Chapter 20 Electrochemistry

Chapter 20 Electrochemistry
Dr. Peter Warburton

Electrochemical cells fall into one of two basic types
Galvanic cells Electrochemical cells fall into one of two basic types Galvanic cells convert chemical energy into electrical energy (batteries) Electrolytic cells convert electrical energy into chemical energy.

Here we put a piece of zinc metal into a Cu2+ ion solution
Here we put a piece of zinc metal into a Cu2+ ion solution. A reaction occurs where we get Zn2+ ions and solid copper deposited on the zinc surface.

This is an oxidation-reduction (redox) process where electrons are transferred from one chemical to another. One chemical loses electrons in a process called oxidation, while the other chemical gains electrons in a process called reduction.

Zinc in Cu2+ solution is spontaneous
Since we actually see this reaction occurring, this reaction must be spontaneous! The reverse reaction, where we put copper metal into a Zn2+ ion solution is non-spontaneous!

we can see that each zinc atom gives away 2 electrons
Redox reaction Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) we can see that each zinc atom gives away 2 electrons to a copper (II) ion to give us a copper atom and a zinc (II) ion in the spontaneous reaction!

We call this the oxidation half-reaction.
Half-reactions To clarify the redox process, we often break a redox reaction down into two separate steps (half-reactions). In one half-reaction, a chemical loses electrons (is oxidized) Zn (s)  Zn2+ (aq) + 2 e- We call this the oxidation half-reaction. Notice that we are effectively treating electrons as a “product” of the half-reaction. In the other half-reaction, we look at the reduction half-reaction, where a chemical gains electrons (is reduced) Cu2+ (aq) + 2 e-  Cu (s)

Half-reactions The sum of these half-reactions must give us the overall reaction of interest. Zn (s)  Zn2+ (aq) + 2 e- Cu2+ (aq) + 2 e-  Cu (s) Zn (s) + Cu2+ (aq) + 2 e-  Zn2+ (aq) + 2 e- + Cu (s) Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

Half-reactions Why do we call them “half-reactions”?
Each half-reaction is written so we can see what is happening to the electrons in the overall reaction. In reality a half reaction CANNOT occur by itself to any great extent. The lost electrons in the oxidation half-reaction MUST go somewhere. The gained electrons in the reduction half-reaction MUST come from somewhere. Two half-reactions ALWAYS work together to give an overall reaction that can occur to a great extent.

Half-reactions The electrons stay on the metal electrode and are NEVER found in solution!

Since the chemicals are in direct contact with each other, the electron transfer occurs directly and we can’t use the electrons to do anything useful. How can we separate the chemicals but allow the electrons to transfer indirectly so we can use them?

Copper in Ag+ solution is spontaneous

to take place in separate containers
The half-reactions to take place in separate containers (called half-cells). Since a half-reaction cannot take place by itself we need to connect the half-cells together. It turns out that we must make a circuit (two connections!) for the entire galvanic cell to work.

The left half-cell has a solid copper electrode in a Cu2+ ion solution, while the right half-cell has a solid silver electrode in a Ag+ ion solution. A wire can connect the two solid electrodes for the electrons to move through. To connect the two solutions so that ions can move between the half-cells requires us to use a salt bridge, which is just another solution of ions.

Oxidation occurs at the copper electrode, which we give the special name
ANODE Since the anode collects the electrons that are lost, it has a negative charge and positive copper ions leave the anode! Cu (s)  Cu2+ (aq) + 2 e-

to the silver electrode through the wire
Electrons move from the ANODE to the silver electrode through the wire We can get them to do something useful, like light a bulb!

Reduction occurs at the silver electrode, which we give the special name
CATHODE Since the cathode collects the positive silver ions so they can gain the electrons, the cathode has a positive charge! Ag+ (aq) + e-  Ag (s)

Positive ions leave the anode while the cathode collects positive ions.
Alternatively, negative ions collect at the anode and move away from the cathode The ions are free to move through the salt bridge and are REQUIRED to complete the circuit!

Overall, negative charges (electrons and negative ions) are moving clockwise
Overall, positive charges (positive ions and electron “holes”) are moving counterclockwise

Note # of e- must balance!
The overall reaction is exactly the same as when we place solid copper in a Ag+ solution, but since we have separated the half-cells, we can look at the separate half-reactions as they occur. Note # of e- must balance! Cu (s)  Cu2+ (aq) + 2 e- 2 x [Ag+ (aq) + 1 e-  Ag (s)] Cu (s) + 2 Ag+ (aq) + 2 e-  Cu2+ (aq) + 2 e- + 2 Ag (s) Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)

Shorthand notation for galvanic cells
Drawing a diagram for a galvanic cell or describing it as we did in the previous problem is too time-consuming to do on a regular basis. We can use a shorthand notation! Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s) Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)

Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)
A single vertical line indicates a change in phase, like that between a solid electrode and the solution its immersed in. A double vertical line indicates a salt bridge. What is not shown in the shorthand (but is always implied) is the wire connecting the two electrodes to complete the circuit.

