General Chemistry Electrochemistry

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is electrochemistry.0003

We are going to start off with a brief introduction in electrochemistry--what it is about.0010

We are going to go ahead and go into the main reactions that are0015

at the core of electrochemistry which are what we call redox reactions.0020

A big application of electrochemistry is what we call a voltaic cell.0025

We are going to take a look at that followed by quantifying aspects in electrochemistry which deal with what we call electrochemical potentials.0028

After we introduce the potentials, we will then go ahead and get into quantitative problems0040

that include current, charge, and time, followed by a brief summary and sample problems.0046

Electrochemistry is the branch of chemistry which deals with harnessing chemical energy and converting it into mechanical energy.0054

Basically it studies the processes in which energy is released during a chemical reaction0064

so that it can be used to perform some type of work.0071

The perfect example of this is the battery which powers everyday machines including vehicles, computers, and phones.0076

Basically a battery is a chemical process that produces energy that is then converted into electricity.0084

Basically in this lecture, we are going to examine both the concepts and equations that are at the core of electrochemistry.0095

An electrochemical reaction is one that involves a complete transfer of electrons from one reactant to another.0104

The reactant which loses the electrons is said to be oxidized or undergo oxidation.0111

The nice little acronym that allows us to remember this is OIL or oxidation-is-loss.0119

The reactant that gains electrons is said to be reduced or undergo reduction.0125

The acronym for that is RIG or reduction-is-gain.0135

OILRIG, oxidation is loss; reduction is gain.0138

An oxidation reduction reaction is also known as a redox reaction for short.0143

Let's now look at how to identify each type of a reactant.0151

To identify which reactant is being oxidized and which is being reduced, it helps to0158

keep track of oxidation states which we learned in the first half of general chemistry.0164

Let's go ahead and consider the following reaction.0169

CH₄ gas plus O₂ gas goes on to form CO₂ gas and H₂O gas.0175

Let's go ahead and balance this guy.0191

We are going to need two of these and two of those.0194

This is the reaction that represents the combustion of methane.0201

We are going to now assign oxidation states to each of the elements.0207

Hydrogen is +1 of course which means overall this carbon is going to be -4.0214

Here in O₂, this is a homonuclear diatomic; each oxygen is 0 here.0222

Here in CO₂, oxygen is -2 each which means carbon is going to be +4.0230

Here oxygen is -2; each hydrogen is +1.0238

We see that there are two elements that experience a change in oxidation state.0243

Carbon starts off as -4; it becomes +4 in the product side.0249

Oxygen starts off as 0; it becomes -2 on the product side.0261

We see that carbon's oxidation state has increased.0269

If the oxidation number goes up, that means it has lost electrons.0275

Oxygen's oxidation state has gone down which means it has gained electrons.0284

If carbon has lost electrons, we say that it has been oxidized.0293

If oxygen gains electrons, we say that it has been reduced.0303

But it is not just the element that undergoes the process.0308

It is that entire compound.0311

To be more accurate, we say that CH₄ gas was oxidized.0313

We say that O₂ gas was reduced.0323

We are going to introduce two more terms here.0330

CH₄ gas was oxidized; if it was oxidized, it performed the reduction.0333

Therefore CH₄ gas is also known as the reducing agent.0341

O₂ gas was reduced.0353

If it was reduced, it performed the oxidation which means that O₂ gas is what we call the oxidizing agent.0357

You definitely want to keep track of these terms because they are somewhat confusing.0369

A reducing agent performs the reduction; but it itself is oxidized.0373

A reducing agent performs the reduction... excuse me.0379

An oxidizing agent performs the oxidation; it itself is reduced.0385

Let's now learn how to balance redox reactions.0389

Redox reactions can occur in neutral solutions and in acidic solutions and basic solutions.0397

We are going to learn how to balance redox reactions in all three of these types of medium.0407

In neutral solutions, let's go ahead and start off.0413

Cl₂ gas can react with Zn²⁺ to go and form Cl^1- aqueous plus zinc solid like that.0422

What we are going to do is we are going to balance these guys.0447

When we balance in neutral solutions, you must make sure that the0452

stoichiometry is balanced and the net charge on both sides of the equation.0456

When we go ahead and look at this, we see that we have one zinc on each side.0467

