Unit 5:Unit 5:Everything You Wanted to Know Everything You Wanted to Know About Electrochemical Cells, But About Electrochemical Cells, But
Were Too Afraid to AskWere Too Afraid to Ask
By : Michael By : Michael
““Chuy el Chulo” BilowChuy el Chulo” Bilow
And “H”Elliot PinkusAnd “H”Elliot Pinkus
Redox ReactionsRedox Reactions
A Redox Reaction features the transfer of A Redox Reaction features the transfer of electrons between ions.electrons between ions.
X + Y X + Y X Xn+n+ + Y + Yn-n-
Oxidation Half-Reaction: X Oxidation Half-Reaction: X X Xn+n+ + e + e--
Reduction Half-Reaction: Y + eReduction Half-Reaction: Y + e-- Y Yn-n-
Redox ReactionsRedox Reactions
Determine Oxidation NumbersDetermine Oxidation NumbersAtoms in a pure element have oxidation number of zeroAtoms in a pure element have oxidation number of zero
A monatomic ion has oxidation number equal to its A monatomic ion has oxidation number equal to its charge.charge.
Sum of oxidation numbers equals overall charge of Sum of oxidation numbers equals overall charge of compound.compound.
Fluorine is always –1 with other elementsFluorine is always –1 with other elements
H is +1 and O is –2 in most compoundsH is +1 and O is –2 in most compounds
Cl, Br, and I are –1 except with Oxygen or Fluorine.Cl, Br, and I are –1 except with Oxygen or Fluorine.
Redox ReactionsRedox Reactions
HH22SOSO4 4
– 1 H = +11 H = +1– 1 O = -21 O = -2– -8 + 2 + S = 0-8 + 2 + S = 0– 1 S = +61 S = +6
CrCr22OO772-2-
– 1 O = -21 O = -2– -14 + 2(Cr) = -2-14 + 2(Cr) = -2– 1 Cr = +61 Cr = +6
Balancing Redox EquationsBalancing Redox Equations
Balance: CrBalance: Cr22OO772- 2- + Cl+ Cl-- Cr Cr3+3+ + Cl + Cl22
Split Equation into half-reactionsSplit Equation into half-reactions– CrCr22OO77
2- 2- 2Cr 2Cr3+3+
– ClCl-- Cl Cl22
Add HAdd H++, then H, then H22O, then eO, then e- - to balance.to balance.– 6e6e-- + 14H + 14H++ + Cr + Cr22OO77
2Cr 2Cr3+3+ + 7H + 7H22OO– 2Cl2Cl-- Cl Cl22 + 2e + 2e--
Combine into overall reactionCombine into overall reaction– 6Cl6Cl- - + 14H+ 14H++ + Cr + Cr22OO77
2Cr 2Cr3+3+ + 7H + 7H22O + 3ClO + 3Cl22
Balancing Redox EquationsBalancing Redox Equations
To balance in a BASIC solution:To balance in a BASIC solution:
Take final answer for Acidic Solution:Take final answer for Acidic Solution:– 6Cl6Cl- - + 14H+ 14H++ + Cr + Cr22OO77
2- 2- 2Cr 2Cr3+3+ + 7H + 7H22O + 3ClO + 3Cl22
Add OHAdd OH- - to cancel Hto cancel H++ and add H and add H22OO
– 6Cl6Cl- - + 7H+ 7H22O + CrO + Cr22OO772- 2- 2Cr 2Cr3+3+ + 3Cl + 3Cl2 2 +14OH+14OH--
What are Electrochemical (EC) Cells?What are Electrochemical (EC) Cells?
An Electrochemical Cell converts chemical An Electrochemical Cell converts chemical energy into electrical energy by reducing energy into electrical energy by reducing one substance and oxidizing another.one substance and oxidizing another.
For example:For example:
Cu+FCu+F22CuCu2+2++2F+2F--
The copper is oxidized and the fluorine is The copper is oxidized and the fluorine is reduced because of a transfer of reduced because of a transfer of electrons, thus creating a current.electrons, thus creating a current.
What are EC Cells?What are EC Cells?
There are two types of EC cells:There are two types of EC cells:
Galvanic cells spontaneously produce Galvanic cells spontaneously produce energyenergy
Electrolytic cells must have work done on Electrolytic cells must have work done on them to go to completion, and are thus them to go to completion, and are thus nonspontaneousnonspontaneous
Electrolytic and Galvanic CellsElectrolytic and Galvanic Cells
In both electrolytic and galvanic cells, In both electrolytic and galvanic cells, oxidation takes place at the anode and oxidation takes place at the anode and reduction takes place at the cathodereduction takes place at the cathode
But, galvanic cells have positively charged But, galvanic cells have positively charged cathodes and negatively charged anodescathodes and negatively charged anodes
And electrolytic cells have negative And electrolytic cells have negative cathodes and positive anodescathodes and positive anodes
WHY?WHY?
