A SIMPLE ELECTROCHEMICAL CELL

Consider once again the copper-zinc electrochemical cell shown in Figure 2.2.

In the right half-cell, the cathode, Cu2+ ions in solution are being reduced to Cu0, indicating solid copper metal deposit on the Cu cathode. In the left half-cell, the anode, Zn0 or solid zinc is being oxidized to Zn2+ ions, which dissolve in solution. The anions—SO42- (aq)—are spectator ions and do not participate in the reaction. The salt bridge is necessary to maintain electrical neutrality and retard polarization.

The short-hand notation for this cell is as follows:

It is understood that the anode cell, where oxidation occurs, is written first and is separated by a double vertical line from the cathode cell, where reduction occurs. Single vertical lines separate the solid electrode from the aqueous solution into which it is immersed. Concentrations of the solutions are expressed in moles per liter in parentheses.

The two half-reactions, followed by the net reaction, can be written as follows:

At Cathode

At Anode

Net Reaction

Or finally,

Note that the electrons from the two half reactions exactly cancel each other, and do not appear in the final net reaction. If this cancellation of electrons does not happen automatically, one or both of the half-reactions must be multiplied by a suitable coefficient (i.e., an integer) or coefficients to obtain the exact cancellation of electrons. Note also that the positive value for Enet° indicates that this reaction proceeds spontaneously and the cell produces 1.100 V under standard conditions.

Also note that the value at the cathode represents a reduction potential and may be written as Eo ге4 while the Eo value at the anode represents an oxidation potential and may be written as Eoox. This convention may be used in other reference manuals or textbooks.

Useful Quantitative Relationships

Example 2.28

Consider the electrochemical cell with the following net reaction, which is observed to proceed spontaneously. All species are in their standard states, that is, concentrations of dissolved species are 1.00 molar, and T = 25°C.

A. Using Table 2.5, write the two half-reactions with their respective E° values.

B. Indicate which species is oxidized and which is reduced.

C. Identify the anode and cathode.

D. Compute Enet°.

E. Confirm that the reaction proceeds spontaneously.

Solution

A. Mg (s)^ Mg2+(aq) + 2e- E°= 2.370 V

Sn2+ (aq) + 2e- ^ Sn (s) E° = -0.140 V

B. Mg (s) is oxidized to Mg2+ (aq), while Sn2+ (aq) is reduced to Sn (s).

C. The anode is the electrode where oxidation takes place, while the cathode is the electrode where reduction takes place. Since Mg is oxidized, the half-cell containing the Mg electrode and Mg2+ (aq) solution must be the anode.

Similarly, the cathode is the half-cell containing the Sn electrode dipped into Sn2+ (aq) solution.

D. Adding the two half-cell potentials determined in A gives Enet° = 2.230 V.

E. Since Enet° is a positive number, the net reaction must proceed spontaneously.

Example 2.29

Consider the reaction in Example 2.28. Instead of standard conditions of 1.00 M concentrations for each solution, assume now that the Mg2+ (aq) solution is 0.850 M and that the Sn2+ (aq) solution is 0.0150 M. Find the Enet° under these conditions, assuming T = 25.0°C.

Solution

Since this is under nonstandard conditions, the Nernst equation applies. However, use the simplified version of this equation, equation 2.38, since T = 25.0°C.

In this case, n = 2, since 2 mol of electrons are exchanged in the net reaction.

Also, by examining the net reaction, Q is found to be:

Enet0 was calculated in D of Example 2.28 as 2.230 V. Substitution now gives:

Example 2.30

Compute the Gibbs free energy change, AG, for the cell in Example 2.29.

Solution

Example 2.31

Consider the following cell reaction, in which all species are standard- state conditions:

A. Predict the effect on the electromotive potential of this cell of adding NaOH solution to the hydrogen half-cell until pH = 7.

B. Compute the number of coulombs required to deposit 4.20 g of Cu (s) in the copper half-cell.

C. How long in seconds will this deposition take, if the measured current is 4.00 A?

Solution

A. Refer to the simplified Nernst equation, equation 2.38. The EMF is

the Enet:

If NaOH (aq) is added, H+ ions (aq) will be neutralized, thereby raising the pH. [H+ (aq)] will decrease, reducing the magnitude of the log term, which is subtracted from Enet°. This, in turn, will increase the Eet or the emf of the cell.

B. First, compute the number of electrons that must be transferred to deposit 4.20 g of Cu (s). Then, recall that 1.00 mol of Cu2+ (aq) is deposited as Cu (s) for every 2.0 mol of electrons used, and that the Faraday constant F = 96,485 C/1.00 mol electrons. Thus, first,

C.

This is about 53 min or a little less than 1.0 hr.

BIBLIOGRAPHY

Petrucci, R. H., Harwood, W. S., and Herring, F. G. 2002. General Chemistry, 8th ed. Upper Saddle River, NJ: Prentice Hall.

Chang, R. and Goldsby, K. A. 2013. Chemistry, 11th ed. New York: McGraw Hill.

Brown, T. L., Lemay, H. E., Bursten, B. E., Murphy, C. J., Woodward, P. M., et al. 2012. Chemistry, the Central Science. 12th ed. Upper Saddle River, NJ: Prentice Hall.

Lide, D. R., Ed. 2006. Handbook Chemistry and Physics, 87th ed. Boca Raton, FL: CRC.

Olia, M., Ed. 2013. Fundamentals of Engineering Exam, 3rd ed. Hauppauge, FL: Barron Educational Series.

Frank, A. 1971. Thermodynamics: Principles and Applications. Hoboken, NJ: John Wiley and Sons.

Wall, F. T. 1965. Chemical Thermodynamics, 2nd ed. San Francisco, CA: W. H. Freeman and Company.

 
Source
< Prev   CONTENTS   Source   Next >