# Answers to Chapter 18 Study Questions

• • Contributed by Delmar Larsen
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1. -3; $$\mathrm{[x + 3(1) = 0]}$$
2. 0
3. +4; $$\mathrm{[x +2(-2) = 0]}$$
4. +5; $$\mathrm{[x + 3(-2) = -1]}$$

1. $$\ce{Br2}$$
2. $$\ce{I}$$
3. $$\ce{Br}$$
4. $$\ce{Br2}$$
5. $$\ce{I-}$$

1. First determine whether substances may be oxidized or reduced, then look up $$\mathrm{E^\circ_{ox}}$$ for reducing agents and $$\mathrm{E^\circ_{red}}$$ for oxidizing agents and list in order.

Oxidizing agents (are reduced): $$\ce{H+}$$ ($$\mathrm{E^\circ_{red} = 0\, V}$$).

Reducing agents (are oxidized): $$\ce{Ni}$$ ($$\mathrm{E^\circ_{ox} = +0.23\, V}$$), $$\ce{Au}$$ ($$\mathrm{E^\circ_{ox} = -1.50\, V}$$), $$\ce{Mg}$$ ($$\mathrm{E^\circ_{ox} = +2.37\, V}$$).

Both: $$\ce{Sn^2+}$$ ($$\mathrm{E^\circ_{red} = -0.14\, V}$$; $$\mathrm{E^\circ_{ox} = -0.15\, V}$$), $$\ce{Fe^2+}$$ ($$\mathrm{E^\circ_{red} = -0.44\, V}$$, $$\mathrm{E^\circ_{ox} = -0.77\, V}$$), $$\ce{Cl2}$$ ($$\mathrm{E^\circ_{red} = 1.36\, V}$$, $$\mathrm{E^\circ_{ox} = -1.47\, V}$$).

Oxidizing agents (list in order of decreasing $$\mathrm{E^\circ_{red}}$$): $$\mathrm{Cl_2 > H^+ > Sn^{2+} > Fe^{2+}}$$

Reducing agents (list in order of decreasing $$\mathrm{E^\circ_{ox}}$$): $$\mathrm{Mg > Ni > Sn^{2+} > Fe^{2+} > Cl_2 > Au}$$

1. ox: $$\ce{[Ag \rightarrow Ag+ + e^- ]\times 2}$$; red: $$\ce{Cu^2+ + 2 e^- \rightarrow Cu}$$
balanced: $$\ce{2 Ag(s) + Cu^2+ (aq) \rightarrow 2 Ag+ (aq) + Cu(s)}$$
$$\mathrm{E^\circ = E^\circ_{ox}(Ag) + E^\circ_{red}(Cu^{2+}) = -0.80\, V + 0.34\, V = -0.46\, V}$$.
Not spontaneous.
2. ox: $$\ce{[Ni(s) \rightarrow Ni^2+ + 2 e^- ] \times 5}$$; red: $$\ce{[5 e^- + 8 H+ + MnO_4- \rightarrow Mn^2+ + 4 H2O] \times 2}$$
balanced: $$\ce{5 Ni(s) + 16 H+ (aq) + 2 MnO_4- (aq) \rightarrow 5 Ni^2+ (aq) + 2 Mn^2+ (aq) + 8 H2O}$$
$$\mathrm{E^\circ = E^\circ_{ox}(Ni) + E^\circ_{red}(MnO_4^-) = +0.23\, V + 1.51\, V = +1.74\, V}$$.
Spontaneous.
3. ox: $$\ce{[2 H2O + Mn^2+ \rightarrow MnO2 + 4 H+ + 2 e^- ] \times 3}$$
red: $$\ce{[3 e^- + 4 H+ + NO3- \rightarrow NO + 2 H2O] \times 2}$$
balanced: $$\ce{2 H2O + 3 Mn^2+ (aq) + 2 NO3- (aq) \rightarrow 3 MnO2(s) + 4 H+ (aq) + 2 NO(g)}$$
$$\mathrm{E^\circ = E^\circ_{ox}(Mn^{2+}) + E^\circ_{red}(NO_3^-) = -1.21\, V + 0.96\, V = -0.25\, V}$$.
Not spontaneous.

1. $$\mathrm{E^\circ = E^\circ_{ox}(Fe) + E^\circ_{red}(Fe^{3+}) = +0.44\, V + 0.77\, V = +1.21\, V}$$.
Spontaneous.
2. $$\mathrm{E^\circ = E^\circ_{ox}(I^-) + E^\circ_{red}(Zn^{2+}) = - 0.54\, V - 0.76\, V = -1.30\, V}$$.
Not spontaneous.

1. $$\mathrm{E^\circ_{ox}(Br^-) = -1.09\, V}$$. An ion will oxidize $$\ce{Br-}$$ if it has an $$\mathrm{E^\circ_{red} > +1.09\, V}$$.
1. $$\mathrm{E^\circ_{red}(Pb^{2+}) = -0.13\, V \Rightarrow}$$ No.
2. $$\mathrm{E^\circ_{red}(H^+) = 0 \Rightarrow}$$ No.
3. $$\mathrm{E^\circ_{red}(Au^{3+}) = +1.50\, V \Rightarrow}$$ Yes.
4. $$\mathrm{E^\circ_{red}(MnO_4^-) = +1.51\, V \Rightarrow}$$ Yes.

Answer: $$\ce{Au^3+}$$ and $$\ce{MnO4-}$$

1. $$\mathrm{E^\circ_{ox}(Al) = +1.66\, V}$$; $$\mathrm{E^\circ_{ox}(Pb) = +0.13\,V}$$. $$\ce{Al}$$ has a greater tendency to be oxidized. Thus, for spontaneous reaction, $$\ce{Al}$$ will be oxidized and $$\ce{Pb}$$ will be reduced:

$$\ce{2 Al(s) + 3 Pb^2+ (aq) \rightarrow 2 Al^3+ (aq) + 3 Pb(s)}$$ 1. lead ($$\ce{Pb}$$); since $$\ce{Pb^2+ \rightarrow Pb}$$

1. Iron ($$\ce{Fe}$$) is the strongest reducing agent because it is always oxidized. It reduces all of the other metals.
2. Gold ($$\ce{Au}$$) is the weakest reducing agent because it is never oxidized.
3. Using the voltages of the other metals with $$\ce{Pb}$$.
 Half-reaction $$\mathrm{E^\circ_{red}}$$ (volts) $$\mathrm{Au^{3+} + 3 e^- \leftrightarrow Au}$$ + 0.80 V $$\mathrm{Pb^{2+} + 2 e^- \leftrightarrow Pb}$$ 0.00 V $$\mathrm{Ni^{2+} + 2 e^- \leftrightarrow Ni}$$ - 0.10 V $$\mathrm{Fe^{2+} + 2 e^- \leftrightarrow Fe}$$ - 0.25 V