Cu (s) | Cu2+ (aq) || Ag+ (aq) | Ag (s)
If we read the shorthand notation from left to right it says: “A solid copper anode is in a solution of copper (II) ions which is connected by a salt bridge to a solution of silver (I) ions into which a solid silver cathode has been placed. The electrodes are connected by a wire.”

We ALWAYS choose to write the cell notation with the oxidation reaction first and then the reduction reaction. This means the leftmost chemical in the notation is ALWAYS the anode, while the rightmost chemical is ALWAYS the cathode. Additionally, the electrons ALWAYS flow from left to right through the wire, which is the way we read the shorthand.

(through the wire connecting the electrodes)

Galvanic cells – Easy as ABC
Anode  Cathode Negative  Positive Oxidation  Reduction Left  Right “The anode is the negative electrode where oxidation takes place. We put it on the left in shorthand notation.” “The cathode is the positive electrode where reduction takes place. We put it on the right in shorthand notation.”

Other shorthand notation considerations
Sometimes gases are involved in galvanic cells. Including them in the shorthand is easy once we realize the gas is just a separate phase and must be separated from other phases by a vertical line.

Cu (s) + Cl2 (g)  Cu2+ (aq) + 2 Cl- (aq)
Other shorthand notation considerations Consider this reaction Cu (s) + Cl2 (g)  Cu2+ (aq) + 2 Cl- (aq) Since we CAN’T use a gas as an electrode we need some solid substance to do that job. In this case we bubble the gas by a carbon rod The cell notation with the carbon acting as the cathode is Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)

Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s)
Other shorthand notation considerations Cu (s) | Cu2+ (aq) || Cl2 (g) | Cl- (aq) | C (s) We can also be more specific by giving concentration and pressure data for any of the aqueous or gaseous chemicals of the system. e.g. Cu2+ (aq, 0.58 M) and Cl- (aq, 0.34 M) and Cl2 (g, 0.89 bar)

Fe (s) + Sn2+ (aq)  Fe2+ (aq) + Sn (s)
Problem Write the shorthand notation for a galvanic cell that uses the reaction Fe (s) + Sn2+ (aq)  Fe2+ (aq) + Sn (s) Fe (s) | Fe2+ (aq) || Sn2+ (aq) | Sn (s)

Problem Pb (s) + Br2 (l)  Pb2+ (aq) + 2 Br- (aq)
Write a balanced equation for the overall cell reaction and give a brief description of the galvanic cell represented by Pb (s) | Pb2+ (aq) || Br2 (l) | Br- (aq) | Pt (s) Pb (s) + Br2 (l)  Pb2+ (aq) + 2 Br- (aq) The reduction of Br2 to Br- occurs on the surface of a Pt cathode

Cell potentials for cell reactions
Electrons move from the copper anode through the wire to the silver cathode because it is energetically favourable for the electrons to move! An electron in a siver atom has less free energy than the same electron in a copper atom. Much like a ball wants to roll down a hill so it ends up where it has the lower potential energy, an electron wants to move to the atom where it has the lower free energy.

Potential The difference in the free energy for the electrons in the anode and the cathode is somewhat like the slope from the top to the bottom of the hill. If the hill is “steep”, the ball experiences more of the force of gravity than it does on a “gentle” hill.

also known as the cell potential (E)
The equivalent of the force of gravity to the difference in the free energy of electrons in different atoms is called the electromotive force (emf) – also known as the cell potential (E) or the cell voltage (V). Like a ball on a steep hill, electrons are under a “greater” force to transfer from the anode to the cathode when the cell potential has a larger magnitude.

Potential Because there is a free energy difference for an electron in the anode as compared to the same electron in the cathode, the electron must lose free energy during the trip, just like a ball loses potential energy (as motion!) as it rolls down the hill. The free energy change is negative and so the movement of the electrons is a spontaneous process!

Potential The free energy change is negative and so the movement of the electrons is a spontaneous process! This occurs when the potential is positive so a positive potential indicates a spontaneous process

Potential We can get energy out of a ball (with its certain mass) rolling down a slope (the experienced gravity), We can get energy out of an electron (with electrical charge) that “rolls down the slope” that is the potential difference of electron free energy between the two electrodes.

