Exp 7 electrochemistry 2022W

.pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

CHEM 154

Subject

Chemistry

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by DeaconFire2734

EXPERIMENT 7A Electrochemistry: Galvanic Cells and the Nernst Equation PRE-LAB WEEK REQUIREMENTS DURING THE PRE LAB WEEK …... 1. Read the entire experiment. 2. Read the Techniques and Resources: • Techniques: Under glossary – read topics related to Electrochemistry These materials are accessed by clicking on Experiment 7 in the Chemistry 154 Online Laboratory Canvas website . 3. Complete the Experimental Design Form: Think carefully about interpretation of your results as you design some steps of your procedure and answers the questions in the EDF quiz on Canvas. Consult the list of Reagents and Apparatus available in the lab. Be sure to submit the form prior to the lab week (Experiment 7B) on canvas. Do not attempt to leave all of this Pre-Lab Week material until the Lab Week as you will be overloaded if you do!   

EXPERIMENT 7 Electrochemistry: Galvanic Cells and the Nernst Equation Electrochemistry: Galvanic Cells and the Nernst Equation Aim: To investigate redox reactions of several metals and ions. OBJECTIVES The objectives of this experiment are to:  Construct galvanic cells.  Design an experiment to arrange metals in an electrochemical series.  Understand the Nernst Equation and apply it to determine an unknown concentration. This Experiment is presented in two halves: Part I and Part II, to be done together Introduction to Electrochemistry and Part I of the Experiment Any chemical reaction involving the transfer of electrons from one substance to another is an oxidation-reduction (redox) reaction. The substance losing electrons is oxidized and the substance gaining electrons is reduced . Redox reactions are quite common and have a large impact on the world around us. The rusting of iron, for example, is a redox reaction where iron is oxidized and oxygen is reduced. We could make use of redox reactions if we found a way to put to work the energy involved in the electron transfer. Consider the redox reaction between metallic zinc and an aqueous solution of lead nitrate: Zn(s) + Pb 2+ (aq) → Zn 2+ (aq) + Pb(s) (1) The nitrate ions are spectator ions and are omitted in the above net ionic equation because they play no part in the redox reaction. Two electrons are transferred from each reacting zinc atom to a lead ion. The metallic zinc is thus oxidized to produce Zn 2+ ions and the lead ions (Pb 2+ ) are reduced to produce metallic lead. This redox reaction can be carried out by simply putting a piece of zinc in an aqueous solution of lead nitrate. Metallic lead immediately begins to plate out on the zinc, the concentration of aqueous Pb 2+ decreases, and the concentration of Zn 2+ in solution increases as time passes. Since reaction is clearly spontaneous, ∆ G for the process is negative. The reverse reaction does not take place spontaneously; nothing happens when a piece of lead is placed in a zinc nitrate solution. Lead ions, therefore, must have a greater tendency to undergo reduction (accept electrons) than do zinc ions . As shown below in Figure 1, the same spontaneous reaction between zinc and lead nitrate can be carried out without ever putting the two reactants in direct physical contact. A strip of zinc metal is dipped into a solution of zinc sulfate (ZnSO 4 ) and a piece of lead into a solution of lead nitrate, Pb(NO 3 ) 2 . These two metal strips are now the two electrodes of an electrochemical cell. Next, the two solutions are connected by a salt bridge , a porous substance (for example, a piece of a filter paper) saturated with a solution of salt, such as KNO 3 . The porous substance allows ions to pass between the solutions, but prevents the solutions themselves from mixing. If the two

electrodes are connected by a wire, electrons will flow from one electrode to the other, allowing the redox reaction to proceed in the electrochemical cell we have created. By convention, the short form for this electrochemical cell is denoted as: Zn(s) | Zn 2+ (aq) || Pb 2+ (aq) | Pb(s) where | represents the boundary between the electrode and the solution; || represents the boundary between the two half-cells. This convention assumes that reduction always takes place at the right-hand electrode and that oxidation always takes place at the left-hand electrode . In our galvanic cell shown above, as zinc is oxidized on the left-side, Zn 2+ ions enter the solution. The following oxidation half-reaction can thus be written: Zn(s) → Zn 2+ (aq) + 2e - oxidation half-reaction (Left-Hand Side) The electrons that are released pass through the external circuit wire from left to right . As the electrons flow into the lead strip, they are picked up by the lead ions at the metal-solution interface, forming lead atoms, and thus causing metallic lead atoms to plate out on the surface of the electrode. The following reduction half-reaction can thus be written: Pb 2+ (aq) + 2e - → Pb(s) reduction half-reaction (Right-Hand Side) • The electrode at which oxidation occurs is called an anode. (Figure 1, Left -hand side) • The electrode at which reduction occurs is called a cathode. (Figure 1, Right -hand side) Since the electrons flow from the anode to the cathode, this process would lead to an increase of positive charge in the left-hand beaker (the anode), and an increase of negative charge in the right- hand beaker (the cathode), and therefore the process could not proceed. The salt bridge, however, permits a compensating flow of ions to complete the circuit. What is the charge on the ions flowing into each beaker? How does this affect the overall charge in each beaker? We have effectively separated reaction (1) into two components (two half-reactions) and placed them into two separate compartments (half-cells). We have seen that metallic zinc is oxidized in the left beaker (anode) while lead ion is reduced in the right beaker (cathode). The zinc plate is the negative electrical pole as electrons are leaving the metallic Zn at this point, while the lead plate is the positive pole as electrons are being accepted by the Pb 2+ ions to form metallic Pb at this plate. Figure 1: a Zinc/Lead galvanic cell Zn → Zn 2+ + 2e - Pb 2+ + 2e - → Pb oxidation reduction anode cathode 1 M Pb(NO 3 ) 2 1 M ZnSO 4