Because reduction is forced in electrolytic Because reduction is forced in electrolytic cells, electrons collect there, giving a cells, electrons collect there, giving a negative charge.negative charge.
And because the oxidation is not favored, And because the oxidation is not favored, the anode develops a positive chargethe anode develops a positive charge
How do You Make a Galvanic Cell?How do You Make a Galvanic Cell?
Many EC cells are made with two metals Many EC cells are made with two metals in a solution of one of their sulfate or in a solution of one of their sulfate or nitratenitrateThe two metal bars are connected by a The two metal bars are connected by a salt bridge. The salt bridge allows anions salt bridge. The salt bridge allows anions to pass through to the oxidized side to to pass through to the oxidized side to restore chargerestore chargeFor example, take zinc and copper in For example, take zinc and copper in solutions of CuSOsolutions of CuSO44 and ZnSO and ZnSO44..
How a Galvanic Cell is made
In this reaction, Zn(s) would be oxidized to Zn2+
(aq) and Cu2+
(aq) would be reduced to Cu(s)
The zinc-copper galvanic cell would look like this:
http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/electrochem.html
How do Galvanic How do Galvanic Cells Produce Electricity?Cells Produce Electricity?
The electron flow from cathode to anode The electron flow from cathode to anode produces a current, and thus electricity.produces a current, and thus electricity.
Over time, the Zn anode will deteriorate as Over time, the Zn anode will deteriorate as it is oxidized to Znit is oxidized to Zn2+2+, and Cu, and Cu2+2+ ions will be ions will be reduced to Cu and leave the solution, reduced to Cu and leave the solution, plating the Cu cathodeplating the Cu cathode
How are Electrolytic Cells Made?
There are many ways to make electrolytic cells, but all require an outside source of energy to force the reaction towards the products
This shows the electrolysis of NaCl(l) to Na(l) and Cl2(g)
How Much Electricity?How Much Electricity?
Cell PotentialsCell Potentials
To find how much electricity is produced or To find how much electricity is produced or needed, you must use the oxidation and needed, you must use the oxidation and reduction potentials of each of the half-reactions reduction potentials of each of the half-reactions that take place in the system. that take place in the system. Reduction Potentials show how much energy is Reduction Potentials show how much energy is either released or needed to cause a reduction either released or needed to cause a reduction half-reaction to occurhalf-reaction to occurSince oxidation is the opposite of reduction, Since oxidation is the opposite of reduction, reduction potentials are the opposite of oxidation reduction potentials are the opposite of oxidation potentials.potentials.
Cell PotentialsCell Potentials
To find a cell’s potential difference To find a cell’s potential difference (voltage), first find its standard oxidation (voltage), first find its standard oxidation and reduction potentials of its half-and reduction potentials of its half-reactions, usually listed as Ereactions, usually listed as Eoo..Then, subtract the standard reduction Then, subtract the standard reduction potential for the oxidized species from the potential for the oxidized species from the standard reduction potential of the standard reduction potential of the reduced species to get:reduced species to get:
EEoocellcell=E=Eoo
redred-E-Eoooxox
Cell PotentialsCell Potentials
Let’s go back to the zinc-copper cellLet’s go back to the zinc-copper cell
Make two half-reactions: ZnMake two half-reactions: Zn(s)(s) Zn Zn2+2+(aq)(aq)
+2e+2e--
And CuAnd Cu2+2+(aq)(aq)+2e+2e-- Cu Cu(s)(s)
The reduction potential for the copper (II) The reduction potential for the copper (II) ion to copper metal is +0.34 Vion to copper metal is +0.34 V
The reduction potential for the zinc (II) ion The reduction potential for the zinc (II) ion to zinc metal is -0.76 Vto zinc metal is -0.76 V
Cell PotentialsCell Potentials
So, ESo, Eoocellcell=E=Eoo
CuCu2+2+-E-Eoo
ZnZn
Or EOr Eoocellcell= .34 V- (-.76)V= .34 V- (-.76)V
Therefore ETherefore Eoocellcell=1.10 V=1.10 V
Remember that oxidation and reduction Remember that oxidation and reduction potentials change, and most are only listed potentials change, and most are only listed for 1for 1MM concentrations of electrolytes at concentrations of electrolytes at 2525ooC and 1 atm of pressure. Changes to C and 1 atm of pressure. Changes to this will result in changes in potentials.this will result in changes in potentials.
Cell PotentialsCell Potentials
Determine the spontaneous cell reaction Determine the spontaneous cell reaction and the cell potential of a cell that has and the cell potential of a cell that has these two half reactionsthese two half reactions
AlAl3+3++3e+3e--Al(s) EAl(s) EooAlAl
3+3+=-1.66V=-1.66V
CuCu2+2++2e+2e--Cu(s) ECu(s) EooCuCu
2+2+=0.34V=0.34V
First determine which species is to be First determine which species is to be oxidized and which to be reducedoxidized and which to be reducedThe oxidized substance in a spontaneous The oxidized substance in a spontaneous cell will always have the lesser potentialcell will always have the lesser potential
Cell PotentialsCell Potentials
Remember to reverse the equation of the Remember to reverse the equation of the oxidized species and balance the total oxidized species and balance the total ionic equations so that no electrons are ionic equations so that no electrons are left over.left over.