Potential In terms of units, we can define one Joule as the energy
we get from a charge of one Coulomb multiplied by the potential of one volt. 1 J = 1 C·V (one Coulomb-volt)

Potential A Coulomb is a very large unit of charge!
The charge on one electron is 1.60 x C, so one Coulomb is the charge of about 6 billion billion electrons!

Faraday Constant or faraday (F)
Potential It is generally easier to talk about the charge of one mole of electrons, which we give the special name of Faraday Constant or faraday (F) 1 faraday = x 1023 mol-1 e- x 1.60 x C 1 faraday = 9.65 x 104 C·mol-1

Potential We can measure the potential between two electrodes with a voltmeter, which should give a positive reading when the positive terminal of the voltmeter is connected to the positive electrode (the cathode), and the negative terminal is connected to the negative electrode (the anode). When the voltmeter gives a positive potential, we have identified the direction of spontaneous change!

Copper in Ag+ solution is spontaneous

Standard cell potentials
Cell potentials depend on many factors other than the chemicals in the system, including the temperature, ion concentrations, and pressure. Like in the thermodynamics chapter, where we defined a standard state of conditions for enthalpy tables, we can do the same to define standard cell potentials E°.

We can only measure a standard cell potential if we have pure solids and liquids (activities of 1), all solution activities are 1 molL-1), all gas activities are 1 bar), and the temperature is specified (usually 25 °C).

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s) we can only measure the STANDARD cell potential if the [Zn2+] and [Cu2+] are both 1 molL-1 , and the copper and zinc electrodes are pure. The E° for this cell is 1.10 V at 25 °C.

Standard electrode potentials
The standard cell potential E for any galvanic cell can be expressed as the difference of the standard electrode potentials for the cathode and the the anode Ecell = E(red),cathode - E(red),anode

The standard electrode potential depends on whether the electrode is acting as the cathode or the anode. However, the process at the cathode (reduction) is the opposite process that would occur if it were happening at the anode (oxidation).

Reversing a process changes the sign of the electrode potential associated with the process. Therefore we choose to report ALL standard electrode potentials as reduction processes because for any specific electrode E(red),cathode = - E(red),anode

It would be nice to create a table of standard electrode potentials for all possible electrodes, then we could find standard cell potentials for any cell. However, there is one problem! We’ve already seen that half-reactions cannot occur without another half-reaction!

We got around a problem like this in thermodynamics by defining the standard enthalpy of formation of elements in their standard states as ZERO. We can do the same for electrode potentials and set the potential for a specific electrode as ZERO and measure all other electrode potentials in comparison to the standard.

Standard hydrogen electrode
The standard electrode for potentials is the standard hydrogen electrode (S.H.E). The electrode consists of hydrogen gas at 1 bar bubbling through a 1 molL-1 (actually activity of 1) solution of H+ past a platinum electrode. Therefore 2 H+ (aq, a = 1) + 2 e-  H2 (g, 1 bar) E(red),cathode = EH+/H2 = 0 V

Standard hydrogen electrode
If the oxidation reaction occurs instead in this half-cell as an anode, the overall reaction is H2 (g, 1 bar)  2 H+ (aq, 1 molL-1) + 2 e- Reversing a reaction changes the sign of the potential. For the S.H.E. E(red),anode = - E(red),cathode = - EH+/H2 = 0 V

Pt (s) | H2 (g) | H+ (aq) || Cu2+ (aq) | Cu (s)
The standard potential for this cell has been measured as V at 25 C, and our anode is the standard hydrogen electrode!

standard reduction potential.
We have defined the standard electrode potential of the reduction of Cu2+ ions to solid Cu! This is also known as a standard reduction potential. Cu2+ (aq) + 2 e-  Cu (s) E(red) = V

Cu (s)  Cu2+ (aq) + 2 e- E(ox) = -0.340 V
If we reverse the half-reaction we’ll get the standard OXIDATION potential for the oxidation of solid Cu to Cu2+ ions… Cu (s)  Cu2+ (aq) + 2 e- E(ox) = V

Zn (s) | Zn2+ (aq) || H+ (aq) | H2 (g) | Pt (s)
The standard potential for this cell has been measured as V at 25 C, and our anode is the zinc electrode!

standard oxidation potential.
We have found the standard potential of the oxidation of solid zinc to zinc ions! This is a standard oxidation potential. Zn (s)  Zn2+ (aq) + 2 e- E(ox) = V

We report this value as the standard electrode potential!
If we reverse the half-reaction we’ll get the standard reduction potential for Zn2+ ions to solid zinc Zn2+ (aq) + 2 e-  Zn (s) E(red) = V We report this value as the standard electrode potential!

