Solutions 1: Electrochemistry

These are solutions to select problems from homework set #1.

Q1.1

Determine the oxidation states of the underlined elements. Hint: Read Module on Oxidation States

1. Using rule #1, Cl isn't combine with any other element other than itself, therefore the oxidation state for Cl is 0.
2. Using rule #2, we know that the sum of the oxidation states for Na and H is 0 because NaH is a neutral compound. Now, we might be tempted to say H has the oxidation state of +1 because of rule #5. However, recall that one must use the rules in order.  Because Na is in Group 1A, according to rule #3, the oxidation state of Na is +1. Thus, H must have the oxidation state of -1 so that the sum of the oxidation state for Na and H is 0.
3. Using rule #2, the sum of the oxidation state is 0. Using rule #5, H has the oxidation state of +1, and using rule #6, O has the oxidation state of -2.  The equation for H2CO:  2(+1) + C + (-2) = 0. -----> 2 + C - 2 = 0  --------> C = 0
4. Using rule #2, the sum of the oxidation state is -2.  Using rule #6, O has the oxidation state of -2. We want to know the oxidation of S, so the equation forS2O3 2-: 2S + 3(-2) = -2  ------> 2S -6 = -2 ------>   2S = +4  ------> S = 4/2 -----> S = 2
5. Using rule #2, the sum of the oxidation state is 0, and using rule #3, the oxidation state for K is +1. Using rule #6, O has the oxidation state of -2. The equation for KMnO4: 1 + Mn + 4(-2) = 0 -----> Mn = -1 + 8 -----> Mn = 7
6. Using rule #2, the sume of oxidation state is 0. Using rule #7, Cl has a -1 charge. The equation for FeCl3 Fe + 3(-1) = 0 -----> Fe -3 = 0 -----> Fe = 3
7. Using rule #1, N isn't combine with any other element other than itself, therefore the oxidation state for N is 0.
8. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1. Using rule #6, O has the oxidation state of -2. The equation for H2SO4: 2(1) + S + 4(-2) = 0 -----> S = -2 + 8 -----> S= 6
9. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1. Using rule #6, O has the oxidation state of -2. The equation for HClO2: 1 + Cl + 2(-2) = 0 -----> Cl = -1 + 4 -----> Cl = 3
10. Using rule #2, the sum of the oxidation state is 0.  Using rule #6, O has the oxidation state of -2. Cu only have one charge which is +2. The equation for CuSO4: 2 + S + 4(-2) = 0 -----> S = -2 + 8 -----> S = 6

Q1.2

Balance the following equations in both acidic and basic environments. Hint: Read Module on balancing REDOX reactions.

1) $$H_2(g) + O_2(g) \rightarrow H_2O(l)$$

• Acidic Answer: $$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$
• Basic Answer: $$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$

2) $$Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)$$

• Acidic Answer: $$2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 16H^+(aq) \rightarrow 4Cr^{3+}(aq) + 2CO_2(g) + 11H_2O(l)$$
• Basic Answer: $$2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 5 H_2O(l) \rightarrow 4Cr^{3+}(aq) + 2CO_2(g) + 16OH^-(aq)$$

3) $$Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)$$

• Acidic Answer: $$MnO_4^-(aq) + 5Fe^{2+}(aq) + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$$
• Basic Answer: $$MnO_4^-(aq) + 5Fe^{2+}(aq) + 4H_2O(l) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 8OH^-(aq)$$

4) $$Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)$$

• Acidic Answer: $$3Zn(s) + 2NO_3^-(aq) + 8H^+(aq) \rightarrow 3Zn^{2+}(aq) + 2NO(g) + 4H_2O(l)$$
• Basic Answer: $$3Zn(s) + 2NO_3^-(aq) + 4H_2O(l) \rightarrow 3Zn^{2+}(aq) + 2NO(g) + 8OH^-(aq)$$

5) $$Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)$$

• Acidic Answer: $$4Al(s) + 3O_2(g) + 10 H_2O(l) \rightarrow 4[Al(OH)_4]^-(aq) + 4H^+(aq)$$
• Basic Answer: $$4Al(s) + 3O_2(g) + 6H_2O(l) + 4OH^-(aq) \rightarrow 4[Al(OH)4]^-(aq)$$

Q1.3

not given. But read Using Redox Potentials to Predict the Feasibility of Reactions if you are confused.

Q1.4

1. $$\mathrm{Al(s)| Al^{3+}|| Zn^{2+}(aq)| Zn(s)}$$

Oxidation (anode):  $$\mathrm{Al(s) \rightarrow Al^{3+} + 3e^- \quad E=-1.676\,V}$$

Reduction (cathode): $$\mathrm{Zn^{2+} + 2e^- \rightarrow Zn(s) \quad E=-0.7618\,V}$$

\begin{align} \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\ & = -0.7618 - (-1.676) \,\mathrm V \\ & = +0.9142\,\mathrm{V} \end{align}

1. $$\mathrm{Pt (s)| Fe^{2+}(aq), Fe^{3+}(aq)|| Cu^{2+}(aq)| Cu(s)}$$

Oxidaiton (anode): $$\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+} \quad E=0.771\,V}$$

Reduction (cathode): $$\mathrm{Cu(s)\rightarrow Cu^{2+} + 2e^- \quad E= 0.3419\,V}$$

\begin{align} \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\ & = 0.3419 - 0.771\,\mathrm V \\ & = 0.4291\,\mathrm{V} \end{align}

