Electrochemistry

INTRODUCTION

Part I

An electrolytic cell uses electricity to bring about a chemical reaction whereas a voltaic cell uses a chemical reaction to produce electricity. The electrolytic cell used in today's experiment involves the following reaction:

M(s) + 2 H⁺ (aq) ® M⁺² (aq) + H₂ (g)

where M represents an unknown metal. As you can see, it is an oxidation -reduction reaction.

By measuring the temperature, pressure and volume of hydrogen, and using the Ideal Gas Law equation, PV = nRT, you will be able to determine the number of moles of hydrogen gas formed. From the balanced equation for the reaction and the mass of the metal that has reacted, you can then calculate the number of moles of metal that has reacted and the atomic weight of the metal.

Also, it would be interesting to see how the moles of hydrogen produced compares with the moles predicted based on the following relationships:

coulombs = amperes x seconds

one mole of electrons = 96,500 coulombs

A coulomb represents the amount of electricity carried in one second by a current of one ampere.

Part II

Any redox reaction can be divided into two half-reactions. For example,

Cu⁺² (aq) + Zn = Cu + Zn⁺² (aq)

can be separated into

Cu⁺² (aq) + 2 e^- = Cu reduction half-reaction

Zn = Zn⁺² (aq) + 2 e^- oxidation half-reaction

Theoretically, it should be possible for the complete reaction to proceed with each half-reaction taking place in separate containers with electrons being transferred through an external wire. Such an apparatus, as shown below, is an example of a voltaic cell.

VOLTAIC CELL DIAGRAM

At the zinc electrode electrons are produced. Zinc is being oxidized. Oxidation always occurs at the anode. At the copper electrode, electrons are picked up by the copper ions. The copper ions are reduced. Reduction always takes place at the cathode. In a voltaic cell the anode is given a negative (-) sign and the cathode a positive (+) sign. The shorthand notation for this cell is:

Zn | Zn⁺² (aq) || Cu⁺² (aq) | Cu {The anode is always represented on the left}

The voltmeter measures the electromotive force ( the cell voltage ) which is a measure of the tendency of the reaction to occur. The electromotive force (emf) is equal to the sum of the two half-cell potentials. For example, the standard electrode potentials ( E^o ) for the Cu⁺² / Cu and Zn⁺² / Zn electrodes are as follows:

Cu ⁺² (aq) + 2 e^- = Cu E^o = 0.34 Volt

Zn⁺² (aq) + 2 e^- = Zn E^o = -0.76 Volt

Note that the standard reduction potentials are the ones that are usually given in a table. To find the cell voltage for the above reaction, the Zn⁺² /Zn half-reaction is reversed and the sign of its potential is changed to give its standard oxidation potential. The half-cell potentials are then added to give the emf of the cell.

Cu⁺² (aq) + 2 e^- = Cu E = 0.34 Volt

Zn = Zn⁺² (aq) + 2 e^- E = -0.76 Volt

Cu⁺² (aq) + Zn = Cu + Zn⁺² (aq) emf = +1.10 Volts

The positive sign for the emf indicates that the reaction will proceed as written.

Part III

The concentration of a solute is a variable that can affect electrode potentials. The Nernst equation can be used to calculate emf values at concentrations other than 1M. The Nernst equation at 25 ^°C is

{Note: 0.0257 becomes 0.0592 if “log” is used}

for the reaction: aA + bB = cC + dD. In this experiment the only reaction species that will appear in the Nernst equation will be the solutes.

For example, the Nernst equation for: Cu⁺² (aq) + Zn = Cu + Zn⁺² (aq)

will be

where

E = the cell potential at the non standard concentrations

E⁰ = the cell potential at standard conditions

n = the number of electrons involved in the cell reaction

( in the above example, n = 2 )

EQUIPMENT 250 mL beaker, buret, hotplate, sand paper, voltmeters, premade cells

CHEMICALS NaCl (Sodium Chloride), 0.5 M Sodium Sulfate, Unknown electrode

SpillOnly : B1, 1 M Sulfuric Acid

PROCEDURE

Part I This part is a demonstration done by the lab instructor

Demonstration setup

1. The unknown metal electrode is cleaned, weighed and then placed in a beaker containing 0.5M Na₂SO₄ solution. It will act as the anode. The buret is suspended in a beaker filled with 1M H₂SO₄ solution. Using an aspirator, the sulfuric acid solution is drawn into the buret to the top of its graduated scale. The bare end of the copper wire at the bottom of the buret will be the cathode. The nichrome wire keeps the solutions separate while completing the circuit.

2. Using a power supply, approximately 50 mL of H₂(g) is collected in the buret. Record the amperage used and the elapsed time during the electrolysis. The data needed for your calculations will be put on the board by your lab instructor.

Part II

1. Using the Volt-Ohmmeter determine the voltage of the four cells given on the data sheet. Do not leave a cell connected too long. The voltage will drop with time.

2. You will be using a red and a black lead to connect the cells to the Beckman volt-ohmmeter. The settings should be as shown below. Make sure that the alligator clips (red & black leads) are attached securely to the metal of the electrodes in the cells.

(Take all voltages as positive numbers).

Part III

1. Determine the voltage of the cell using two copper electrodes with a 0.001M Cu⁺² solution in one half-cell and a 1 M Cu⁺² solution in the other half-cell.