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)
The standard cell potential for this cell can be calculated if we know the anode is the zinc electrode and the cathode is the copper electrode!

Standard electrode potentials E(red)

Using standard electrode potentials
Using tabulated standard electrode potential data is accomplished much like a Hess’s Law problem with one very important difference! Let’s consider Zn (s) | Zn2+ (aq) || Ag+ (aq) | Ag (s) which has the balanced equation 2 Ag+ (aq) + Zn (s)  2 Ag (s) + Zn2+ (aq)

Using standard electrode potentials
Oxidation Zn (s)  Zn2+ (aq) + 2 e- Reduction 2 [Ag+ (aq) + 1 e-  Ag (s)] 2 Ag+ (aq) + Zn (s)  2 Ag (s) + Zn2+ (aq) Ecell = V

However, we DO NOT multiply the potential for either half-reaction.
Like Hess’s Law, we look up the standard electrode potential reactions for both our sets of chemicals and then reverse the half-reaction for the set undergoing oxidation while changing the sign of the electrode potential (- E(red),anode!) . However, we DO NOT multiply the potential for either half-reaction. Why?

Recall the potential is
like the slope of a hill. A hill does not change its slope if we have two (or more!) balls rolling downhill instead of one ball. Therefore the potential of an electrode does not change if we multiply to get the right number of electrons!

Al (s) | Al3+ (aq) || Cr2+ (aq) | Cr (s)
Problem The standard cell potential for the following galvanic cell is 0.78 V Al (s) | Al3+ (aq) || Cr2+ (aq) | Cr (s) The standard electrode potential for the Al electrode is V. Calculate the standard electrode potential for the Cr electrode.

Problem Use the data from Table 20.1 to determine the Ecell for the redox reaction in which Fe2+ (aq) is oxidized to Fe3+ (aq) by MnO4- (aq) in acidic solution. Also provide the overall reaction. Answer: Ecell = 0.74 V 5 Fe2+ (aq) + MnO4- (aq) + 8 H+ (aq)  5 Fe3+ (aq) + Mn2+ (aq) + 4 H2O (l)

Free energy and electrical work
We’ve already seen for any system the energy free to do work is given by DG = DH - TDS at standard conditions, or DG = DH – TDS at non-standard conditions.

Not all work has to be expansion (PDV) work. There are other types of work, one of which is electrical work! There must be a connection between DG and electrical work done by a galvanic cell.

We’ve seen for a spontaneous process that DG < 0. We’ve also seen for a galvanic cell the overall reaction is spontaneous, and the cell potential is positive to indicate spontaneity.

Therefore for a spontaneous process in a galvanic cell, the change in free energy (the electrical work done) must be directly proportional to the negative of the potential. DG  -Ecell or DG = welec = -kelec Ecell

The constant of proportionality kelec must depend on two things. First, it depends on how many electrons we have moved through the wire. Twice as many electrons should mean twice as much work… We will usually measure numbers of electrons in moles and symbolize it by n.

The constant of proportionality kelec must also depend on the charge of each electron moving through the wire. Since we are already talking about moles of electrons, we should talk about the charge of one mole of electrons. We’ve already seen that the faraday (F) = 9.65 x 104 C·mol-1

DG = welec = -kelec Ecell DG = welec = -nFEcell and at standard conditions DG = welec = -nFEcell

Do the units match those for work?
DG = welec = -nFEcell = -(mol)(Cmol-1)(V) = CV = J Yes! The units match those for work.

Problem Use the given electrode potential data to determine DG for the reaction 2 Al (s) + 3 Br2 (l)  2 Al3+ (aq, 1 M) + 6 Br- (aq, 1 M) Al3+ (aq) + 3 e-  Al (s) EAl3+/Al = V Br2 (l) + 2 e-  2 Br- (aq) EBr2/Br- = V Answer: Since Ecell is 2.741V and the rxn involves 6 mol of e-, then DG is kJ.

Problem The hydrogen-oxygen fuel cell is a galvanic cell with a reaction 2 H2 (g) + O2 (g)  2 H2O (l) Using the given data calculate Ecell for this reaction: DGf (H2O) = kJmol-1 DGf (H2) = 0.00 kJmol-1 DGf (O2) = 0.00 kJmol-1 Answer: Since DG is kJ and the rxn involves 4 mol of e-, then Ecell is V .

