Unit 4: Electrochemistry

Unit 4: Electrochemistry

Introduction Electrochemistry involves the study of electron movement or electric current in chemical reactions. Electrochemical reactions involve the transfer of electrons between reactants.

Terms electrochemical cell oxidation electrolytic cell reduction
electrolysis electroplating fuel cell pyrometallurgy hydrometallurgy oxidation reduction oxidizing agent reducing agent half-reaction oxidation number redox reaction

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 occurs when hydrogen is gained the Gain of Electrons in a reaction is Reduction (GER) a substance is reduced when it gains electrons

GER LEO

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

Oxidation Is Loss Reduction Gain

Identify the species being oxidized and the species being reduced below: Fe(NO3)3 + Al → Al(NO3)3 + Fe Pb + SnCl2 → Sn + PbCl2

Why is this NOT an example of an electrochemical reaction? CaCl2 + 2 AgNO3 → Ca(NO3)2 + 2 AgCl

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) Ag(s)

Net Ionic Equations (NIE)
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) Ag(s) Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) → Cu2+(aq) + 2 NO3-(aq) + 2 Ag(s) Cu(s) Ag+(aq) → Cu2+(aq) Ag(s)

Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) NET IONIC EQUATION:
Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) NET IONIC EQUATION: Cu(s) 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) NaI(aq) → NaCl(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 Agents
eg. Identify the OA and the RA in the following redox reactions Cu(s) Ag+(aq) → Cu2+(aq) + 2 Ag(s) Cl2(g) 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 Reactions eg. Write the oxidation and reduction half reactions for Cu(s) Ag+(aq) → Cu2+(aq) + 2 Ag(s) oxidation: Cu(s) →Cu2+(aq) + 2e- reduction: 2 Ag+(aq) + 2e- → 2 Ag(s)

Half Reactions Write 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) Br -(aq) → 2 F-(aq) + Br2(s) Reduction: F2(g) + 2 e -(aq) → 2 F-(aq) Oxidation: 2 Br -(aq) → Br2(s) + 2e-

Half Reactions Write the overall equation for: Al3+ + 3e- →Al(s)
Mg(s) → Mg2+(aq) + 2e-

Text Exercises: - p #’s 1-4 p #’s 5-7 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+ ion Mg atom C in CH4 oxidation # -1 +2 -4

Determining Oxidation Numbers (See p. 724)
Pure elements → oxidation # = 0 eg. Cl2(g), Mg(s), P4(s), Fe(s) Simple ions → oxidation # = ion charge eg. Cl- → oxidation # = -1 Mg 2+ → oxidation # = +2

Determining Oxidation Numbers
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 Oxygen has an oxidation number of -2 except in: - peroxides such as H2O2 - and in OF2.

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

The sum of all oxidation numbers in a compound is ZERO. eg. CH4 H → C must be ???

The sum of all oxidation numbers in a polyatomic ion equals the ion charge. eg. CO32- O → C + 3(-2) = -2 C = +4 Textbook: p #’s 9 – 12

Extra Practice (oxidation #’s)
1. Determine the oxidation number of: a) S in SO2 d) Cr in Cr2O72- 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 NH4Cl

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 -2 -1

eg. Using oxidation numbers, explain why this is NOT a redox reaction: CaCO3(s) → CaO(s) + CO2(g) Text: p #’s 13 – 15

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)

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)

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 Solution
eg. Balance each ½ reaction in acidic solution. a) MnO4-(aq) → Mn2+(aq) b) HNO2(aq) → N2O(g)

MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) 5 e H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

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

- write and balance each half reaction IO3-(aq) → I2(s) NO2(g) → NO3-(aq)

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 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) H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 e- (Oxidation: RA is NO2(g) )

Oxidation: H2O(l) + NO2(g) → NO3-(aq) + 2 H+(aq) + 1 e- Reduction: 10 e H+(aq) + 2 IO3-(aq) → I2(s) + 6 H2O(l) 10 H2O(l) + 10 NO2(g) → 10 NO3-(aq) + 20 H+(aq) + 10 e- Overall equation: 4 H2O(l) + 10 NO2(g) + 2 IO3-(aq) → I2(s) + 10 NO3-(aq) + 8 H+(aq)

Extra Practice (balance in acidic conditions)
CH3OH(l) + MnO4-(aq) → CH2O(l) + Mn2+(aq) CH3OH(l) → CH2O(l) CH3OH(l) → CH2O(l) + 2 H+(aq) + 2 e- MnO4-(aq) → Mn2+(aq) 5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) x 5 x 2

p. 739 # 27 5 CH3OH(l) → 5 CH2O(l) + 10 H+(aq) + 10 e-
10 e H+(aq) MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) __ CH3OH(l) + __ H+(aq) + __MnO4-(aq) → __ CH2O(l) + __Mn2+(aq) + __ H2O(l) p # 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- SO42-(aq) → H2SO3(aq) SO42-(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-(aq)

