Gatewaylectrochemistry: Using Standard Cell Potentials


Table of Contents

Icon Legend
Ox amp; Red


Voltaic Cells

Cell Voltage

Calc. Potentials


Electrolytic Cells

Calculating the Cell Potential:

The net reaction of a voltaic cell constructed from a standard zinc electrode and a standard copper electrode is obtained by adding the two half-reactions together:

oxidation  Zn(s)Zn2+(aq) + 2e E° = 0.763 V
reduction  2 [ 1 e + Ag+(aq) Ag (s) ] E° = 0.337 V

net  Zn(s) + 2 Ag+(aq) 2 Ag(s) + Zn2+(aq) E° = 1.100 V

Note: even though we had to multiply the silver reduction by a factor of 2 so that the electrons consumed would balance with those produced by the zinc oxidation, we do NOT multiply the silver reduction potential by this factor. In manipulating potentials, we should only change the signs of the values, not the magnitude.

How do we know which metal will become oxidized and which metal ion reduced?

By looking at a table of standard reduction potentials!

One of the half-reactions must be reversed to yield an oxidation. Reverse the half-reaction that will yield the highest (positive) net emf for the cell. Remember that when one reverses a reaction, the sign of (+ or –) for that reaction is also reversed. Consider again the following table of standard reduction potentials:

In aqueous solutions at 25 °C
2e + Hg2+ (aq) Hg (l) E° = 0.86 V
2e + I2 (s) 2 I(aq) E° = 0.54 V
2e + 2 H+ (aq) H2 (g) E° = 0 V
4e + Zr4+ (aq) Zr (s) E° = –1.53 V
1e + Rb+ (aq) Rb (s) E° = –2.93 V

Let's calculate the potential generated in by a cell constructed from standard Zr and I2 electrodes: From the table, we write a balanced reduction half-reaction for each electrode and copy down the reduction potentials:

2e + I2 (s) 2 I(aq) E° = 0.54 V
4e + Zr4+ (aq) Zr (s) E° = –1.53 V

Reversing which reaction will yield most positive standard reduction potential? Let's try both!

Case 1: reversing the iodine reduction:

net = ox(I2) + red(Zr)

net = (– 0.54) + (– 1.53)

net = – 2.07 V

Case 2: reversing the zirconium reduction:

net = red(I2) + ox(Zr)

net = ( 0.54) + ( 1.53)

net = 2.07 V

Note: in the first case, the calculated net potential is not even positive! This reaction would not occur spontaneously as proposed.

The highest positive potential is found by using the Zr oxidation half-reaction. The cell would therefore proceed spontaneously in Case 2. Notice that we did not multiply the value for the reduction potential of I2 by a factor of 2, even though the iodine reduction equation would be multiplied by this factor to balance the number of electrons produced and consumed.

Review: On a sheet of paper, derive the balanced net equation for this voltaic cell. When finished, click the question mark to reveal the answer.

Zr + I2

Use the previous table of standard reduction potentials to calculate the voltage generated by the following voltaic cells at standard conditions--two decimal places, please!:

electrode 1 electrode 2            Eºnet ( *standard hydrogen electrode )
Hg / Hg2+ SHE*
Zr / Zr4+ Rb / Rb+
Rb / Rb+ Hg / Hg2+

Nicely done!

A negative voltage represents a non-spontanous reaction! Try again.

 0.86 (mercury electrode, reduction) + 0.00 (hydrogen electrode, oxidation) = ?

 -1.53 (zirconium electrode, reduction) + 2.93 (rubidium electrode, oxidation) = ?

 2.93 (rubidium electrode, oxidation) + 0.86 (mercury electrode, reduction)= ?

Before moving to the next page, you should how to calculate the net potential of a voltaic cell by using a table of standard reduction potentials.

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