Unit 6 Electrochemistry

Chemistry 020, R. R. Martin

ElectrochemistryElectrochemistry is the study of the interconversion of electrical and chemical energy. We can use chemistry to generate electricity.... this is termed a Voltaic (or sometimes) Galvanic Cell Or we can use electricity to bring about chemical changes. This would be termed an Electrolytic cell. We will consider both as separate topics, though the underlying theory is the same in each case. Consider the following reaction that you can do in a beaker on the bench: Add zinc metal to a solution of copper sulphate. We will observe that the blue colour associated with the copper sulphate fades as a reddish brown precipitate is formed. Simultaneously, the zinc metal dissolves.

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The equation for the reaction is:

Zn(s) + CuSO4(aq) → Zn SO4(aq) + Cu(s)

The brown precipitate is copper metal. In terms of the ions involved ( SO4

2- is a spectator) : 1) Zn → Zn2+ + 2e-

2) Cu2+ + 2e- → Cu These are so-called half-reactions (remember balancing redox equations) in 1) zinc is oxidised (the oxidation # goes from 0 to 2+) . Zinc is acting as the reducing agent. in 2) copper ions are reduced to copper (gain electrons) metal. Copper ions are acting as the oxidising agent.

This is a redox reaction that occurs spontaneously

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If the reaction is allowed to proceed in the beaker, the liberated energy will appear as heat. If we make each of the half reactions happen in separate, but electrically connected containers, the two e- can be made to travel along a ‘wire’ and do electrical work. Some definitions: Reduction reactions occur at the cathode (red cat). Oxidation reactions occur at the anode (an ox) Thus: Cations are the +ve species that

move to the cathode, Anions are the -ve species that

move to the anode The movement of electrons through a wire requires an unbalanced electrical force. This force called the electromotive force (emf) is the electrical driving force that pushes the electrons generated in the oxidation half-reaction towards the reaction where reduction takes place.

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The difference in potential energy between the reactants and products (the emf) of an electrochemical reaction is measured in volts (V). The greater the potential energy difference between the two electrodes, the larger the emf. Charge is measured in coulombs, the charge on one electron = 1.602 x 10-19 coulombs. A difference of one volt causes a charge of one coulomb to acquire kinetic energy equal to one joule 1 V = 1 J / C We measure the emf of a cell with a voltmeter . The potential energy difference, commonly referred to as the potential of the cell is designated Ecell. In our setup above, as this reaction occurs, there is a build-up in cation concentration at the Zn anode, and a decrease in cation concentration at the Cu cathode. So anions migrate through the salt bridge from Cu side to Zn side and cations pass in the opposite direction to preserve the charge balance.

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We write the Daniel Cell as: Zn / Zn2+ // Cu2+ / Cu Boundary salt bridge boundary between between Zn & or partition Cu & solution solution By convention, the oxidation side is shown on the left and the reduction side is shown on the right. When we connect a voltmeter to measure the cell emf, we connect the red or +ve connection to the right or reduction electrode: Remember: RED, RIGHT, REDUCTION: this is the side of the cell where electrons are RECEIVED If the cell is connected in this manner, a positive indicated voltage shows that the reaction is proceeding in the expected spontaneous direction.

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Experimentally it is found that the voltage and direction of current flow generated by a galvanic cell depend upon the electrode materials, ion concentration and temperature, T. We must define standard conditions: Pure metals for solids, 1 M concentration for solutions, 25 0C and 1 atmosphere pressure. Under any other conditions, the cell voltage will change (We do not deal with the effect of concentration in this course. If you are interested read about the Nernst equation) For the Daniell cell: Zn / Zn2+ // Cu2+ / Cu Eo

cell = 1.10v Eo refers to standard conditions

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Eo refers to standard conditions But we should ask, how much of that voltage is ‘generated’ at the oxidation side and how much at the reduction side? Suppose we consider cells with the same anode reaction, but different cathode reactions, what might we expect? The LHS side reaction remains the same and must be associated with the same potential. Cells with the same anode reaction, but different cathode reaction exhibit different Eo. Zn / Zn2+ // Cu2+ / Cu Voltage = 1.10 V Zn / Zn2+ // Cd2+ /Cd Voltage = 0.36 V Zn / Zn2+ // Ag+ / Ag Voltage = 1.56 V Since the tendency of the Zn to oxidize remains the same, the tendency of the other half-reaction to reduction must be different. Since the RHS involves reduction, we can say that the easier the reduction is, the larger the voltage difference.