Spontaneous change in redox reactions
We’ve already related the free energy change to the cell potential DG = -nFEcell and we also know a spontaneous process MUST HAVE DG < 0 WHICH MEANS Ecell > 0 for ALL spontaneous electrochemical (oxidation-reduction) processes.

Problem When sodium metal is added to seawater, which has [Mg2+] = M, no magnesium metal is obtained. According to the data below, should this reaction occur? What reaction does occur? Na+ (aq) + 1 e-  Na (s) ENa+/Na = V Mg2+ (aq) + 2 e-  Mg (s) EMg2+/Mg = V 2 H2O (l) + 2 e-  H2 (g) + 2 OH- (aq) EH2O/H2 = V

2 Na (s) + Mg2+ (aq)  2 Na+ (aq) + Mg (s)
Problem answer For the reaction 2 Na (s) + Mg2+ (aq)  2 Na+ (aq) + Mg (s) Ecell = V and the reaction should be spontaneous. However, the reaction of sodium with water is 2 Na (s) + 2 H2O (l)  2 Na+ (aq) + H2 (g) + 2 OH- (aq) and has Ecell = V and this reaction should also be spontaneous. Since this reaction is “more spontaneous” (higher Ecell) sodium preferentially reacts with water and not magnesium ions!

Problem Without using the data for a detailed calculation, explain why Sn2+ solutions must be protected from oxygen. One way to protect them is to add metallic (solid) tin. Sn4+ (aq) + 2 e-  Sn2+ (aq) ESn4+/ Sn2+ = V Sn2+ (aq) + 2 e-  Sn (s) ESn2+/ Sn = V O2 (g) + 4 H+ (aq) + 4 e-  2 H2O (l) EO2/H2O = V

Problem answer For both possible reactions the reduction of oxygen is the cathode half-cell reaction. Since Ecell = E(red),cathode - E(red),anode, then the anode half-cell reaction that is more negative will give the higher (“more spontaneous”) Ecell reaction that will preferentially occur.

Metals and acids Some metals will react with acidic solutions to form H2 gas and metal ions in solution while others will not. We now know that those metals that do react with acid do so because the reaction is spontaneous while those that do not react do not because the reaction is non-spontaneous.

E(red),cathode = EH+/H2 = 0 V
Metals and acids In MOST cases the reduction reaction of metals in acidic solutions is 2 H+ (aq) + 2 e-  H2 (g) E(red),cathode = EH+/H2 = 0 V This is the standard hydrogen electrode half-reaction !

E(red),anode < 0 Metals and acids (0 V) - E(red),anode > 0
IF this is the preferred reduction (cathode) reaction, and since a spontaneous process must have a positive potential then for a metal to react with the H+ of an acid means Ecell = E(red),cathode - E(red),anode > 0 (0 V) - E(red),anode > 0 E(red),anode < 0

Na (ENa+/Na = -2.713 V) or Al (EAl3+/Al = -1.676 V)
Metals and acids The metals that CAN REACT with H+ are the ones with a negative Ered value like Na (ENa+/Na = V) or Al (EAl3+/Al = V) or Pb (EPb2+/Pb = V)

The metals that CAN NOT REACT with H+ have a positive Ered value like
Metals and acids The metals that CAN NOT REACT with H+ have a positive Ered value like Ag (EAg+/Ag = V) or Au (EAu3+/Au = V) or Cu (ECu2+/Cu = V)

E(red),cathode = ENO3-/NO
Metals and acids Some acids, like HNO3 have a different preferred reduction (cathode) reaction. For example NO3- (aq) + 4 H+ (aq) + 3 e-  NO (g) + 2 H2O (l) E(red),cathode = ENO3-/NO = V

E(red),anode < 0.956 V Metals and acids
IF this is the preferred reduction (cathode) reaction, and since a spontaneous process must have a positive potential then for a metal to react with the NO3- of nitric acid means Ecell = E(red),cathode - E(red),anode > 0 (0.956 V) - E(red),anode > 0 E(red),anode < V

Metals and acids In nitric acid all the metals that usually react with acids will still react, but now Ag (EAg+/Ag = V) WILL ALSO REACT!

DG = -RT ln Keq DG = -nFEcell
Ecell and Keq We have two equations relating free energy to Keq and Ecell DG = -RT ln Keq and DG = -nFEcell

-RT ln Keq = -nFEcell Ecell = (RT/nF) ln Keq Ecell and Keq
By setting the equations equal to each other we find the relationship between cell potential and the thermodynamic equilibrium constant -RT ln Keq = -nFEcell Ecell = (RT/nF) ln Keq

Ecell = (RT/nF) ln Keq Ecell and Keq
We have two constants (R and F) in this equation and we often perform reactions at K. With these three fixed values we can simplify this equation (but we DON’T HAVE TO!)

This form ONLY applies at 298.15 K!
Ecell and Keq Ecell = ( V/n) ln Keq using R = JK-1mol-1 BE CAREFUL! This form ONLY applies at K!

Everything is connected!
We COULD also put kinetics and rate constants and how they relate both to thermochemistry and equilibrium in this diagram to show ALL the possible connections in the chemistry you’ve seen!

2 Al (s) + 3 Cu2+ (1 M)  3 Cu (s) + 2 Al3+ (1 M)
Problem Should the reaction of solid Al with Cu2+ ions go to completion at 25 C if Ecell for the reaction is V? 2 Al (s) + 3 Cu2+ (1 M)  3 Cu (s) + 2 Al3+ (1 M) Answer: The reaction involves 6 moles of electrons, so Keq = e471 = very large, so the reaction goes to completion.

Problem Should the reaction of solid Sn with Pb2+ ions go to completion at 25 C? Pb2+ (aq) + 2 e-  Pb (s) EPb2+/Pb = V Sn2+ (aq) + 2 e-  Sn (s) ESn2+/Sn = V Answer: Since Ecell = V and 2 moles of electrons are involved in the process Keq = 2.5 and the reaction does not go to completion.

Ecell as a function of concentration
Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s) We’ve seen that Ecell is V for this reaction at standard conditions. However, what happens to the cell at non-standard conditions? Zn (s) | Zn2+ (aq, 0.10 M) || Cu2+ (aq, 2.0 M) | Cu (s) If we set up this cell and measure the potential, then Ecell is V.

Since the actual reaction is Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) then Le Chatalier’s Principle tells us that decreasing [Zn2+] from 1 M to 0.10 M should shift the reaction towards products and increasing [Cu2+] from 1 M to 2.0 M should shift the reaction towards products as well.

Both new concentrations serve to make the reaction more complete (or more spontaneous), and so we expect a more positive potential! Recall that DG = DG + RT ln Qeq and DG = -nFEcell

By substituting we see -nFEcell = -nFEcell + RT ln Qeq or Ecell = Ecell - (RT/nF) ln Qeq

Sometimes we prefer log to ln! Since ln x = log x we can change Ecell = Ecell - (RT/nF) ln Qeq into the Nernst Equation Ecell = Ecell (RT/nF) log Qeq

For this reaction Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) Qeq = aZn2+ / aCu2+ and so Ecell = Ecell (RT/nF) log{aZn2+/aCu2+}

If we plot Ecell versus log Qeq we should get a straight line with a slope of [- ( RT/nF)] and y-intercept of Ecell

Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) Qeq = aZn2+ / aCu2+

Ecell = Ecell - ( RT/nF) log Qeq Since R ( JK-1mol-1) and F (9.65 x 104 C·mol-1) are constants, and if we choose the temperature to be K, then we can replace these three fixed values as we have done before (slide 93)

It makes the most sense to memorize the Nernst
Nernst equation at K Ecell = Ecell – ( V/n) log Qeq It makes the most sense to memorize the Nernst Equation and substitute rather than remembering this form for one temperature!

Problem Calculate Ecell for the for the following galvanic cells at K. Will the reactions be spontaneous? Al (s) | Al3+ (0.36 M) || Sn4+ (0.086 M), Sn2+ (0.54 M) | Pt(s) Pt(s) | Cl2 (1 atm) | Cl- (1.0 M) || Pb2+ (0.050 M), H+ (0.10 M) | PbO2(s)

PbO2 (s) + 4 H+ (aq) + 2 e-  Pb2+ (aq) + 2 H2O (l)
Problem data Sn4+ (aq) + 2 e-  Sn2+ (aq) ESn4+/Sn2+ = V Al3+ (aq) + 3 e-  Al (s) EAl3+/Al = V PbO2 (s) + 4 H+ (aq) + 2 e-  Pb2+ (aq) + 2 H2O (l) EPbO2/Pb2+ = V Cl2 (g) + 2 e-  2 Cl- (aq) ECl2/Cl- = V

Problem answer 3 Sn4+ (0.086 M) + 2 Al (s) 
3 Sn2+ (0.54 M) + 2 Al3+ (0.36 M) Ecell = V and Ecell = V PbO2 (s) + 4 H+ (0.10 M) + Cl2 (1 atm)  Pb2+ (0.050 M) + 2 H2O (l) + 2 Cl- (1.0 M) Ecell = V and Ecell = V Since both Ecell values are positive, both reactions will be spontaneous at the given conditions.