1. $$\mathrm{Pt (s)| Cr^{3+}(aq), Cr_2O_7^{2-}(aq)|| Ag^+(aq)| Ag(s)}$$

Oxidation (cathode): $$\mathrm{ 2 Cr^{3+} + 7H_2O(l) \rightarrow Cr_2O_7^{2−} + 14H^+ + 6e^- \quad E=1.36\,V}$$

Reduction (anode): $$\mathrm{Ag(s) \rightarrow Ag^+ + e^- \quad E=0.7996\,V}$$

\begin{align} \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\ & = \mathrm{0.7996\,V-1.36\,V} \\ & = -0.56\,\mathrm{V} \end{align}

1. $$\mathrm{O_2^-(aq)| O_2(g)|| H^+(aq)|H_2(g)| C(s)}$$

Oxidation (anode): $$\mathrm{O_2^-+(aq) \rightarrow O_2(g) + e^- \quad E = -0.33\,V}$$

Reduction (cathode): $$\mathrm{ 2H^+ + e^- \rightarrow H_2 \quad E=0\,V}$$

\begin{align} \mathrm{E} & = \mathrm{E_{cathode} - E_{anode}} \\ & = 0 - (-0.33)\,\mathrm V \\ & = +0.33 \,\mathrm{V} \end{align}

Q1.5

1. The standard reduction potential of the reduction half reaction $$Ag^+(aq) + e^- \rightleftharpoons Ag(s)$$ is 0.7996 V
The standard reduction potential of the oxidation half reaction $$Cu^{2+} + 2e^− \rightleftharpoons Cu(s)$$  is +0.3419 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = 0.7996 - 0.3419 = 0.46$ Since $$\mathrm{E^\circ_{cell}}$$ is positive, the reaction will occur spontaneously as drawn.
2. The standard reduction potential of the reduction half reaction $$\ce{Fe^3+(aq)}$$ is 0.77 V
The standard reduction potential of  the oxidation half reaction $$\ce{Na(s)}$$ is 2.71 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = 0.77 + 2.71 = 3.48$ $$\mathrm{E^\circ_{cell}}$$ is positive, Yes it will happen spontaneously as drawn.
3. The standard reduction potential of the reduction half reaction $$Zn^{2+}(aq) + 2e^− \rightleftharpoons Zn(s)$$ is –0.7618 V
The standard reduction potential of the oxidation half reaction $$I_2(aq) + 2e^− \rightleftharpoons 2I^-(s)$$ is 0.5355 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} =-0.7618 - 0.5355 = -1.2973$ Since $$\mathrm{E^\circ_{cell}}$$ is negative, the reaction will not happen spontaneously as drawn.

Q1.6

1. $$\ce{2Ag+(aq) + Cu(s) \rightarrow 2Ag(s) + Cu^2+(aq)}$$
The standard reduction potential of the reduction half reaction $$\ce{Ag+(aq)}$$ is +0.8 V
The standard reduction potential of the oxidation half reaction $$\ce{Cu(s)}$$ is +0.34 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = \mathrm{0.8 -0.34 = 0.46 V}$
2. $$\ce{2Fe^3+(aq) + Cu(s) \rightarrow 2Fe^2+(aq) + Cu^2+(aq)}$$
The standard reduction potential of the reduction half reaction $$\ce{2Fe^3+(aq)}$$ is 0.77 V
The standard reduction potential of the oxidation half reaction $$\ce{Cu(s)}$$ is +0.34 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = \mathrm{0.77 -0.34 = 0.43 V}$
3. $$\ce{PbO2(s) + Mn^2+(aq) \rightarrow Pb^2+(aq) + MnO2(s)}$$.
The standard reduction potential of the reduction half reaction $$\ce{PbO2(s)}$$ is 1.46 V
The standard reduction potential of the oxidation half reaction $$\ce{Mn^2+(aq)}$$ is +1.229 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = \mathrm{1.229 - 1.46 = -0.231 V}$ As drawn, this cell is non-spontaneous. The cell diagram should be flipped to represent a galvanic cell.
4. $$\ce{Cu(s) + O2(g) + 2H3O^+(aq) \rightarrow H2O2(aq) + Cu^2+(aq)}$$
The standard reduction potential of the reduction half reaction $$\ce{O2(g)}$$ is 0.695 V
The standard reductio npotential of the oxidation half reaction $$\ce{Cu(s)}$$ is +0.34 V
$\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}$ $\mathrm{E^\circ_{cell}} = \mathrm{0.695 -0.34 = 0.355 V}$

not given

Q1.8

a)  -3;  [x + 3(1) = 0]          b)  0         c)  +4;  [x +2(-2) = 0]          d)  +5;  [x + 3(-2) = -1]

Q1.9

a)  Br2         b)  I                  c)  Br               d)  Br2              e)  I-

Q1.10

First determine whether substances may be oxidized or reduced, then look up Eoox for reducing agents and Eored for oxidizing agents and list in order.

• Oxidizing agents (are reduced):  H+ (Eored = 0 V).
• Reducing agents (are oxidized):  Ni (Eoox = +0.23 V), Au (Eoox = -1.50 V), Mg (Eoox = +2.37 V).

Both: Sn2+ (Eored = -0.14 V; Eoox = -0.15 V), Fe2+ (Eored = -0.44 V, Eoox = -0.77 V), Cl2 (Eored =  1.36 V, Eoox  =  -1.47 V,).

• Oxidizing agents (list in order of decreasing Eored):  Cl2  >  H+  >  Sn2+  > Fe2+
• Reducing agents (list in order of decreasing Eoox):  Mg  >  Ni  >  Sn2+  >  Fe2+  >  Cl2 > Au