2}(g) + 2 e^{-} → 2F^{-}(aq)$	2.87
	$H_{2}O_{2}(aq) + 2 H^{+}(aq) + 2 e^{-} → 2H_{2}O(l)$	1.78
	$MnO^{-}_{4}(aq) + 8 H^{+}(aq) + 5 e^{-}→ Mn^{2+}(aq) + 4 H_{2}O(l)$	1.51
	$Cl_{2}(g) + 2 e^{-} → 2Cl^{-}(aq)$	1.36
	$Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(l)$	1.36
	$O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} → 2 H_{2}O(l)$	1.23
	$Br_{2}(aq) + 2 e^{-} → 2 Br^{-}(aq)$	1.09
	$Ag^{+}(aq) + e^{-} → Ag(s)$	0.80
	$Fe^{3+}(aq) + e^{-} → Fe^{2+}(aq)$	0.77
	$O_{2}(g) + 2 H^{+}(aq) + 2 e^{-} → H_{2}O_{2}(aq)$	0.70
	$I_{2}(s) + 2 e^{-} → 2 I^{-}(aq)$	0.54
	$O_{2}(g) + 2 H_{2}O(l) + 4 e^{-} → 4 OH^{-}(aq)$	0.40
	$Cu^{2+}(aq) + 2 e^{-} → Cu(s)$	0.34
	$Sn^{4+}(aq) + 2 e^{-} → Sn^{2+}(aq)$	0.15
	$2 H^{+}(aq) + 2 e^{-} → H_{2}(g)$	0
	$Pb^{2+}(aq) + 2 e^{-} → Pb(s)$	-0.13
	$Ni^{2+}(aq) + 2 e^{-} → Ni(s)$	-0.26
	$Cd^{2+}(aq) + 2 e^{-} → Cd(s)$	-0.40
	$Fe^{2+}(aq) + 2 e^{-} → Fe(s)$	-0.45
	$Zn^{2+}(aq) + 2 e^{-} → Zn(s)$	-0.76
	$2 H_{2}O(l) + 2 e^{-} → H_{2}(g) + 2 OH^{-}(aq)$	-0.83
	$Al^{3+}(aq) + 3 e^{-} → Al(s)$	-1.66
	$Mg^{2+}(aq) + 2 e^{-} → Mg(s)$	-2.37
	$Na^{+}(aq) + e^{-} → Na(s)$	-2.71
	$Li^{+}(aq) + e^{-} → Li(s)$	-3.04

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$Pb^{2+}(aq)$ is above Al(s) in the table but below Cu(s). Therefore, $Pb^{2+}(aq)$ can oxidize Al(s) but can’t oxidize Cu(s). To confirm these predictions, calculate E° values for the overall reactions.

For the oxidation of Al by $Pb^{2+}$ , E° is positive (1.53 V), and the reaction is therefore spontaneous:

\begin{array}{rc} \begin{matrix} 3 × [Pb^{2+}(aq) + 2 e^{-} → Pb(s)] \\ 2 × [ Al(s) → Al^{3+}(aq) + 3 e^{-} \end{matrix} & \begin{matrix} E° = -0.13  V \\ E° = 1.66  V \\ \end{matrix} \\ \hline \begin{matrix} 3 Pb^{2+}(aq) + 2 Al(s) → 3 Pb(s) + 2 Al^{3+}(aq) \\ \end{matrix} & \begin{matrix} E° = 1.53  V \\ \end{matrix}\end{array}

Note that we have multiplied the $Pb^{2+} /Pb$ half-reaction by a factor of 3 and the $Al/Al^{3+}$ half-reaction by a factor of 2 so that the electrons will cancel, but we do not multiply the E° values by these factors because electrical potential is an intensive property. For the oxidation of Cu by $Pb^{2+}$ , E° is negative (-0.47 V) and the reaction is therefore nonspontaneous: