Unit 4: Electrochemistry · Introduction Electrochemistry involves the study of electron movement...

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Unit 4: Electrochemistry

Chemistry 3202

Introduction● Electrochemistry involves the study of

electron movement or electric current in chemical reactions.

● Electrochemical reactions involve the transfer of electrons between reactants.

● Some of these reactions produce electricity for our use (batteries)

● Others need electricity to occur (electrolysis of water)

● Many reactions happen w/o any electricity produced or required (Na + 1 Cl ---> NaCl) 2

● The transfer of electrons may be illustrated using simple Bohr models to show formation of Na+ from Na and Cl- from Cl and the resulting electron loss and gain.

oxidation

reduction

oxidizing agent

reducing agent

half-reaction

oxidation number

redox reaction

electrochemical cell

electrolytic cell

electrolysis

electroplating

fuel cell

pyrometallurgy

hydrometallurgy

Your 1st electrochemical reaction

Ag(s) + CuCl(aq) → Cu(s) + AgCl(aq)

Which species loses an electron?

Which species gains an electron?

Which species has no electron change?

Your 1st electrochemical reaction

Ag(s) + CuCl(aq) → Cu(s) + AgCl(aq)

Which species loses an electron?

Which species gains an electron?

Which species has no electron change?

Oxidation and Reduction

● OXIDATION occurs when an apple turns brown or steel rusts upon exposure to air (reaction with oxygen)

● the Loss of Electrons in a reaction is Oxidation (LEO)

● a substance is oxidized when it loses electrons

REDUCTION ● REDUCTION occurs when hydrogen is

gained ● the Gain of Electrons in a reaction is

Reduction (GER)● a substance is reduced when it gains

electrons

Zinc metal is added to a solution of copper (II) sulfate.

Zn(s) + CuSO4(aq) ----> Cu(s) + ZnSO

Copper metal forms on the surface of the zinc, zinc metal disappears, andthe blue color of the Cu2+ ions fades.

● What happened to the Cu2+ ions?● They become Cu metal

● What is the equation for this?● Cu2+---> Cu(s)

● How does this occur?● Cu2+ gains two electrons

● Is the equation balance?● Nope, it needs two electrons to become balanced

● Cu2+ + 2e- ---> Cu● This is one half reaction

● Where do these electrons come from?● The Zn as it changes to Zn2+ in solution

● What is the second half reaction?● Zn ---> Zn2+ + 2e-

GERLEO

Oxidation

Reduction

● Oxidation-Reduction reactions are usually referred to as redox reactions.

Identify the species being oxidized and the species being reduced below:

Fe(NO3)3 + Al → Al(NO3)3 + Fe

Pb + SnCl2 → Sn + PbCl2

Zn + CuSO4 ----> Cu + ZnSO

Identify the species being oxidized and the species being reduced below:

Fe(NO3)3 + Al → Al(NO3)3 + Fe

Pb + SnCl2 → Sn + PbCl2

● Al is being oxidized since it is losing electrons to become Al3+

● Fe3+ is being reduced since it is gaining electrons to become Fe● The electrons it is gaining are the electrons that Al is

losing

● Pb is being oxidized since it is losing electrons to become Pb2+

● Sn2+ is being reduced since it is gaining electrons to become Sn● The electrons it is gaining are the electrons that Pb is

losing

● Zn is being oxidized and Cu2+ is being reduced

Why is this NOT an example of an electrochemical reaction?

CaCl2 + 2 AgNO3 → Ca(NO3)2 + 2 AgCl

● The reason is because there is no transfer of electrons between reactants

● There is no change in charge on anything

Net Ionic Equations (NIE)● Net Ionic Equations show only the species that

change during a chemical reaction.● Ions not directly involved in the chemical

change are called spectator ions.

eg. Write the net ionic equation for the reaction of Cu(s) metal in AgNO3(aq)

Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

Net Ionic Equations (NIE)

+ 2 Ag+(aq) + 2 NO3

-(aq)Cu(s) → Cu2+

(aq) + 2 NO3-(aq) + 2 Ag(s)

Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + 2 Ag(s)

NET IONIC EQUATION:Cu(s) + 2 Ag+

(aq) → Cu2+(aq) + 2 Ag(s)

● Cu loses 2 electrons – Cu is oxidized● Ag+ gains 1 electron – Ag+ is reduced● The spectator ion (ion not changed by this

reaction) is NO3-(aq)

eg. Write the NET IONIC EQUATION for the reaction between Cl2(g) and NaI(aq). Identify the species being oxidized and the species being reduced.

Cl2(g) + 2NaI(aq) → 2NaCl(aq) + I2(s)

Oxidizing and Reducing Agents● the substance that is oxidized in a redox

reaction is called the reducing agent (RA).

● the substance that is reduced in a redox reaction is called the oxidizing agent (OA).

Oxidizing and Reducing Agentseg. Identify the OA and the RA in the

following redox reactions

Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + 2 Ag(s)

Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s)

Oxidizing and Reducing Agentseg. Identify the OA and the RA in the

following redox reactions

Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + 2 Ag(s)

Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s)

Half Reactions

● A redox reaction can be separated into two parts called half-reactions.

● The oxidation half-reaction shows the loss of electrons.

● The reduction half-reaction shows the gain of electrons.

Half Reactionseg. Write the oxidation and reduction half

reactions for

Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + 2 Ag(s)

oxidation: Cu(s) →Cu2+(aq) + 2e-

reduction: 2 Ag+(aq) + 2e- → 2 Ag(s)

Half ReactionsWrite the half-reactions for:Sn4+

(aq) + Pb2+(aq) → Sn2+

(aq) + Pb4+(aq)

Reduction: Sn4+(aq) + 2e-

(aq) → Sn2+(aq)

Oxidation: Pb2+(aq) → Pb4+

(aq) + 2e-

F2(g) + 2 Br -(aq) → 2 F-

(aq) + Br2(s)

Reduction: F2(g) + 2 e -(aq) → 2 F-

Oxidation: 2 Br -(aq) → Br2(s) + 2e-

Half Reactions

Write the overall equation for:Al3+ + 3e- →Al(s)

Mg(s) → Mg2+(aq) + 2e-

Half Reactions

Write the overall equation for:Al3+ + 3e- →Al(s)

Mg(s) → Mg2+(aq) + 2e-

● A redox reaction is the combination of a reduction half reaction and an oxidation half reaction.

● Remember that oxidation, reduction, oxidizing agent and reducing agent only apply to reactants

● For you to do:

- p. 715 #’s 2-4- p. 716 #’s 7- Electrochemistry #1: #’s 1 - 3

Oxidation Numbers● The oxidation number is the actual or

hypothetical charge on an element or ion using an assigned set of rules.

species Cl- ion Mg2+ ionMg

atomC in CH4

oxidation # -1 +2 0 -4

Determining Oxidation Numbers

(See p. 724)

1. Pure elements → oxidation # = 0

eg. Cl2(g), Mg(s), P4(s), Fe(s)

2. Simple ions → oxidation # = ion charge

eg. Cl- → oxidation # = -1

Mg 2+ → oxidation # = +2

3. Hydrogen has an oxidation number of +1 except in metal hydrides.

eg. H is +1 in NH3, H2O, and HCl

H is -1 in NaH and CaH2

4. Oxygen has an oxidation number of -2 except in: - peroxides such as H2O2

- and in OF2.

5. If H or O are not in a covalent compound, the atom with the highest electronegativity has the oxidation number that matches its charge.

eg. Determine the oxidation # on nitrogen in: a) NF3 b) NI3

6. The sum of all oxidation numbers in a compound is ZERO.

eg. CH4

H → +1

C must be ???

C + 4(+1) = 0 C = -4

7. The sum of all oxidation numbers in a polyatomic ion equals the ion charge.

eg. CO32-

O → -2

C + 3(-2) = -2

C = +4

Extra Practice (oxidation #’s)

1. Determine the oxidation number of:a) S in SO2 d) Cr in Cr2O7

b) Cl in HClO4 e) I in MgI2

c) S in SO42-

2. What is the oxidation number of nitrogen in each of the following?N2O NO NO2 NH3

N2H4 NaNO3 N2 NH4Cl38

Identifying Redox Reactions

● Redox reactions can be identified by looking at the oxidation numbers of the species involved.

● Changes in oxidation numbers indicate a reaction is a redox reaction.

● An increase in oxidation number indicates oxidation

(ie. the reducing agent)● A decrease in oxidation number indicates a

reduction

(ie. the oxidizing agent)

eg. Use oxidation numbers to show why this is a redox reaction:

CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)

-4 -20 -1

eg. Using oxidation numbers, explain why this is NOT a redox reaction:

CaCO3(s) → CaO(s) + CO2(g)

Extra Practice (identifying redox)

1. Determine whether the following are redox reactions. For the redox reactions, identify the OA and the RA.

a) Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

b) Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)

c) Br2(l) + 2 NaI(aq) → 2 NaBr(aq) + I2(s)

Extra Practice (identifying redox)

d) 2 NaCl(l) → 2 Na(l) + Cl2(g)

e) HCl(aq) + NaOH(aq) → HOH(l) + NaCl(aq)

f) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

g) 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

Work on the rest of worksheet 1 si vous plait!

Balancing Redox Equations

Balancing Redox Equations● Redox equations are written by first balancing the

½ reactions and then combining the balanced ½ reactions.

● Three ways to balance redox equations:– reactions in acidic solution– reactions in basic solution– standard reduction potentials table

Reactions in Acidic Solution (p. 732)● balance each ½ reaction using the 5 rules from

p. 732

● multiply the ½ reactions to cancel electrons

● combine the ½ reactions

CHARGE AND # OF ATOMS MUST BE BALANCED

Reactions in Acidic SolutionSteps to balance:● Write the half reactions (no electrons)● Balance all atoms except oxygen & hydrogen● Balance oxygen by adding H

2O to the side that

needs it● Balance hydrogen by adding H

2 to the side that

needs it● Balance charge by adding electrons● Make electrons gained equal electrons lost● Add the half reactions to get overall equation50

Reactions in Acidic Solution

eg. Balance each ½ reaction in acidic solution.

a) MnO4-(aq) → Mn2+

b) HNO2(aq) → N2O(g)

MnO4-(aq) → Mn2+

(aq) + 4 H2O(l)

8 H+(aq) + MnO4

-(aq) → Mn2+

(aq) + 4 H2O(l)

5 e- + 8 H+(aq) + MnO4

-(aq) → Mn2+

(aq) + 4 H2O(l)

HNO2(aq) → N2O(g)

2 HNO2(aq) → N2O(g)

2 HNO2(aq) → N2O(g) + 3 H2O(l)

4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l)

4 e- + 4 H+(aq) + 2 HNO2(aq) → N2O(g) + 3 H2O(l)

(Reduction: OA is HNO2(aq) )

p. 732 #’s 19, 20

eg. Balance the following redox equation in acidic solution.

IO3-(aq) + NO2(g) → NO3

-(aq) + I2(s)

Solution:

- write and balance each half reaction

IO3-(aq) → I2(s) NO2(g) → NO3

IO3-(aq) → I2(s)

2 IO3-(aq) → I2(s)

2 IO3-(aq) → I2(s) + 6 H2O(l)

12 H+(aq) + 2 IO3

-(aq) → I2(s) + 6 H2O(l)

10 e- + 12 H+(aq) + 2 IO3

-(aq) → I2(s) + 6 H2O(l)

(Reduction: OA is IO3- )

NO2(g) → NO3-(aq)

H2O(l) + NO2(g) → NO3-(aq)

H2O(l) + NO2(g) → NO3-(aq) + 2 H+

(aq) + 1 e-

(Oxidation: RA is NO2(g) )

Oxidation:

(l) + NO

2 (g) → NO

(aq) + 2H+

(aq) + 1e-

Reduction:

10e- + 12 H+(aq) + 2 IO3

-(aq) → I2(s) + 6 H2O(l)

(l) + 10NO

2 (g) → 10NO

(aq) + 20H+

(aq) + 10e-

Overall:

(l) + 10NO

2 (g) + 2 IO3

-(aq) → I2(s) + 10NO

(aq) + 8H+

Extra Practice (balance in acidic conditions)

CH3OH(l) + MnO4-(aq) → CH2O(l) + Mn2+

CH3OH(l) → CH2O(l)

MnO4-(aq) → Mn2+

+ 2 e-+ 2 H+(aq)

+ 4 H2O(l)8 H+(aq)5 e- +

5 CH3OH(l) → 5 CH2O(l) + 10 H+(aq) + 10 e-

10 e- + 16 H+(aq) + 2 MnO4

-(aq) → 2 Mn2+

(aq) + 8 H2O(l)

CH3OH(l) + H+(aq) + MnO4

-(aq) → CH2O(l) + Mn2+

(aq) + H2O(l)

p. 739 # 27

Balance the following redox equation in acidic solution.

