## Calculating Voltages

You can also use the standard oxidation potential list to calculate the voltage or tendency of a redox reaction to occur.

### Examples

Here we have the reaction of zinc metal and copper ion becoming zinc ion and copper metal.

Because of the relative positions of the chemicals on the oxidation potential list we know that this reaction will occur spontaneously (because zinc is on the left side of the oxidation potential list and copper ion is on the right side and below the zinc).
 Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-     -0.34v

To figure the potential or voltage for a redox reaction, add the voltages for the two half-reactions.

In this case, the oxidation half-reaction is Zn Zn2+ + 2e- and the voltage for this half-reaction on the oxidation potential list is 0.76 volts.

The reduction half-reaction is Cu2+ + 2e- Cu . That is the reverse of the way that it is written on the S.O.P. list. Reversing the reaction will also reverse the voltage. Instead of being -0.34 volts, it will be +0.34 volts.

You then combine those two voltages, add them together, to get +1.10 volts for the reaction.

 Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-      -0.34v Zn Zn2+ + 2e-      0.76v Cu2+ + 2e- Cu     +0.34v ____________________________ Zn + Cu2+ Zn2+ + Cu    1.10v
An alternate approach, but one that accomplishes the same thing, is to leave both half-reactions written as oxidation reactions but subtract the voltage of the lower one because it is going in the reverse direction.
 Zn Zn2+ + 2e-      0.76v Cu Cu2+ + 2e-      -(-0.34v) ____________________________ Zn + Cu2+ Zn2+ + Cu    1.10v

Next we have zinc metal reacting with iron(II) ion, which is just a little bit below the zinc on the list. In this case the zinc metal becomes zinc ion with a voltage of +0.76. Iron(II) ion becoming iron metal is the reverse of the reaction as written on the standard oxidation potential list. Thus, it will have a voltage of -0.44. The two half-reactions together then would have a voltage of those two added together (+0.76 and  -0.44) giving a voltage of +0.32 volts.
 Zn Zn2+ + 2e-      0.76v Fe Fe2+ + 2e-     0.44v Zn Zn2+ + 2e-      0.76v Fe2+ + 2e- Fe     -0.44v ____________________________ Zn + Fe2+ Zn2+ + Fe    +0.32v

### Positive and Negative Voltages

These two examples are reactions which occur spontaneously. They are reactions with positive voltages. Redox reactions with negative voltages do not occur spontaneously. They must be forced.

For example, if you consider reacting zinc metal with sodium ion, the voltage for that reaction has a negative value and does not occur. So, positive voltages correspond to redox reactions which do occur and negative voltages correspond to redox reactions which do not occur without help.
 Na Na+ + e-      2.71v Zn Zn2+ + 2e-      0.76v 2 Na+ + 2 e- 2 Na      -2.71v Zn Zn2+ + 2e-      0.76v ______________________________ Zn + 2 Na Zn2+ + 2 Na   -1.95v

Incidentally, you do not have to take into account how many electrons are given off when you calculate the voltage. That has already been taken care of and worked into the value.

### Practice

Determine the voltages for the following reactions or combinations of chemicals.

 Al with Cr3+ Pb with Ag+ 2 I- + Br2 I2 + 2 Br- 2 Cr + 3 Cu2+ 2 Cr3+ + 3 Cu

Here are the voltages for the following reactions and combinations of chemicals.

 Al with Cr3+  0.92v Pb with Ag+  0.93v 2 I- + Br2 I2 + 2 Br-  0.53v 2 Cr + 3 Cu2+ 2 Cr3+ + 3 Cu  1.08v

If you did not get these values, first check to see that you properly dealt with the negative voltages that some of these half-reactions have. If you are still having troubles check with an instructor.

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