3Cu3Cu2+ 2+ +2Al+2Al2Al2Al3+3++3Cu+3Cu
Now, find the ENow, find the Eoocellcell
EEoocellcell=E=Eoo
redred-E-Eoooxox
EEoocellcell=E=Eoo
CuCu2+2+-E-Eoo
AlAl3+3+
Cell PotentialsCell Potentials
EEoocellcell=.34-(-1.66)=.34-(-1.66)
EEoocellcell= 2.00 V= 2.00 V
Note that the reduction potentials are not Note that the reduction potentials are not multiplied by the coefficients in the multiplied by the coefficients in the equation.equation.
What else?What else?
The SI unit of electric current is the The SI unit of electric current is the ampere (A) and the SI unit of charge is the ampere (A) and the SI unit of charge is the coulomb (C).coulomb (C).
1A= 1 coulomb per second1A= 1 coulomb per second
It has been determined that the charge of It has been determined that the charge of one mole of electrons is 9.65x10one mole of electrons is 9.65x1044 C, which C, which is referred to as Faraday’s constant and is referred to as Faraday’s constant and symbolized Fsymbolized F
Faraday’s ConstantFaraday’s Constant
From this, we can determine how much From this, we can determine how much anode material is used up or how much is anode material is used up or how much is produced at the cathodeproduced at the cathode
For example, how many grams of copper For example, how many grams of copper will be deposited on the cathode of an will be deposited on the cathode of an electrolytic cell if a current of 4.00 A is run electrolytic cell if a current of 4.00 A is run through a solution of CuSOthrough a solution of CuSO44 for 10.0 min? for 10.0 min?
Faraday’s ConstantFaraday’s Constant
First, convert the minutes to seconds to First, convert the minutes to seconds to coulombscoulombs
10.0 min*60.0sec*min10.0 min*60.0sec*min-1-1*4.00A=2.40x10*4.00A=2.40x1033 C C
Then coulombs to moles of electronsThen coulombs to moles of electrons
2.40x102.40x1033 C*1mol e C*1mol e--/9.65x10/9.65x1044 C=.0249 mol e C=.0249 mol e--
To grams of copper. Remember that it takes 2 To grams of copper. Remember that it takes 2 mol emol e-- to reduce 1 mol Cu to reduce 1 mol Cu2+2+
.0249 mol e.0249 mol e--*63.55 g Cu/2 mol e*63.55 g Cu/2 mol e--=.791 g Cu=.791 g Cu
Gibbs Free EnergyGibbs Free Energy
The maximum amount of work that can be The maximum amount of work that can be done is the opposite of done is the opposite of ΔΔG, the change in G, the change in Gibbs Free EnergyGibbs Free Energy
Since 1 Volt= 1 joule/1coulomb, and the Since 1 Volt= 1 joule/1coulomb, and the joule is the SI unit of work, we getjoule is the SI unit of work, we get
ΔΔG=-nFEG=-nFEcellcell
Where n is the moles of electrons Where n is the moles of electrons transferred and F is Faraday’s constanttransferred and F is Faraday’s constant
Equilibrium Constants Equilibrium Constants and Cell Potentialsand Cell Potentials
To find the equilibrium constant of an To find the equilibrium constant of an equation from its equation from its EEoo
cellcell, the equation is:, the equation is:
EEoocellcell=RTln(K=RTln(Kcc)/nF)/nF
Where R=8.314 J/molK, T is temperature Where R=8.314 J/molK, T is temperature in Kelvin, ln(Kin Kelvin, ln(Kcc) is the natural logarithm ) is the natural logarithm
(log base e) of the equilibrium constant, n (log base e) of the equilibrium constant, n is the number of moles of electrons is the number of moles of electrons transferred, and F is faraday’s constant.transferred, and F is faraday’s constant.
The Nernst EquationThe Nernst Equation
The Nernst Equation relates the calculated The Nernst Equation relates the calculated potential of a cell to its potential at a certain potential of a cell to its potential at a certain time.time.
EEcellcell=E=Eoocellcell-RT/nFln(Q)-RT/nFln(Q)
Where R,T, n, and F are the same as above Where R,T, n, and F are the same as above and Q is the mass-action constant of the and Q is the mass-action constant of the equation, which equals the concentrations equation, which equals the concentrations of products that can change concentration of products that can change concentration to their coefficient’s power, divided by to their coefficient’s power, divided by reactants that act similarly.reactants that act similarly.
The EndThe End
