1. Anode

2. It is important in calculating half-cell potentials because it serves as a reference. Without this electrode, there would be no basis to calculate values of cell potentials.

3. The left is the anode and the right is the cathode.

4. It is important to use an inert electrode in this situation because it will not react or participate in the reaction in the cell, just provide a surface area for the reaction to occur.

5. 0 volts.

6.  a. i) Ba²⁺_(aq) + 2e- ? Ba_(s)     E^o = -2.92 V        Anode
Cu_²⁺(aq) + 2e- ? Cu_(s)     E^o = +0.340 V     Cathode
a. ii)Al³⁺_(aq) 3e^-? Al_(s)           E^o = -1.66 V        Anode          Sn²⁺_(aq) ? Sn_(s) +2e^-      E^o = -0.137 V     Cathode

b.i)    Ba²⁺_(aq) | Ba_(s) || Cu_(s) | Cu_²⁺(aq)

b. ii)    Al_(s) | Al³⁺_(aq) || Sn²⁺_(aq) | Sn_(s)

c. i)    E^o_cell = 0.34 - (-2.92) = 3.26 V

c. ii)    E^o_cell = -0.137 - (-1.66) = 1.523 V

7. oxidation: Sn_(s) ? Sn²⁺_(aq) + 2e^-_(aq)  
reduction: Pb²⁺_(aq) + 2e^-_(aq) ? Pb_(s)     
cell diagram: Sn_(s) | Sn²⁺_(aq) || Pb²⁺_(aq) | Pb_(s)
8. Cu_(s) + 2Ag⁺_(aq) ? 2Ag_(s) + Cu²⁺_(aq)
9. Because the half cell containing the Cu electrode in Cu2+ solution is the cathode, this is the half cell where reduction is taking place.  Therefore, this half cell has a higher potential to be reduced.
10. The redox reaction.
11. Cell potential is measured in Volts (=J/C). This can be measured with the use of a voltmeter.
12. We can divide the net cell equation into two half-equations.

• Oxidation: {Al(s) + 4OH^-(aq) ? [Al(OH₄)]^-(aq) + 3e^-} x4;   -E^o= ? This is what we are solving for.
• Reduction: {O₂(g) + 2H₂O(l) + 4e^- ? 4OH^-(aq)} x3 E^o= +0.401V
• Net: 4Al(s) + 3O₂(g) + 6H₂O(l) + 4OH^-(aq) ? 4[Al(OH)₄]^-(aq) E^o_cell = 2.71V

E^o_cell= 2.71V= +0.401V - E^oox = -2.31V

13. Cl^- chlorine H⁺ hydrogen

Cl^- chlorine Cu²⁺ copper

I^- iodine H⁺ Hhydrogen

14. 12 mol e^– is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e^–/Cr). Then the Faraday constant can be used to find the quantity of charge. 

n_Cr  n_e– Q

Q = 1.386 mol Cr ×  ×  = 8.024 × 10⁵ C

15. The product of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, n_e–. From the first half-equation we can then find the amount of peroxydisulfuric acid, and the second  leads to n_H2O2 and finally to m_H2O2.

                = 05666  × g H₂O₂ = 0.5666 g H₂O₂

16. 0.259  mol e^-

17. d

18.d

19. b

20. d

21. Write the half-reactions for each process.

Zn(s) ? Zn²⁺(aq) + 2 e^-

Cu²⁺(aq) + 2 e^- ? Cu(s)

Look up the standard potentials for the reduction half-reaction.

E^o_{reduction of Cu2+} = + 0.339 V

E^o_{reduction of Zn2+} = - 0.762 V

Determine the overall standard cell potential.

E^o_cell = + 1.101 V

22. +1139.68 kJ

23. 6.4 x 10²⁵

24.+0.195 V