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Thus the tendency towards reduction increases in the order:

Ag+ > Cu2+ > Cd2+

Conversely, the tendency to oxidation increases in the order:

Cd > Cu > Ag

Thus we might say that Cd2+ wants to stay as Cd2+ more than Ag+ wants to stay as Ag+

Anthropomorphic: Attribution of human motivation, characteristics, or behavior to inanimate objects, animals, or natural phenomena. For large EMF’s we want: Negative electrode (LH electrode) with a strong tendency to oxidize

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Positive electrode (RH electrode) with strong tendency to reduce The cell voltage is determined by the difference in the tendencies of the two electrode metals to oxidize, or difference in the tendencies of the two ions to be reduced

Standard Electrode Potentials. The voltmeter in our picture of the Daniell cell measures the difference between the potentials of the two electrodes. We cannot measure the potential of each electrode individually because the electrons have nowhere to go or to come from. Thus, we need a reference system. We arbitrarily assign a potential value to a particular half reaction:

2 H+( 1 Molar aq) + 2e- → H2 (g, 1 atm )

This is the Standard Hydrogen Electrode (SHE), it has :

Eo = 0.0 V

By definition.

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We can now construct a cell that combines the standard hydrogen electrode (SHE) with any other standard ‘half-cell’ and the observed voltage is then all attributed to the reaction in the second half-cell. In this way we can construct a table of standard electrode potentials.

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Some things to note in the table: 1) In all cases, standard conditions are used. 2) All of the reactions are shown as proceeding in the direction of reduction no matter how likely or unlikely that is.... 3) The reduction potentials E0

RED in the table have a negative sign if the SHE in the cell is found to be the positive electrode. (reduction is taking place at the SHE, oxidation at the other electrode) 4) To construct a cell from any two of these reactions, one of the half reactions must be reversed and written as an oxidation and so the sign of the potential is reversed. But which one????

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Consider the following reaction: Fe(s) + 2Ag+ (aq) → Fe2+(aq) + 2Ag(s) We can write this in ‘cell’ form... Fe/Fe2+ // Ag+/Ag We can ask: Is this reaction spontaneous in the direction as indicated, and what would be the cell emf under standard conditions? From the table: 1) Ag+(aq) + e → Ag(s) E0 = +0.8 V 2) Fe2+(aq) + 2e → Fe(s) E0 = -0.41V But in the reaction, we are looking for the reverse of 2) ie Fe(s) → Fe2+(aq) + 2e E0 = +0.41V Since the oxidation half-reaction produces two moles of electrons per mole of Fe we need to reduce two

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moles of Ag+ so multiply equation 1 by 2. Note E0 remains the same since it reflects only a difference in energy between the two states. 2Ag+(aq) +2 e → 2Ag(s) E0 = +0.8 V Now we have: 2Ag+(aq) +2 e → 2Ag(s) E0 = +0.8 V Fe(s) → Fe2+(aq) + 2e E0 = +0.41V Adding these two yields: Fe(s) + 2Ag+ (aq) → Fe2+(aq) + 2Ag(s) E0

cell = +1.21V The sign is positive and therefore the reaction is spontaneous in the direction shown, and the cell emf would be + 1.21 volts

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Q. Is the following reaction spontaneous?

Fe/Fe2+//Zn2+/Zn

From the tables

1. Zn2+ + 2e → Zn E0 = -0.762v

2. Fe2+ + 2e → Fe E0 = -0.409v But we are looking for the reverse of 2. Fe → Fe2+ + 2e E0 = +0.409v

Zn2+ + 2e → Zn E0 = -0.762v Fe → Fe2+ + 2e E0 = +0.409v

Sum Zn2+ + Fe → Fe2+ + Zn E0 = -0.300v

The negative sign of the cell tells us that the reaction is not spontaneous in the indicated direction, and in fact that the reverse reaction will take place:

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Zn + Fe2+ → Zn2+ + Fe and that the emf of this cell under standard conditions would be +0.300v. Q. Calculate the emf of a cell which has the following set up. Assume all reactants in standard state. Write a balance equation for the reaction.

Al / Al3+ // Ag+ / Ag

Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s) From the table:

Al3+ + 3e- → Al E0 = - 1.68 V Ag+ + e- → Ag E0 = 0.8 V

But we want the reverse of the first equation:

Al → Al3+ + 3e E0 = +1.68v

We now have:

Al → Al3+ + 3e E0 = +1.68v

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3Ag+ + 3e- → 3Ag E0 = 0.8 V

Adding yields:

Al + 3Ag+ → Al3+ + 3Ag E0

cell = +2.48v Lets take another look our table of reduction potentials. What sorts of things might we predict from looking at the Eo values?

