WS 17 Key

Chem 1B
Dr. White
WS 18 Key Galvanic Cells and Strength of Oxidizing/Reducing Agents (You may have
to look at a table of standard reduction potentials to solve many of these problems!)
1. A galvanic cell is to be constructed using the Ag/Ag+ half cell and the Pb/Pb+2 half-cell.
Measurement shows that the silver electrode is positive.
a) Write balanced half-reactions and the overall spontaneous reaction, without
using a table of half-cell potentials.
Since the silver electrode is positive, that means it is the cathode in this voltaic cell,
and so Ag+ is reduced, thus the reactions must be
2Ag+ (aq) + 2 e-
Pb (s)
2 Ag (s)
Pb+2 (aq) + 2e-
2Ag+ (aq) + Pb (s)
Pb+2 (aq) + 2 Ag (s)
a) Diagram the cell, labeling electrodes as anode and cathode, labeling the salt
bridge, showing what ions are in solution, and showing the direction of electron
flow in the circuit and of cation and anion flow in the salt bridge.
c) Pb (s) ⏐Pb2+ (aq)⏐⏐Ag= (aq)⏐Ag (s)
d) Using a table of half-cell potentials, calculate the standard cell potential for the cell you
have diagrammed above. Did you choose your electrodes correctly?
2Ag+ (aq) + 2 ePb (s)
2 Ag (s)
Pb+2 (aq) + 2e-
2Ag+ (aq) + Pb (s)
Pb+2 (aq) + 2 Ag (s)
E˚cell = E˚cathode - E˚anode = 0.80 V – (- 0.13 V) = + 0.93 V
Chem 1B
Dr. White
Yes, we have chosen the electrodes correctly, as we ended up with a +
E˚cell, as we should for a voltaic cell
2. In principle, a battery (galvanic cell) could be made from aluminum metal and chlorine
a) Write a balanced equation for the reaction that would occur in a battery using
Al (s) /Al+3(aq) and Cl2(g)/Cl- (aq) half-reactions.
A battery, just like a voltaic cell, must produce a +E˚cell, so use table
of reduction potentials to see which way half-reactions would be
used. Al has a more negative reduction potential than Cl2, so it will
be oxidized.
2 Al (s) + 3 Cl 2 (g) ----> 2 Al +3 (aq) + 6 Cl - (aq)
b) Tell which half-reaction occurs at the anode and which at the cathode. What are
the polarities of these electrodes?
The oxidation of Al (s) occurs at anode, negative polarity:
Al (s) ----> Al +3 (aq) + 3eThe reduction of Cl 2 (g) occurs at the cathode, positive polarity :
Cl 2 (g) + 2e- ----> 2 Cl - (aq)
c) Calculate the standard potential, E˚cell for this battery.
E˚cell = E˚cathode - E˚anode = + 1.36 – (-1.66) = 3.02 V
d) What other metal besides Al could conceivably be used in conjunction with the
Cl2(g)/Cl-(aq) half-reaction to produce a battery with a larger voltage output? Leave
the Cl2/Cl- as whatever half-reaction you decided it should be above. Show calculation
of E˚cell to justify your choice.
Could use any metal that has a more negative reduction potential
than Al on the table of reduction potentials, this will ensure a larger
voltage output.
For example, Mg (s)
Mg(s) ----> Mg+2 (aq) + 2 eCl 2 (g) + 2e- ----> 2 Cl - (aq)
E˚cell = E˚cathode – E˚anode = 1.36 – (-2.37) = 3.73 V
Chem 1B
Dr. White
3. Cathodic protection is the protection of a metal from being oxidized by another metal.
Name two metals you could you use to cathodically protect Sn metal from being oxidized, and
explain your answer.
You could use any number of metals to cathodically protect Sn – any
metal that has a more negative reduction potential than Sn will be
preferentially oxidized and keep Sn from oxidizing. For example:
Sn+2 + 2eZn+2 + 2e-
E˚ = -0.14 V
E˚ = -0.76 V
So Zn will be oxidized preferentially in presence of Sn
4. Look at the table of standard reduction potentials and suggest a reason why dentists use
metals like gold or mercury for fillings in your cavities rather than something like copper or
I don’t know if this is really why dentists use gold or mercury for fillings,
but it makes sense that since these metals have very high reduction
potentials compared to copper and zinc, they won’t oxidize in your mouth,
You wouldn’t want your fillings to corrode, would you?