The electric potential difference between the two plates in Figure 1 is the motivating force that causes the electrons to flow from the left beaker to the right beaker. This potential difference , ΔE, is called a cell potential ( E cell = ΔE = E right - E left ) or electromotive force (emf) and can be measured by a voltmeter. The electrochemical cell described above operates spontaneously (negative ∆ G) and is called a galvanic cell . Such a cell allows a net conversion of chemical energy into electrical energy, which can then be used to perform work. Quantifying and Calculating Cell Potentials: Different metals, such as zinc and lead, have different tendencies to oxidize; similarly, their ions have different tendencies to undergo reduction. The cell potential of a galvanic cell is due to the difference of the two metals' tendencies to be oxidized (lose electrons), or their ions' tendencies to be reduced (gain electrons). Commonly, a reduction potential , a tendency to gain electrons, is used to represent the relative tendency for a given metal ion to undergo reduction. The potential measured in the cell is the result of the two half-reactions and the magnitude of the potential depends on the ions' concentrations, the temperature, and gas pressures; standard conditions exist when ionic species have a concentration of 1 M and gases have a pressure of 1 atm. When all the ionic concentrations in the zinc/lead system are 1.0 M and the temperature is 25 ° C, the cell voltage is 0.63 volts. It would be a monumental task to assemble a list of all possible cells and report a cell voltage so instead we use the potential of the half-reactions. We cannot measure the half-cell potential directly so we pick another half reaction, call it the standard, construct a cell, measure the cell voltage and report the potential relative to the standard. The standard that has been chosen by convention is: 2 H + + 2e - → H 2 E ° = 0.00 V Here the notation E ° is called the standard electrode potential and the pressure of the hydrogen gas is 1 atmosphere. The measured cell voltage using the standard hydrogen electrode as one of the half cells is, therefore, the potential of the other half reaction. In a teaching laboratory, it is difficult to work with hydrogen gas. Hence, for the purposes of this experiment, we will use Sn/ Sn 2+ as the reference electrode because it has an electrode potential close to zero. 2 Sn 2+ + 2e - → Sn E ° = -0.14 V

Tables of standard half-reaction potentials have been made. The reactions by convention are written as reductions and hence the tables are called tables of standard reduction potentials . A brief excerpt is shown below in Table 1. The greater the tendency of the ion to gain electrons and undergo reduction, the more positive the reduction potential of the ion. Since lead has a greater reduction potential than zinc, the metallic lead plates out on the cathode. Standard reduction potentials can be used to calculate the standard cell potential of any combination of half-cells. Recall that the reaction: Zn(s) + Pb 2+ (aq) → Zn 2+ (aq) + Pb(s) (1) is for the galvanic cell that is denoted as: Zn(s) | Zn 2+ (aq) || Pb 2+ (aq) | Pb(s) (2) By convention, standard cell potentials are calculated from the standard reduction potentials of the constituent half-cells according to eqn (3), where RH means Right-Hand side and LH means Left- Hand side of cell (2) above. E ° cell = E ° RH - E ° LH (3) As by convention the cathode is placed on the right while the anode is placed on the left, equation (3) can also be written as: E ° cell = E ° cathode - E ° anode (4) Using the data in Table 1, we can calculate the standard cell potential for the zinc/lead cell: E ° cell = E ° RH - E ° LH = - 0.13 – (- 0.76) = 0.63 V Table 1: Standard Reduction Potentials at 25 ° C Reduction Half-reaction Potential (V) Cu 2+ + 2e - → Cu(s) +0.34 2H + + 2e - → H 2 (g) 0.00 Pb 2+ + 2e - → Pb(s) -0.13 Zn 2+ + 2e - → Zn(s) -0.76 Mg 2+ + 2e - → Mg(s) -2.37 Li + + e - → Li(s) -3.05

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version