That is good to go.0472

We have two chlorines on the left and only one on the right.0473

Let's go ahead and put the two chlorines there.0476

My net balanced charge here is 2+.0478

My net charge here is going to be 2-.0482

Let's go ahead and see what the half reactions are.0492

For the half reactions, we have Cl₂ gas going to become 2Cl^1- aqueous.0501

That means that each chlorine is gaining an electron.0516

I am going to need two electrons on the left side.0520

Zn²⁺ is becoming zinc solid.0524

That means I am going to need two electrons on this side.0530

When we look at the half reactions, these are both reduction.0550

But you can't have two reduction half reactions.0558

You always have to have one oxidation; you always have to have one reduction.0561

It turns out that the chlorine one is going to be the one that reacts.0566

What is going to happen is we are going to flip this one.0577

That is going to be zinc solid going to form Zn²⁺ plus two electrons.0589

This immediately told me that we have done something wrong because my net charge on each side is not the same.0599

My overall reaction is going to be Cl₂ gas plus zinc solid going on to form 2Cl^1- aqueous and Zn²⁺ aqueous.0608

As you can see, now my overall charge is fixed.0624

This is 0 on the left.0628

This is 0 on the right side for our balanced redox reaction in neutral solutions.0629

Let's now jump into solutions where we have acidic or basic conditions.0637

For example, in acid, silver solid plus nitrate aqueous goes on to form NO₂ gas and Ag⁺¹ aqueous.0644

Step one is to break the unbalanced equation into your half reactions.0652

My half reaction here is going to be silver solid going to Ag¹⁺ aqueous.0658

NO₃^1- aqueous going on to form NO₂ gas.0671

Those are our half reactions.0681

Step two, we are going to balance each of the half reactions by being careful to follow each of these following steps.0683

Balance all elements except oxygen and hydrogen by stoichiometry.0692

For the silver half reaction, we have one of each on side.0699

We are good to go on that.0703

For nitrate going to nitrogen dioxide, we have one nitrogen on each side.0705

We are good on that; part A is done.0711

Part B, we are going to balance oxygen by adding water to the side that is deficient in oxygen.0715

Here for nitrate, I have three oxygens; for nitrogen dioxide, I only have two oxygens.0721

I am going to add H₂O to the side that is deficient in oxygen.0727

Let's go ahead and move on now to step number three.0737

If in acidic solution, we are going to add hydronium for0742

every missing hydrogen to the side that is deficient in H0745

and the same number of water molecules to the other side.0749

Our half reaction again was NO₃^1- aqueous going on to form NO₂ gas.0752

We added water last time; we are going to follow part C.0762

We are going to add hydronium for every missing hydrogen to the side that is deficient in hydrogen.0767

On the right side of this equation, I have the two hydrogens here.0773

I am going to add two hydroniums on the left side.0776

Remember you are adding one hydronium for every hydrogen.0780

At the same time, you are going to add water, the same number of water molecules to the other side.0784

Let me go ahead and do that in blue; plus two more H₂Os.0789

If this was basic solution, we would add one water for0796

every missing hydrogen to the side that is deficient in H0799

and the same number of hydroxide to the other side.0803

What we are going to do right now is we are going to go on to part E.0807

We are going to balance the net charge by adding electrons to the side that is deficient in negative charge.0812

Let's go ahead and look at this.0818

Right now we have two hydroniums aqueous plus NO₃^1- aqueous.0820

That goes on to form NO₂ gas plus 3H₂O liquid.0829

We have a net charge of +1 on the left side.0842

We have a net charge of 0 on the right side which means that0851

we have to add an electron to the left side here to get us to 0.0856

We will then multiply the half reactions by integers that will lead to cancellation of electrons.0875

In other words, each half reaction should have an identical number of electrons.0880

Ag solid going to Ag¹⁺ aqueous.0887

It was now NO₃^1- aqueous going to NO₂ gas.0895

I believe this was three waters; this was two hydroniums aqueous.0906

Let's go ahead and fix the electrons; I am going to need one electron here.0916

I am only going to need one electron right there.0922

Because the number of electrons is the same in each reaction, we are good to go.0930

We are going to now recombine the half reactions to get the net balanced equation.0939