Sn (s) | Sn2+ (aq) || Pb2+ (aq) | Pb (s)
Problem For what ratio of [Sn2+] / [Pb2+] will the given cell reaction NOT be spontaneous in either direction? Sn (s) | Sn2+ (aq) || Pb2+ (aq) | Pb (s) Pb2+ (aq) + 2 e-  Pb (s) EPb2+/Pb = V Sn2+ (aq) + 2 e-  Sn (s) ESn2+/Sn = V

Qeq = [Sn2+] / [Pb2+] = Keq = 2.5
Problem answer The reaction is NOT spontaneous in either direction ONLY at equilibrium, where Ecell = 0. Since for this cell Ecell = V the equilibrium occurs when Qeq = [Sn2+] / [Pb2+] = Keq = 2.5 (see slide 96)

The mixing is a spontaneous process!
Concentration cells We know if we mix two solutions of the same chemical but with different concentrations, then the final solution will have a single uniform concentration. The mixing is a spontaneous process!

We can set up the mixing process as an electrochemical cell!
Concentration cells We can set up the mixing process as an electrochemical cell! The different concentrations in the two half-cells will lead to a non-zero Ecell

We can set up the mixing process as an electrochemical cell!
Concentration cells We can set up the mixing process as an electrochemical cell! The different concentrations in the two half-cells will lead to a Ecell different from Ecell

Concentration cells for H+
Pt (s) | H2 (1 atm) | H+ (x M) || H+ (1 M) | H2 (1 atm) | Pt (s) H2 (g, 1 atm)  2 H+ (x M) + 2 e- 2 H+ (1 M) + 2 e-  H2 (g, 1 atm) Net reaction: 2 H+ (1 M)  2 H+ (x M) Ecell will ALWAYS be zero for a concentration cell…

Net reaction: 2 H+ (1 M)  2 H+ (x M) Ecell = Ecell (RT/nF) log Qeq Ecell = (RT/nF) log x2/12 Ecell = (2RT/nF) (-log x)

Ecell = (2RT/nF) (-log x) if x = [H+] then (-log x) = pH At K we can replace our three fixed values R, F, and T to give Ecell = V (pH) This is the basis for electronic pH meters!

Concentration cells for finding Ksp
Ag (s) | Ag+ (sat’d AgI) || Ag+ (0.100 M) | Ag (s) Ag (s)  2 Ag+ (sat’d AgI) + e- Ag+ (0.100 M) + e-  Ag (s) Net reaction: Ag+ (0.100 M)  Ag+ (sat’d AgI) Ecell will ALWAYS be zero for a concentration cell…

Net reaction: Ag+ (0.100 M)  Ag+ (sat’d AgI) We measure Ecell for this concentration cell and we find it to be V Since Q = [Ag+] / [Ag+] and Ecell = Ecell – ( V/n) log Qeq and n = 1

Net reaction: Ag+ (0.100 M)  Ag+ (satd AgI) Ecell = Ecell - ( V) log Qeq 0.417 V = 0 - ( V) log Qeq log Qeq = V / ( V) log Qeq = Qeq = [Ag+]/(0.100 M) = 9.04 x 10-8 So [Ag+] = 9.04 x 10-9 M

Since the [Ag+] came from a saturated AgI solution, then [Ag+] = [I-] = 9.04 x 10-9 M and Ksp = [Ag+] [I-] Ksp = (9.04 x 10-9) (9.04 x 10-9) Ksp = 8.3 x 10-17

Problem Answer: Ecell = 0.23 V
If Ksp = 1.8 x for silver chloride then what would be Ecell for Ag (s) | Ag+ (sat’d AgCl) || Ag+ (0.100 M) | Ag (s) Answer: Ecell = 0.23 V

Problem Ecell = 0.0567 V Answer: Ksp = 7.1 x 10-9
Calculate the Ksp for lead iodide with the given concentration cell information Pb (s) | Pb2+ (sat’d PbI2) || Pb2+ (0.100 M) | Pb (s) Ecell = V Answer: Ksp = 7.1 x 10-9

Electrolysis and electrolytic cells
The reverse of every spontaneous chemical reaction is non-spontaneous. If we apply electric current to a chemical system, it is possible to force non-spontaneous chemical reactions occur in a process called electrolysis, in what we call electrolytic cells.

Electrolysis and electrolytic cells
The potential we apply to the electrolytic cell must be greater than that for the spontaneous reaction, and must be applied in the opposite direction. Ebattery > -Ecell

Since we actually see this reaction occurring, this reaction must be spontaneous! The reverse reaction, where we put copper metal into a Zn2+ ion solution is non-spontaneous!

