chem 1412 lab 9 dr. pahlavan electrochemistry redox reactions

CHEM 1412 Lab – 9 Dr. Pahlavan

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Electrochemistry – Redox Reactions

Purpose

To prepare a set-up containing both types of electrochemical cells.

To predict formula mass of the metal Lead, from the solid deposited on one electrode of the

electrolytic cell.

To build several electrochemical cells.

Equipment and Materials (Chemicals):

250-mL beakers (2) glass U-tube for the salt bridge cotton plugs for the salt bridge

steel wool to clean electrodes metal electrodes: copper, zinc, lead DC voltmeter

copper wires (2) insulated, with alligator clips metal electrodes: copper, zinc, lead, and tin

Solutions:

1.0 M Cu (NO3)2 1.0 M Zn (NO3)2 1.0 M Pb (NO3)2 1.0 M SnCl2

0.5M KNO3- for the salt bridge

Safety

Follow normal lab safety guidelines. There are no specific safety hazards for this lab.

Metal electrodes: copper, zinc, lead, and tin

INTRODUCTION: The Double-Cell Set-Up for Electrochemistry:

Electrochemistry is the area of chemistry that deals with the relation between chemical changes and electrical

energy. Chemical reactions can be used to produce electrical energy in voltaic (galvanic) cells. Electrical

energy, on the other hand can be used to bring about chemical changes in what are termed electrolytic cells. In

this experiment, you will investigate some of the properties of voltaic cells.

Oxidation- reduction reactions, redox relations, are those that involve the transfer of electrons from one

substance to another. A redox reaction is the simultaneous occurrence of the two components or half

reactions. Oxidation occurs when a chemical species loses or gives up electrons to another chemical species.

Reduction occurs when a chemical species receives or gain electrons. The oxidation process provides the

electrons necessary for reduction to occur. Therefore, the oxidized species is the reducing agent. and the

reduced species is the oxidizing agent.


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For example, if a piece of zinc metal is immersed into a solution containing copper (II) ions, zinc will be

oxidized by the copper (II) ions. Zinc loses electrons and it is oxidized while copper (II) ions gain electrons and

are reduced. We can conveniently express these two processes by the following two half-reactions, which add

to give the overall redox reaction.

Half-reactions:

Zn(s) Zn2+ (aq) + 2e- (Oxidation)

Cu2+(aq) + 2e- Cu(s) (Reduction)

___________________________________________

Overall redox reaction:

Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)

The function of a voltaic cell is based upon reactions similar to the one illustrated in equation (1). A salt bridge

must be used to avoid polarization of the electrodes by facilitating the circulation of ions from both cell

compartments.

Electrochemical Cell

The cell voltage, or the electromotive force (abbreviated emf), is indicated on the voltmeter in volts. The cell

emf is also called the cell potential. The magnitude of the emf is a quantitative measure of the driving force or

thermodynamic tendency for the reaction to occur. In general, the emf of a voltaic cell depends upon the

substances that make up the cell as well as on their concentrations. Hence, it is common practice to compare

standard cell potentials, symbolized by Eocell .


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These potentials correspond to cell voltages under standard state conditions-gases at 1 atm pressure, solutions at

1 M concentration, and temperature at 25oC.

Just as the overall cell reaction may be regarded as the sum of two half-reactions the overall cell emf can be

thought of as the sum of two half-cell potentials, that is, the sum of the voltage of the oxidation half-reaction

(Eoox) and the voltage of the reduction half-reaction (E

ored).

Eocell = Eo

ox + Eored

Because it is impossible to measure directly the potential of an isolated half-cell, the standard hydrogen half-

reaction has been selected as a reference electrode and has been assigned a standard reduction potential of

exactly 0.00 V;

Let us look again at the Daniell cell described above. The standard half-cell potentials are:

(Anode) Zn(s) Zn2+ (aq) + 2 e- (oxidation), Eo(ox) = 0.76 V

(Cathode) Cu2+(aq) + 2e- Cu(s) (reduction), Eo( red) = 0.34 V

Therefore, cell reaction will be

Eocell = Eo

ox + Eored Eo

cell = 0.76 + 0.34 Eocell = 1.10 V

(4)

The complete notation scheme for a galvanic cell is written with the anode on the left and a double vertical line

denoting the presence of a porous plate or salt bridge. Cell equation for the reaction presented as:

Zn | Zn2+

|| Cu2+

| Cu.