N2O(g) + SO42-

(aq) → HNO2(aq) + H2SO3(aq)

N2O(g) → HNO2(aq)

3 H2O(l) + N2O(g) → 2 HNO2(aq) + 4 H+(aq) + 4 e-

(aq) → H2SO3(aq)

(aq) + 4 H+(aq) + 2 e- → H2SO3(aq) + H2O(l)

Reactions in Basic Solution

● balance each ½ reaction using the 7 rules from p. 733

● multiply the ½ reactions to cancel electrons● combine the ½ reactions

CHARGE AND # OF ATOMS MUST BE BALANCED

eg. Balance this ½ reaction in basic solution

ClO-(aq) → Cl-

ClO- → Cl- + H2O

2 H+ + ClO- → Cl- + H2O

2 OH- + 2 H+ + ClO- → Cl- + H2O + 2 OH-

2 H2O + ClO- → Cl- + H2O + 2 OH-

1 H2O + ClO- → Cl- + 2 OH-

2e- + H2O + ClO- → Cl- + 2 OH-

eg. Balance the following redox equation under basic conditions.

(aq)+ NO(g) → SO32-

(aq) + NO2-(aq)

2 e- + 2 H2O + SO42-

→ SO32-

+ H2O + 2 OH-

2 e- + H2O + SO42-

→ SO32-

+ 2 OH-

→ SO32-

2 H+ +

2 e- +

+ 2 OH- + 2 OH-

2 OH- + H2O + NO → NO2- + 2 H2O+ 1 e-

2 OH- + NO → NO2- + H2O + 1 e-

NO → NO2-H2O + + 2 H+

2 OH- ++ 2 OH-

+ 1 e-

2 OH- + NO → NO2- + H2O + 1 e-

Multiply by 2 to cancel electrons!!

4 OH- + 2 NO → 2 NO2- + 2 H2O + 2 e-

2 e- + H2O + SO42-

→ SO32-

+ 2 OH-

2 OH- + 2 NO + SO42-

→ SO32-

+ 2 NO2- + H2O

ClO− + CrO2− → CrO4

2− + Cl2CrO2

− → CrO42−

ClO− → Cl2

ClO− + CrO2− → CrO4

2− + Cl2CrO2

− → CrO42−

4 OH- + CrO2− → CrO4

2− + 2 H2O + 3 e-

ClO− → Cl2

2 e- + 2 H2O + 2 ClO− → Cl2 + 4 OH-

2 H2O + + 4 H+ + 3 e- + 4 OH-

4 OH- +

2 + 2 H2O 4 H+ + 2 e- + + 4 OH-

+ 4 OH-

Electrochemical Cells (p. 757)

● An electrochemical cell changes chemical energy into electricity

● AKA: Galvanic cell or Voltaic cell

● These cells require spontaneous redox reactions

Electrochemical Cells

● The ½ reactions must be separate such that electron transfer occurs through an external circuit or wire.

● A salt bridge containing an electrolyte solution connects the ½ cells to complete the circuit.

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Zn(s) → Zn2+(aq) + 2e-

Oxidation ½ rxn

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Cu2+(aq) + 2 e- → Cu(s)

Reduction ½ rxn

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

KEY POINTS● Oxidation occurs at the anode● Reduction occurs at the cathode● Electrons move through the wire from

the anode to the cathode● Cations move toward the cathode● Anions move toward the anode

Electrochemical Cell Notationanode|anodic ion(s)|| cathodic ion(s)|cathode

eg. Use cell notation to represent this galvanic cell.

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s)

Zn(s) | Zn2+(aq)

| | Cu2+(aq)

| Cu(s)

phase boundary phase boundarysalt bridge

Electrochemical Cells

Sample Problem:

Draw a labeled electrochemical cell for the reaction below. Indicate the cathode, anode, direction of electron flow and the direction of ion flow. Use cell notation to represent the cell.