Be sure to cancel like terms.0943

When I add these two reactions together, the electrons cancel out.0946

We get silver solid plus 2H₃O¹⁺ aqueous plus NO₃^1- aqueous0950

going on to form Ag¹⁺ aqueous plus NO₂ gas and 3H₂O liquid.0961

This should be liquid here.0971

Let's go ahead and double check if everything is balanced.0973

On the left side, I have one silver.0977

On the right side, I have one silver; that is good to go.0979

On the left side, I have six hydrogens.0982

On the right side, I have six hydrogens.0984

On the left side, I have one nitrogen.0987

On the right side, I have one nitrogen.0989

On the left side, I have a total of five oxygens.0992

On the right side, I have a total of five oxygens.0996

Stoichiometry is good to go; let's do a final check on the charge.0999

I have a total charge of 1+ on the left side.1003

I have a total charge of 1+ on the right side.1010

Yes, this is our overall balanced redox reaction.1012

Let's go ahead now and look at how a simple battery works.1017

The first battery was referred to as a voltaic cell, also known as a galvanic cell shown below.1023

Let's go ahead and look at the redox reaction for one we already looked at.1031

This was going to be then Cl₂ gas plus zinc solid going on to form 2Cl^1- aqueous and Zn²⁺ aqueous.1040

Basically the redox reaction was the following; we can have a metal electrode here.1070

The metal electrode is going to be dipped in a salt solution that contains its ion.1091

If let's say the metal electrode was represented by A solid, the salt solution1099

could be for example A⁺ and maybe X^1- for our salt solution.1107

Again I have another metal electrode on the right side of a different identity.1121

This could be B solid for example; this would be B⁺ and X^1-.1125

These two electrodes are going to be connected by a conductive wire here.1133

We get a flow of electrons from A to B.1143

The flow of electrons is going this way.1147

Anytime you have a flow of electrons, that generates an electric current which can then be captured to power some type of device.1151

What we see here is that A solid is going to be losing electrons to form A¹⁺ aqueous like so.1162

That is its half reaction.1175

B solid is going to be gaining the electrons to form B^1- aqueous.1177

We need something called a salt bridge to help us complete the circuit and to balance the charge.1195

Here because we are forming a lot of A⁺, some type of ion has to come in and balance the positive charge.1205

Maybe some more X^- comes in from the salt bridge.1218

The salt bridge can be for example A⁺ and X^1-.1222

Because in the B side we are forming a lot of negative charge,1232

some positive cation is going to come in from the salt bridge also.1240

We see that here that this is going to be representing the oxidation half reaction.1249

Here on the right side, this is going to be the reduction half reaction.1255

We can now also look at the charges.1272

If the flow of electrons is from A to B, that means that...1275

Remember that electrons are going to go away from a negative side.1280

This is the negative charge.1284

They are going to go toward the positive side which is going to be where the electrons gravitate toward.1287

If you ever look at a battery, a battery always has a negative sign and a positive sign representing these two electrode terminals.1294

The negative electrode is what we call the anode.1303

The positive terminal is what we call the cathode.1308

A nice little trick to remember this is when you cross the t in the word cathode, it looks like a positive sign.1315

We can also have something we call cell notation.1324

Cell notation typically looks like this.1328

On the left side goes the anode; on the right side goes the cathode.1344

The single lines represent changes in physical state.1357

The double line just simply represents a salt bridge.1367

Here we can easily write in A solid going to A¹⁺ aqueous.1372

Here we can write in B¹⁺ aqueous going to B solid.1382

Let me go ahead and correct this then.1399

This is really B¹⁺ aqueous plus an electron going to be solid.1402

Again this is what we call cell notations.1409

The anode half reaction is written on the left.1411

The cathode half reaction is written on the right side.1415

Another thing about the voltaic cell is that we see that the anode here is getting consumed.1421

The solid is becoming ionized.1431

We see here that the cathode, we are actually depositing more solid onto the electrode.1434

This is going to become heavier, become larger in mass.1442

Why does a battery die?1453

It is essentially because one of the terminals, the anode gets consumed to the point where no more reduction or oxidation can occur.1456

If you ever have a rechargeable battery, you know that a rechargeable battery you have to plug in.1472

What does that do?1479

When we put a rechargeable battery into the wall outlet and charge it overnight, we are basically driving the reverse reaction.1480

The reverse reaction is then replenishing the anode while consuming the cathode.1488

But that doesn't always go on forever; we cannot use a rechargeable battery infinitely.1494

It also too is going to eventually die.1500

That is because no process is 100 percent efficient.1504

You will never ever regain all of your anode.1508

It is going to become less and less efficient as time goes on.1514

Let's now go ahead and look at the quantitative part of electrochemistry.1518

All half reactions have a potential measured in volts, when you purchase for example a 1.5 volt battery.1524

This is just like a chemical reaction having a change in energy associated with it.1531

These potentials are all relative to the hydrogen reduction half reaction which is arbitrarily assigned 0 volts.1536