Li+ + e- → Li Eo = - 3.040 V

Mg2+ + 2e- → Mg Eo = - 2.357 V

2H+ + 2e- → H2 Eo = 0.000 V

Ag+ + e- → Ag Eo = 0.799 V

F2 + 2e- → 2F- Eo = 2.889 V

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For the reaction Li+ + e- → Li the large negative value of the reduction potential tells us that it is unlikely for this reaction proceed in the indicated direction. We are aware that Li metal is extremely reactive and wants to be in the form of positive (cat-) ions. That is to say, Li+ is a VERY weak oxidising agent, but conversely, Li (metal) is a POWERFUL reducing agent In fact we can conclude that Li metal must be the MOST powerful reducing agent and Li can reduce any of the species in the left hand column of the table. We can predict that: 2Li + Mg2+ → 2Li+ + Mg will proceed as written.

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At the other end of the table, the large positive value of reduction potential for F2 + 2e → 2F- indicates that this reaction is VERY likely to proceed as written. F2 is an extremely reactive gas that wants to be in the form of negative (an-) ions. That is to say, F2 is a very powerful oxidising agent and conversely F- is a very weak reducing agent. In fact we can conclude that F2 must be the most powerful oxidising agent and F2 is capable of oxidising any of the species in the right hand column of the table. We can predict that: F2 + 2Ag → 2Ag+ + 2F-

will proceed as written. As an aside, notice that H2O is above F- in the right hand column. Yes, Fluorine can oxidise water. Possible add-on for me to use as an in-class exercise:

Look at the full table of reduction potentials and find: any four species that can reduce Al3+ to Al

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any metal which can reduce Sn4+ to Sn2+ but not Cr3+ to Cr2+

all the metals which will dissolve in aqueous acids a metal ion which can reduce Pb2+ to Pb but not Al3+

to Al ELECTROLYTIC CELLS In an electrolytic cell, a non-spontaneous redox reaction is made to go by applying a large enough voltage (larger than the spontaneous voltage and opposite in sign) to the cell and pumping electrical energy into the system. Thus: for the reaction:

Zn + Cu2+ → Zn2+ + Cu Eo = 1.10 V

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If we apply a reverse voltage > 1.10 V to the Cu electrode, the reverse reaction should occur:

Zn2+ + Cu → Zn + Cu2+

If we pass a current through a molten mixture of KHF2 and anhydrous HF we achieve the following process:

2HF(l) → H2(g) + F2(g) There is no chemical means of achieving this result (WHY?)

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Quantitative relationshipsAs always the stoichiometry of a balanced reaction as written implies the molar ratios involved Thus the reaction: Cu2+ + 2 e- → Cu(s) Shows that one mol of Cu2+ reacts with 2 mol of electrons. We now need to know the total charge on a mol of electrons. Unit of charge is the Coulomb (C). One electron has a charge of 1.602 x 10-19 C therefore 1 mol e- carries a charge of 1.602 x 10-19 x 6.022 x 1023 = 96490 C mol-1

The charge on a mole of electrons ( approx 96,500 coulombs) is called Faraday’s Constant or one Faraday. Electrical current is defined in terms of the rate of flow of charge per unit time thus:

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1 Amp = 1 coulomb per second and Total charge = current(amps) x time (seconds) eg, in our copper purification experiment, suppose a current of 0.2A passed for 30 min. How many grams of copper are deposited at the cathode? We know that the number of moles of Cu deposited will be one half the number of moles of electrons supplied, (Cu2+ + 2 e- → Cu(s) ). Now the number of moles of electrons supplied will be the total charge delivered divided by the charge on a mole of electrons and: Total Charge (in C) = Current (in amps) x Time

(seconds)

Total charge = [0.2 C/sec] x [ 30 min x 60 sec/min]

= 360 C

Moles e- = (360 C) / (96,490 Cmol-1)

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= 0.00373

Finally:

Moles Cu deposited = (moles e- supplied)/2

Moles Cu deposited = 0.00373/2

Mass Cu = moles x MM

Mass Cu = (0.00373/2) x 63.54 = 0.1185 g Chromium metal can be electroplated from an acidic solution of CrO3 1. How many g of Cr will be plated by 1 x104 C? 2. How long will it take to plate one g of Cr using a current of 6 A? The reaction is:

CrO3(aq) + 6 H+ + 6 e- → Cr = H2O 1 mol CrO3 requires 6 mol e-

1 mol CrO3 contains 1 mol of Cr or 52.0 g Cr 1. If 1 x 104 C are supplied;