Pb2+(aq) + 2 Ag(s) → Pb(s) + 2 Ag+

Write the NIE and draw a labeled electrochemical cell for:

Sn(s) | Sn2+(aq) || Tl+(aq) | Tl(s)

Draw a labelled electrochemical cell for:Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g)

Inert Electrodes● Some redox reactions involve substances that

cannot serve as electrodes(gases, mixtures of ions)

● These cells must use an inert electrode - an electrode made from a material that is neither a reactant nor a product.

● Common inert electrodes include graphite - C(s) - and platinum - Pt(s)

eg. Write the cell notation for the reaction:

Pb(s) + 2 Fe3+(aq) → Pb2+

(aq) + 2 Fe2+(aq)

Pb(s) |Pb2(aq)|| Fe3+

(aq), Fe2(aq)|Pt(s)

p. 761 #’s 1 - 4

anode cathodesalt bridge

Standard Reduction Potentials

● The difference in the potential energy of the electrons at the anode and the cathode is the electric potential, E, of the cell

● electric potential is usually called cell voltage or cell potential

● E is measured in volts

Standard Reduction Potentials● The potentials in the table use hydrogen as

a reference electrode ● The potential of each ½ cell is found by

connecting each to the hydrogen ½ cell.● The table of standard reduction potentials

gives the voltage for each ½ reaction in the standard state

(ie. 1 mol/L and 25 °C)

Standard Reduction Potentials● All cell voltages may be determined using

the half cell voltages from the Standard Reduction Potentials table.

DO NOT MULTIPLY VOLTAGES when multiplying equations to cancel

electrons!!

eg. Calculate the standard cell potential for these electrochemical cells:

Cd2+(aq) + Cr(s) → Cd(s) + Cr 2+

Cd2+(aq) + 2 e- → Cd(s)

Cr(s) → Cr 2+(aq) + 2 e-

-0.40 V

+0.91 V

+0.51 V

eg. Calculate the standard cell potential for these electrochemical cells:

2 I- (aq) + Br2(l) → I2(s) + 2 Br - (aq)

Br2(l) + 2 e- → 2 Br - (aq)

2 I- (aq) → I2(s) + 2 e-

+1.07 V

-0.54 V

+0.53 V

eg. Calculate the standard cell potential for the electrochemical cell below:

2 Fe3+(aq) + Sn2+

(aq) → Sn4+(aq) + 2 Fe2+

2 Fe3+(aq) + 2 e -

→ 2 Fe2+(aq)

Sn2+(aq) → Sn4+

(aq) + 2 e-

+ 0.77 V

- 0.15 V

+ 0.62 V

Standard Reduction Potentials● Standard reduction potentials can be used to

predict whether a reaction is spontaneous

Spontaneous redox reactions have a

POSITIVE cell potential

eg. Predict the cell voltage for:

a) Cu(s) | Cu2+(aq)

| | Ag+(aq)

| Ag(s)

b) 2 Fe2+(aq) + Sn2+

(aq) → 2 Fe3+(aq) + Sn(s)

p. 773 #’s 5 - 7

Electrolytic Cells

● An electrolytic cell converts electrical energy to chemical energy

● The process that takes place in an electrolytic cell is called electrolysis

● These cells use electricity to force non-spontaneous reactions to occur

(ie. negative cell voltages)

Electrolytic cell

Carbon electrode

Zn(s) + Cu2+(aq) →

Eo = +1.10 V

Electrochemical Cell

Zn2+(aq) + Cu(s)

p. 780

Electrolytic Cell

Zn2+(aq) + Cu(s) →

Eo = -1.10 V

Zn(s) + Cu2+(aq)

(connected to positive)

(connected to negative)

FIX p. 780

Predicting Redox

All redox reactions occur between the strongest oxidizing agent (SOA) and the strongest reducing agent (SRA).

Spontaneous reactions will produce the calculated voltage.

Non-spontaneous reactions will require more than the calculated voltage.

