determination of an electrochemical series

The following data were measured using a nickel electrode as the standard.
Potential, Volts
Cu2+ (aq) + 2e-  ? Cu (s)      + 0.62
Ni2+ (aq) + 2e-  ? Ni (s)      + 0.00
Fe2+ (aq) + 2e-  ?Fe (s)        - 0.15
Al3+ (aq) + 3e-  ? Al (s)      - 1.38

1. which ion is most easily reduced?
2. which metal is most easily oxidized?
3. the copper and aluminum electrodes are connected in a battery.
a) which is the anode?
b) Which is oxidized?
c)          What will the battery voltage be?
d) Write a balanced net ionic equation for the reaction that takes place.
4. A solution is prepared in which a slight amount of Fe2+ is added to a much larger amount of solution in which the
[OH-] is 1.0 x 10-2 M. Some Fe(OH)2 precipitates. The value of Ksp of Fe(OH)2 = 8.0 x 10-10.
a. Assuming that the hydroxide ion concentration is 1.0 x 10-2 M, calculate the concentration of Fe2+ in the solution.
b. A battery is prepared in which the above solution with an iron wire dipping into it is one half-cell. The other half-cell  is the standard nickel electrode.  Write the balanced net ionic equation for the cell reaction.
c. Use the Nernst equation to calculate the potential of the above cell.

Reduction is the gain of electrons, the 1/2 reactions are all reduction 1/2 reactions and so the Cu+2 (with the largest reduction potential of +.62) will be the easiest to reduce.

Oxidation is the loss of electrons.  Oxidation is the reverse of reduction and the sign of the oxidation potential has the same magnitude as the reduction potential but is opposite in sign.

Al  -->  Al+3  +  3e-      +  1.38

Oxidation occurs at the anode, reduction occurs at the cathode (An Ox and a Red Cat!)
The total cell potential must be positive, so
Aluminum will be the anode, and the copper will be the cathode.

The total cell voltage is the sum of the oxidation volts plus the reduction volts.

2Al  +  3 Cu2+  -->  2 Al3+    +  2 Cu

mindy wrote:

The following data were measured using a nickel electrode as the standard.
Potential, Volts
Cu2+ (aq) + 2e-  ? Cu (s)      + 0.62
Ni2+ (aq) + 2e-  ? Ni (s)       + 0.00
Fe2+ (aq) + 2e-  ?Fe (s)        - 0.15
Al3+ (aq) + 3e-  ? Al (s)       - 1.38

4. A solution is prepared in which a slight amount of Fe2+ is added to a much larger amount of solution in which the
[OH-] is 1.0 x 10-2 M. Some Fe(OH)2 precipitates. The value of Ksp of Fe(OH)2 = 8.0 x 10-10.
a. Assuming that the hydroxide ion concentration is 1.0 x 10-2 M, calculate the concentration of Fe2+ in the solution.
b. A battery is prepared in which the above solution with an iron wire dipping into it is one half-cell. The other half-cell  is the standard nickel electrode.  Write the balanced net ionic equation for the cell reaction.
c. Use the Nernst equation to calculate the potential of the above cell.

The Ksp expression for Fe(OH)2 is
Ksp = [Fe2+][OH-]2
Since the problem provides the [OH-] and the value of the Ksp, you can solve for the [Fe2+].

To have a positive total cell potential the Fe must be oxidized and the Ni will be reduced.
Ni2+   +   2e-   -->   Ni       0.00 volts
Fe     -->  Fe2+  + 2e-      +  .15 volts

Add the two 1/2 cell reactions together to get the net ionic equation (and the standard  total potential Eo required for the next part).

The Nernst equation is   E = Eo  -  .0591/n x log Q  where n is the number of electrons transferred (n =2 in this case) and Q is the reaction quotient ([Fe2+]/[Al3+] )in this case.
We calculated Eo and the [Fe2+] above.  In a standard 1/2 cell, the [Al3+] is 1.0 Molar.

Plug and chug!!