Each electrode is connected to the voltmeter by alligator clips and metal wiring. The voltmeter measures the

voltage generated by the redox reaction. The voltage reading will be positive when the electrodes are connected

properly for spontaneous reaction. A spontaneous redox reaction occurs when the species with higher

reduction potential is connected as the cathode. Otherwise, the voltage reading will be negative. The meter

reading will be positive when the cathode is connected to the (+) outlet and the anode is connected to the (-)

outlet.


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Physically, a negative voltage reading means that you have connected the wrong electrode as cathode. This is

equivalent to reversing equation. When equation is reversed, the measured cell potential difference becomes

Eocell = - 1.10 V. The absolute value of ∆E

ocell is the same in both cases, but the sign is different. The sign of

∆Eocell is positive for a spontaneous reaction and negative for non-spontaneous reaction.

The Nernst Equation:

For measurements taken under standard conditions (1 atm, 1 M solutions), ∆Eocell measures the electric

potential difference between the half- cells. For measurements taken under non-standard conditions (the usual

laboratory situation), the Nernst equation is used to calculate ∆Ecell. The Nernst equation gives us the

relationship between the overall cell potential difference for a redox reaction ∆Ecell and the concentrations of

the metal-ion solutions. The Nernst equation is;

Where F is Faraday constant. R the universal gas constant, n the number of electrons transferred, and Q is the

reaction quotient. When natural (ln) is converted to base ten (log) and (RT/nF) are evaluated using R=8.315 JK-

1mol

-1, T= 295.15 K, and F = 96,485 C mol

-1 the equation becomes;

In this form, galvanic cells are used to determine the concentration of the metal ions present under conditions

other than standard conditions. If we apply the Nernst, equation will get


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Example 1 – Calculate the potential at 25oC for the following cell.

Cu | Cu2+(0.024 M) || Ag+(0.0048 M) | Ag

Solution - We start by translating the line notation into an equation for the reaction.

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

We then look up the standard-state potentials for the two half-reactions from the table and calculate the

standard-state cell potential.

Oxidation half reaction: Cu Cu2+ + 2 e- , Eoox = - 0.3402 V

Reduction half reaction: Ag+ + e- Ag , Eored = + 0.7996 V

Eo cell = Eoox + Eored = (-0.3402V) + (+0.7996 V ) Eo cell = 0.4594 V

We now set up the Nernst equation for this cell, noting that n is 2 because two electrons are transferred in the

balanced equation for the reaction.

Substituting the concentrations of the Cu2+ and Ag+ ions into this equation gives the value for the cell potential.

Example 2 - Find the cell potential of a galvanic cell based on the following reduction half-reactions where

[Ni2+] = 0.030 M and [Pb2+] = 0.300 M.

Ni2+ + 2 e- → Ni, E0 = -0.25 V

Pb2+ + 2 e- → Pb, E0 = -0.13 V

Solution - First, find the electromotive force for the standard cell, which assumes concentrations of 1 M.

In order for this reaction to run spontaneously (positive Eo cell) the nickel must be oxidized and therefore its

reaction needs to be reversed. The added half-reactions with the adjusted E0 cell are:

Pb2++Ni→Ni2++Pb, Eo =0.12 V

The number of moles of electrons transferred is two and Q is;

Q = [Ni2+]/ [Pb2+]

E= Eo− 0.0257(n) ln [Ni2+] / [Pb2+] E= 0.12 − 0.0257(2) ln (0.030)/ (0.30) E=0.15 V


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Procedure

1. Each half-cell will be created by placing a metal electrode in an electrolytic solution containing the same

metal's ions. For example, the copper electrode will be placed in a copper (II) nitrate solution.

For each electrochemical cell you create, you will require two half-cells. Set these cells beside each

other - they will be connected by the U-tube.

2. Fill a beaker about two-thirds full of the electrolytic solution. Clean the electrode using the steel wool (if

available), then place the electrode in its appropriate solution.

3. Fill the porous cup containing a zinc strip with 1.0 M zinc nitrate solution.

4. Clip one end of each copper wire to the two electrodes using the alligator clips.

5. Insert the salt bridge (filter paper dipped in KNO3 solution) bridging both cell compartments. Finally

connect the electrodes with alligator clips and wire to the DC voltmeter.

6. Switch on the voltmeter to proper setting.

7. Touch the other end of the copper wires to the voltmeter terminals. If the indicator on the voltmeter

deflects in the wrong direction, switch the wires on the terminals. Read the highest voltage reading

obtained - you will need to do this quickly after connecting the wires to the voltmeter.