Reduction! Oxidation! Cathode is –ve! Anode is +ve! If we want to get zinc from this cell, we must force the non-spontaneous reaction to occur by applying a potential in the direction opposite that for the spontaneous process! Reduction ALWAYS occurs at the cathode!

Electrolysis as coupled reactions
When we are doing electrolysis we are using the spontaneous battery reaction to drive the non-spontaneous electrolysis reaction. battery reactants  battery products Ebattery > 0 so DG < 0 electrolysis reactants  electrolysis products Ecell < 0 so DG > 0

Electrolysis as coupled reactions
In our setup we are coupling (adding) the reactions which means we add their potentials (or free energies). battery reactants + electrolysis reactants  electrolysis products + battery products Esum = Ebattery + Ecell > 0 so DGsum < 0

Complicating factors in electrolysis
While adding the potentials to get the potential for the coupled reaction is straightforward in theory, in practice there are complicating factors!

Overpotentials The electrolytic cell has electron transfers occurring at the surface of the electrodes. If solutions are involved then there is generally a good contact to the electrode. However, if gases are contacting the electrode the contact is problematic. As the contact to the electrode gets worse we often need to apply an overpotential (extra Eoverpotential) to make up for this problem.

Ebattery > -(Ecell + Eoverpotential)
Overpotentials Ebattery > -(Ecell + Eoverpotential) For example, a solid platinum electrode generally has a near-zero volt Eoverpotential while the formation of H2 gas on the surface of a mercury cathode has Eoverpotential  1.5 V

Esum = Ebattery + Ecell > 0
Competing reactions If we set up an electrolytic cell expecting Esum = Ebattery + Ecell > 0 will give us the reaction we want we may be surprised when we get a completely different reaction because Ebattery + Eother > Ebattery + Ecell

Ebattery + Eother > Ebattery + Ecell
Competing reactions Ebattery + Eother > Ebattery + Ecell We saw in slides that if we have competing reactions, then the one that is “more spontaneous” will preferentially occur.

Competing reactions Often, but not always, when we do electrolysis in aqueous solutions we get the competing reactions 2 H2O (l) + 2e-  H2 (g) + 2 OH- (aq) at the cathode and 2 H2O (l) O2 (g) + 4 H+ (aq) + 4 e- at the anode.

When we do electrolysis in aqueous solution
Competing reactions When we do electrolysis in aqueous solution we must identify which of the two possible reduction reactions is “more spontaneous” when forced and we must identify which of the two possible oxidation reactions is “more spontaneous” when forced. See pages and Example Problem in the text for more info on this very important topic

Non-standard conditions
Industrially we try to maximize product with minimum energy and money input. This often means that we do electrolysis on cells at non-standard conditions, which means Ecell  Ecell

Electrodes Platinum is an inert electrode that only provides a surface for the true reactants to transfer electrons. An active electrode is an actual reactant in the half-cell reaction. Using a different electrode on one side of the electrolytic cell might change the half-cell reaction on that side!

Quantitative aspects of electrolysis

If we pass 1 mole of electrons through the cell, from the balanced equation Na+ (l) + e-  Na (s) we see we will get 1 mole (23.0 g) of solid sodium out. At the other electrode, where 2 Cl- (l)  Cl2 (g) + 2 e- we see that one mole of electrons is enough to give us one-half a mole (35.5 g) of Cl2.

How many electrons pass through the cell depends on the current, which is charge per unit time, (the ampere A, which is a C/s) and the time the current was allowed to pass though the cell… Charge (C) = Current (C/s) x time (s) Charge (C) = Current (A) x time (s)

We saw earlier that one mole of electrons has a charge equal to one Faraday 1 F = 9.65 x 104 Cmol-1 moles of e- = Charge (C) / Faraday moles of e- = (Current x time) 9.65 x 104 Cmol-1

The flowchart shows how to find the amount of substance that comes from electrolysis based on a known current and time. If we want to know the current or time we used to get a certain amount of substance, we reverse the order of the flowchart.

Molar mass of Al is 26.9815 gmol-1
Problem How many kilograms of aluminum can be produced in 8.00 h by passing a constant current of 1.00 x 105 A for an electrolytic cell with the following half reaction at the cathode? Al e-  Al Molar mass of Al is gmol-1 Answer: 268 kg

Molar mass of silver is 107.868 gmol-1
Problem A layer of silver is electroplated (an electrolytic process) on a coffee server using a constant current of A. How much time is required to deposit 3.00 g of silver? Molar mass of silver is gmol-1 Answer: 7.45 hours

Chapter 20 Electrochemistry

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