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moles e- = Charge/Faraday = 1 x 104 C/ (96,490 C/mol) = .1036 moles e- Moles Cr = (Moles e-)/6 = 0.0172 Mass Cr = moles x MM = 0.0172 x 52 = 0.894g 2. Time to deposit 1 g Cr: moles Cr = mass/MM = 1/52 = 0.01923 moles moles e- required = moles Cr x 6 = 0.01923 x 6 = 0.1154 moles Charge on a mole of e- = Faraday = 96, 490 C Total charge required = moles e- x Faraday = 0.1154 x 96,490 = 1.113 x 104 C Finally: Total Charge = Current x time (sec) 1.113 x 104 C = 6 C s-1 x time time = 1.85 x 103 sec = 30.8 minutes C. Reversibility: Electrolytic Cells

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In Voltaic cells, we converted the chemical energy of a spontaneous reaction into electricity (e.g. using a battery) However, we can often do the reverse: use electrical energy to force a non-spontaneous reaction (e.g. recharge a battery). In this context, where we apply electrical energy, we have an electrolytic cell. Consider the lead-acid battery, which can be used in both voltaic- and electrolytic-cell modes (discharge and recharge). There are separate Pb and PbO2 lead plates • As a voltaic cell, the we have the following reactions o Anode E

0

ox = +0.36 V

Pb(s) + SO4

2−

PbSO4(s) + 2 e−

o Cathode E

0

red = +1.69 V PbO2(s) + 4 H+

+ 2 e−

+ SO4

2−

PbSO4(s) + 2 H2O o Cell notation Pb │PbSO4 ║ PbO2 │PbSO4

o E0

cell = approx +2 V, and six of these are connected in series to provide for a car battery of 12 V o Some interesting notes: . As the car battery gets discharged, H2SO4

decreases, and so does the density of the electrolyte. Some batteries have a green “eye” that lets you estimate its condition… when the electrolyte is dense, the eye floats up and appears green. PbSO4(s) coats the cathode as the battery

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discharges. Thus, the more discharged the battery, the thicker the coating, and the less contact with the electrolyte. If you have a dead battery and wait a few minutes, you can crank your engine again… the electrolyte needs time to diffuse through the coating. . Remember that cell potential does not tell you the rate of the chemical reaction. In the automotive industry, battery makers aim for the highest number of Cold Cranking Amps, a measure of how many Coloumbs per second can be produced. • To recharge the battery, the products of the discharge needs to be converted back to starting materials (electrolytic cell). o Cathode E

0

red = −0.36 V

PbSO4(s) + 2 e−

Pb(s) + SO4

2−

o Anode E0

ox = −1.69 V PbSO4(s) + 2 H2O PbO2(s) + 4 H

+

+ 2 e−

+ SO4

2−

o Cell notation PbSO4 │PbO2 ║ PbSO4 │Pb o Note that in recharge mode (electrolytic cell), the anode and cathode have switched places! Nonetheless, the anode is always the electrode where oxidation occurs. o E

0

cell = approx −2 V, which indicates the reaction is not spontaneous. This is expected! o To recharge, we must apply an external voltage of at least 2 V to force this undesirable reaction to occur. When a redox reaction is driven by electricity, it is called electrolysis. The recharging of a battery is just one example of

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electrolysis, but they all rely on the same principles. • Example: Industrial electrolysis of molten (not aqueous) NaCl for the generation of Na(s) and Cl2

o A battery or power source drives the reaction. A small amount of CaCl2 is also added to help lower melting point o Anode 2 Cl

−

(l) Cl2(g) + 2 e−

o Cathode 2 Na+

(l) + 2 e−

2 Na(l) o Try and see if you could calculate the minimum external voltage needed for this electrolysis o Industrial problem: how do you keep Cl2(g) from reacting with the Na(s) product? o Solution: A brilliant piece of engineering that incorporates an inverted funnel to isolate the chlorine gas product (known as a Downs electrolysis cell) o Another question: Why do we need to use molten NaCl? Why not simply use NaCl dissolved in water? • Electrolysis of NaCl in water results in the same oxidation of Cl

−

to Cl2, but H2O is reduced much more easily than Na

+

o Anode 2 Cl−

(aq) Cl2(g) + 2 e−

o Possible cathode reactions: 2 Na

+

(aq) + 2 e−

2 Na(s) E0

re

2 Hd = −2.71 V

2O + 2 e−

H2(g) + 2 OH− E

0

red = −0.83 V o Reduction of water is much more favourable 2 Cl

−

(aq) + 2 H2O Cl2(g) + H2(g) + 2 OH−

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o So, we don’t have any Na(s) being formed. Is this bad? No! All of Cl2, H2, and NaOH are industrially useful, so the electrolysis of salt water is very important!

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D. Quantitative Analysis (now some physics!)

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