8. Repeat the experiment for other combinations of half-cells. Record you reading in the report form.


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Calculation:

1. For each electrochemical cell you created:

a. Write out the two half-reactions for each electrochemical cell you created.

b. Identify each half-reaction as oxidation, reduction, cathode, and, cathode.

c. Identify each half-reaction as the anode and cathode.

d. Indicate the direction of the flow of electrons

2. Using a Table of Standard Reduction Potentials, calculate the theoretical voltage (calculated vale) for

each cell.

3. Compare the voltages you obtained with the theoretical voltage for each cell. What are some reasons

that would account for any differences?

4. Compare the two values, calculated and measurement, for Eocell and calculate the percentage error.

5. Write and balance the equations for all the reactions occurring at the anode and the cathode of each

electrochemical cell.

Write and balance the equations for all the reactions occurring at the anode and the cathode of each

electrochemical cell.

Compare the two values, calculated and measurement, for Eo

cell and calculate the percentage error.

Half-cells

Oxidation half Reduction half

1) Zn|Zn2+ Cu|Cu2+

2) Pb|Pb2+ Cu|Cu2+

3) Cu|Cu2+ Sn|Sn2++


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Electrochemistry – Redox Reactions

REPORT FORM Name _______________________________

Instructor ___________________________

Date ________________________________

1. Cell Equation: (Zn|Zn2+ and Cu|Cu2+)

Cell Half Reactions

Cell Overall Reaction

Eo(ox) , From table

Eo(red) , From table

Eocell , calculated value

Ecell, cell potential

(calculation)

Spontaneity. (S), (NSP)

Eocell , Experimental value

(voltmeter reading)

% Error

Calculation


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2. Cell Equation: (Pb|Pb2+ and Cu|Cu2+)

Cell Half Reactions


Eo(ox) , From table




(calculation)



(voltmeter reading)

% Error

Calculation

3. Cell Equation: (Cu|Cu2+ and Sn|Sn2+)

Cell Half Reactions


Eo(ox) , From table




(calculation)



(voltmeter reading)

% Error

Calculation


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Lab 9: Electrochemistry – Redox Reactions

Pre Laboratory Review Questions and Exercises Due before lab begins. Answer on a separate sheet of paper.

Name __________________________

1. Define the followings:

Oxidation –

Reduction -

2. What is the criterion for spontaneous chemical change based on cell potentials? Explain.

3. E°cell = 1.47 V for the voltaic cell, V(s) | V2+(1 M) || Cu2+(1 M) | Cu(s). Determine the value of E°(V2+/

/V).

4. Let's try one more example of setting up an electrochemical cell, using (Al|Al3+) and (Pb|Pb2+) half-cells.

a) Determine the two half-reactions involved, and which reaction will undergo oxidation and which one

will be reduced. Determine the voltage of the cell.

b) Diagram the set-up of the electrochemical cell, including the following items:

the two half-cells, including the electrodes and electrolytic solutions

the external circuit, showing the direction of electron flow

the salt bridge with an electrolyte, including movement of ions

label anode, cathode, positive and negative posts

http://drfus.com/electrochemistry-lab-experience/attachment/electrochemical-cell


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Lab 9: Electrochemistry – Redox Reactions Post- laboratory Questions and Exercises Due after completing lab. Answer in the space provided.

Name: _______________________

1. Define the followings;

Galvani cell (Voltaic cell) -

Electrolytic cell -

2. What is the function of the salt bridge in an electrochemical cell?

3. E°cell = 1.47 V for the voltaic cell, V(s) | V2+(1 M) || Cu2+(1 M) | Cu(s). Determine the value of E°(V2+/

/V).

4. 5.77 g of zinc is deposited at the cathode when a current of 7.1 amperes passes through an electrolytic cell

for 40. minutes. What is the oxidation state of the zinc in the aqueous solution?

5. A voltaic cell is constructed in which the reaction is, 2Cr(s) + 3Sn4+ (aq) 2Cr (aq) + 3Sn2+ (aq)

If the [Cr3+] = 1.00 M, [Sn4+] = 0.500 M and the [Sn2+] = 0.025, calculate Ecell at 25 ºC. Calculate [Cr3+]

for the reaction if the [Sn2+] is 1.00 M and [Sn4+] is 1.00 M and the measured cell potential is +0.994 volts.

chem 1412 lab 9 dr. pahlavan electrochemistry redox reactions

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