Electrolysis

● Molten Salts (p. 776)● Water ● Aqueous NaCl● Aqueous NiCl2(aq)

Electroplating

● Electrolytic cells may be used to electroplate a thin layer of a more desirable metal over a less desirable metal

eg. gold plated jewelry

tin cans

chrome bumpers

ANODEOxidation

CATHODEReduction

(+) 107

Electroplating

● object to be plated or covered is the cathode – (connected to the negative terminal)

● metal to be plated or form a thin layer is the anode – (connected to the positive terminal)

Pretty to positive – ugly to negative

Faraday’s Law● Faraday’s Law can be used to determine the

amount of substance produced or consumed in an electrolytic cell

● The amount of substance is directly proportional to the quantity of electricity that flows through the cell (or the electric charge).

Faraday’s Law● The Coulomb (C) is the unit used to measure

the quantity of electricity (Q) flowing in a circuit (or the electric charge).

● The charge on one mole of electrons is 96,500 Coulombs.

● This charge is called Faraday’s constant:

F = 96,500 C/mol

Faraday’s Law

FORMULA:

Q = neF

Q = charge in coulombs (C)

ne = # of moles of electrons (mol)

F = 96,500 C/mol

Faraday’s Law● one coulomb is the quantity of electric

charge that flows through a circuit in one second if the current is one ampere.

ie. 1 A = 1 C/sec

FORMULA: Q = I tQ = charge in coulombs (C)I = current in amperes (A)

t = time in seconds (s)

Faraday’s Law

From Q = I t and Q = neF

I t = neF

eg. A spoon was copper plated using CuSO4(aq)

in an electrolytic cell that has 3.10 amps of electricity passing through it for 2.50 h. What mass of copper will be deposited?

I t = neF

(3.10)(2.5 x 60 x 60) = ne x (96,500)

0.289 mol = ne

Cu2+(aq) + 2 e- → Cu(s)

nCu = 0.289 mol e- x 1 Cu = 0.145 mol Cu

mCu = 0.145 mol Cu x 63.55 g/mol

= 9.21 g Cu

Reaction:

Faraday’s Law

eg. 0.423 g of solid Au is deposited on a bracelet using a solution of gold(III) nitrate and 2.00 A of current.

For how long was the current applied?

n = 0.423g /196.97 g/mol = 0.00215 mol Au

Au3+(ag)

+ 3e- → Au(s)

ne = 0.00215 x 3 e-/1 Au

= 0.00645 mol e-

I t = neF

(2.00)(t) = (0.00645)(96,500) t = 311 s

An electrolytic cell has a zinc strip anode and a zinc strip cathode placed in a solution of zinc sulfate. A current of 0.500 A is supplied for 900.0 seconds.

What mass of zinc is electroplated? (0.152g)

In an electrolytic cell, 0.061 g of Zn(s) was plated in 10.0 minutes from a solution of ZnCl2(aq). What current was used? (0.30A)

p. 793 #’s 21 - 24

The titanium cathode in an electrolytic cell increases in mass by 2.35 g in 36.5 min at a current of 6.50 A.

What is the charge on the titanium ion? (+3)

Some last definitions● Pp. 764 – 766, 787, 788● Primary battery – a disposable/non rechargeable

battery – Dry cells (electrolyte in a paste thickened with starch)

such as with flashlights, remote control

– Alkaline batteries – improved longer lasting version of the dry cell

– Button cell battery – smaller alkaline batteries such as the type in watches

Some last definitions● Pp. 764 – 766, 787, 788● Secondary batteries – rechargeable batteries

– Lead storage – car batteries

– Nickel-cadmium – rechargeable version of the alkaline battery found in things that can be recharged

Redox Stoichiometry● A redox titration could be used to find an

unknown concentration of an OA or RA.● The colors of the reactants and products serve

as indicators● Calculations are the same as those in acid-

base stoichiometry.

Redox Stoichiometry(see p. 742)

5 H2O2 2 MnO4-6 H→ 5 O2 2 Mn28H2O

- purple MnO4- changes to the colorless Mn2

until all of the H2O2 reacts- the endpoint occurs when the purple color remains

Redox StoichiometrySample Problem:

25.00 mL of a Fe2+ solution was titrated with acidic 0.02043 mol/L K2Cr2O7 solution. The endpoint was reached when 35.55 mL of K2Cr2O7 solution had been added. What was the molar concentration of Fe2+

Cr2O72Fe2→ Cr3Fe3

(balanced??)

Unit 4: Electrochemistry · Introduction Electrochemistry involves the study of electron movement...

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