That is going to be 2H⁺ aqueous plus two electrons going on to form H₂ gas.1544

This potential is what we symbolize as E standard.1555

That is going to be 0.00 volts.1559

It is very important to be able to use the following table of standard reduction potentials.1561

Here I have listed for you five half reactions from this table.1570

You see that these are all reduction half reactions.1575

On the right side are the reduction potentials; E standard values.1584

It is important that we understand how to decipher this.1590

Point number one is a large E₀ means that the reduction is more likely to occur.1595

Again a large E₀ is reduction more likely to occur.1606

For example, the reduction of Cl₂ has a potential of +1.36 volts.1611

The reduction of Zn²⁺ has a potential of -0.76 volts which means that Cl₂ is more likely to be reduced than Zn²⁺.1622

In fact if Cl₂ is going to be...1638

If we combine the two, Cl₂ gas plus two electrons going to 2Cl^1-1644

and Zn²⁺ aqueous plus two electrons going on to zinc solid,1654

which one is going to be the reduction half reaction?1664

Which one is going to be the oxidation half reaction?1667

Because this half reaction for chlorine is more positive, that is going to remain reduction.1672

Zinc solid is going to become oxidized.1677

The combination of Cl₂ gas plus two electrons going on to form 2Cl^- and1682

zinc solid going on to form Zn²⁺ plus two electrons for a balanced redox reaction of1690

Cl₂ gas plus zinc solid going on to form Zn²⁺ aqueous and 2Cl^1- aqueous.1697

In other words, Cl₂ gets reduced; zinc is going to be oxidized.1708

This is very important because when we read this table, we can always draw an arrow in the following direction.1720

Anything that follows this direction, you are going to have a successful redox reaction.1729

Basically we can summarize it the following way, what this line means.1738

Any element or ion can oxidize any other species...1743

Any element or ion can oxidize any other species to the bottom and right of it, to the bottom right of it.1765

For example, Cl₂ can oxidize silver; it can oxidize H₂.1780

It can oxidize nickel; it can oxidize zinc.1786

Cl₂ itself will be reduced because its reduction potential is so positive.1790

We see therefore that when we go this direction, these are going to be very strong oxidizing agents.1798

When we go this direction on the right side, these are going to be stronger reducing agents.1812

In this table, we see that Cl₂ is the strongest oxidizing agent.1822

It itself is most likely to be reduced.1827

Zinc solid is the strongest reducing agent.1830

It itself is going to be more likely to be oxidized.1832

Let's now look at calculating the potential when not at standard conditions.1837

To calculate the potential for a voltaic cell when not at standard conditions, we use what is called the Nerst equation.1843

This is just like ΔG is equal to ΔG⁰ minus RT natural log of Q; very very similar.1851

Once again Q is going to be the reaction quotient.1857

The temperature is in kelvin of course.1864

R is going to be 8.314 joules per k mole.1868

n is going to be the moles of electrons.1875

You get that from the balanced redox reaction; moles of electrons from balanced redox reaction.1878

F is what we call Faraday's constant.1890

Faraday's constant, you should always ask your instructor if you have to know it or not.1898

It is basically 9.65 times 10⁴.1904

The units, I am going to draw in red, are coulombs for every mole of electron.1909

Once again coulombs per moles of electron or C over mole.1918

Again you are going to use this equation anytime you are dealing with nonstandard conditions1924

which is basically not 1 atm of pressure and not 1 molarity of concentration.1929

It can also be shown that at equilibrium, E is going to be 0.00 volts1937

which means that E₀ is equal to RT over nF times the natural log of K.1943

Again this is any of the K values--K_a, K_b, K_f, K_sp, etc.1948

Again these are the two equations that are heavily used in electrochemistry.1960

Finally it can be shown that we can relate Gibbs free energy to electrochemistry.1968

The equation is ΔG⁰ is equal to ?nFE₀.1977

This equation tells us that if E₀ is large...1982

If E₀ is large, that means the potential is very high which means the redox reaction is likely to occur.1988

Remember what we said if something is likely to occur>2000

If something is likely to occur, ΔG⁰ is negative.2003

Look, that makes perfect sense because if E is positive... we know F is positive.2008

We know n is positive making that whole side of the equation negative giving us a negative Gibbs free energy.2013

Basically if E₀ is very large, that means the reaction is in the forward direction.2020

It is highly product favored which as we have seen translates to a lower energy state.2026

Let's now see how to factor in the lifetime of a battery.2032

A battery basically has three main parameters--the voltage of the cell,2038

the current that is drawn from it by the device,2043

and the total amount of time the battery can run at the specified current.2045

A simple equation relates all of these parameters.2050

n is equal to It over F where I is your current in amps.2053

Time, t is your time in seconds.2063

Of course F is your Faraday's constant, 9.65 times 10⁴ coulombs per mole of electrons.2069

In this equation, n is not going to be a whole number.2079

n is going to be the mole of electrons that is determined stoichiometrically.2082

Let's go ahead and summarize this before we jump into our sample problems.2091

Electrochemistry we saw studies how chemical energy can be produced and converted into2095

mechanical energy which in turn can perform work such as powering an electrical device.2100

A potential is produced when there is a transfer of electrons2107

from the reducing agent to the oxidizing agent which as2110

we have seen can happen in neutral, acidic, or basic media.2114

Finally the Nerst equation allows for determination of the cell potential when not at standard conditions.2118

Sample problem number one, basically here...2128

This is a very common type of problem by the way.2132

You are given a list of half reactions here with their reduction potentials.2134

E₀ is right here.2140

You are just asked to answer the following questions, all qualitative problems.2146

What is the strongest oxidizing agent?--what is the strongest reducing agent?2150

Remember that the stronger oxidizing agents are the ones that are going to be2154

most likely to be reduced which corresponds to a very high reduction potential.2158

Here the strongest oxidizing agent is Cl₂ gas.2163

The strongest reducing agent therefore is going to be zinc solid.2168

Part B, list the species that silver solid can reduce if any.2173

Silver solid is not going to reduce species.2181

But instead it is going to oxidize species.2187

Silver solid is going to only... excuse me.2189

Silver solid can reduce anything to the top and left of it.2198

That is only going to be Cl₂ gas; again reduction occurs to top and left.2205

List the species that zinc solid can reduce if any.2216

Zinc solid because it is such a strong oxidizing agent, it can reduce any species that is to the top and left of it.2220

That is going to be Ni²⁺, H⁺, Ag¹⁺, and Cl₂ gas.2231

Finally part D, what species is most easily oxidized?2240

The species that is most easily oxidized is going to be the strongest reducing agent.2244

Therefore this is going to be zinc solid.2254

What species is most easily reduced?2257

That is going to be the strongest oxidizing agent which is going to be Cl₂ gas.2259

You see, I just want to point out a couple things.2264

That part A and part D, these are really the same question that are asked differently.2266

Again there is going to be multiple ways of phrasing, of posing the same question.2272

Sample problem number two, if a battery contains 0.355 grams of cadmium and excess2280

nickel hydroxide, how long can it operate if it draws 1.25 milliamps of current?2288

You are given the balanced redox reaction below.2294

Because this question mentions current and how long, there is only one equation that deals with that.2299

That is n is equal to It over F.2305

We know that I is amps; that is going to be 0.00125 amps.2308

We know Faraday's constant already, 96500 coulombs per mole of electrons.2316

The question is asking us to solve for time.2324

t is going to be equal to n times F over I.2327

All we need is n.2332

Remember that n is the mole of electrons that is determined stoichiometrically.2334

Let's go ahead and get it.2339

You are told that you have an excess amount of the nickel precursor.2340

The 0.355 grams of cadmium is going to be our limiting reactant here.2345

0.355 grams of cadmium times 1 mole of cadmium divided by 112.41 grams.2350

That is going to be equal to 0.00316 moles of cadmium solid.2365

The redox reaction is right here; that is cadmium zero going to cadmium hydroxide.2374

In this reaction, cadmium is 2+.2388

For cadmium to go to cadmium 2+, it has to give up two electrons.2395

My mole to mole ratio of the electrons to cadmium is a 2:1 ratio.2401

0.00316 moles of cadmium times 2 moles of electrons for every 1 mole of cadmium.2408

That is going to give me the value of n which is 0.00632 moles.2419

When all is said and done, you should get a time of 487904 seconds which is approximately 136 hours.2428

That is an awfully nice lifetime for a battery to operate, 136 hours.2441

If you look at these elements, it is nickel cadmium.2450

I am sure you have heard of batteries that are called nickel cadmium or cadmium nickel batteries2453

which are very nice batteries that help to power devices such as your smartphones and cameras.2459

That is our lecture from electrochemistry; I want to thank you for your time.2469

I will see you next time on Educator.com.2474