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 OH- 2e- + H2O + ClO- → Cl OH-

eg. Balance the following redox equation under basic conditions. SO42-(aq)+ NO(g) → SO32-(aq) + NO2-(aq)

2 H+ + SO42- → SO32- + H2O 2 OH- + + 2 OH- 2 H+ + SO42- → SO32- + H2O 2 H2O + SO42- → SO32- + H2O + 2 OH- H2O + SO42- → SO OH- 2 e- + H2O + SO42- → SO OH-

H2O + NO → NO2- + 2 H+ 2 OH- + + 2 OH- H2O + NO → NO2- + 2 H+ 2 OH- + H2O + NO → NO H2O 2 OH- + NO → NO2- + H2O 2 OH- + NO → NO2- + H2O + 1 e-

2 OH- + NO → NO2- + H2O + 1 e- Multiply by 2 to cancel electrons!! 4 OH- + 2 NO → 2 NO H2O + 2 e- 2 e- + H2O + SO42- → SO OH- 2 OH- + 2 NO + SO42- → SO NO2- + H2O

ClO− + CrO2− → CrO42− + Cl2 CrO2− → CrO42− ClO− → Cl2 p #’s 23, 24 p #’s 27, 28

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.

Demo: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation ½ rxn Zn(s) → Zn2+(aq) + 2e-

Cu2+(aq) + 2 e- → Cu(s) Reduction ½ rxn

-

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 Notation
anode|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) salt bridge phase boundary phase boundary

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+(aq)

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) ADD to Handout

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)

Pb(s) |Pb2+(aq) || Fe3+(aq), Fe2+(aq) |Pt(s)
eg. Write the cell notation for the reaction: Pb(s) Fe3+(aq) → Pb2+(aq) Fe2+(aq) Pb(s) |Pb2+(aq) || Fe3+(aq), Fe2+(aq) |Pt(s) p. 761 #’s 1 – 3 & Electrochemistry #4 anode salt bridge cathode

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

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)

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+(aq) 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+(aq) 2 Fe3+(aq) + 2 e - → 2 Fe2+(aq) Sn2+(aq) → Sn4+(aq) + 2 e- V V V

Standard reduction potentials can be used to predict whether a reaction is spontaneous Spontaneous redox reactions have a POSITIVE cell potential OR OA is higher in the table than RA

eg. Predict the cell voltage for: Cu(s) | Cu2+(aq) | | Ag+(aq) | Ag(s) 2 Fe2+(aq) + Sn2+(aq) → 2 Fe3+(aq) + Sn(s) p #’s 5 - 7

Identify the anode and the cathode
in this electrochemical cell. [June 2004] # 47

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)

(-) (+) Electrochemical Cell electrons → Anode Cathode cations →
anions ←

(-) (+) Electrolytic Cell Battery - electrons  Anode Cathode cations
→ anions ←

Electrolytic Cells An electrolytic cell uses an external source of electricity such as a battery rather than a voltmeter. The battery forces a non-spontaneous reaction to occur

Electrolytic cell Carbon electrode Carbon electrode Na+ Cl-

Zn(s) + Cu2+(aq) → Eo = V Zn2+(aq) + Cu(s) V - + Electrochemical Cell

p. 780

- + Zn2+(aq) + Cu(s) → Eo = -1.10 V Zn(s) + Cu2+(aq) + -
Electrolytic Cell

FIX p. 780 (connected to positive) (connected to negative)

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 Commercial Electroplating

- + CATHODE Reduction (+) ANODE Oxidation (-)

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 t Q = 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) mol = ne Cu2+(aq) + 2 e- → Cu(s) nCu = mol e- x 1 Cu = mol Cu 2 e- mCu = mol Cu x g/mol = 9.21 g Cu Reaction:

Faraday’s Law eg 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 / g/mol = mol Au Au3+(ag) + 3e- → Au(s) ne = x 3 e-/1 Au = mol e- I t = neF (2.00)(t) = ( )(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 A is supplied for seconds. What mass of zinc is electroplated?

In an electrolytic cell, 0. 061 g of Zn(s) was plated in 10
In an electrolytic cell, g of Zn(s) was plated in 10.0 minutes from a solution of ZnCl2(aq). What current was used? p. 793 #’s

The titanium cathode in an electrolytic cell increases in mass by 2
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? (Show workings.)

Applications Pp. 764 – 766, 787, 788 Dry cell Battery
Primary battery Secondary battery Button cell battery Car battery (lead acid) Voisey’s Bay STSE

Last one!! 5 CH3OH(l) + 2 MnO4-(aq) + 6 H+(aq) → 5 CH2O(l) + 2 Mn2+(aq) + 8 H2O(l)

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 MnO H+ → 5 O2 + 2 Mn2+ + 8H2O - purple MnO4- changes to the colorless Mn2+ until all of the H2O2 reacts - the endpoint occurs when the purple color remains

Redox Stoichiometry 0.174 mol/L Sample Problem:
25.00 mL of a Fe2+ solution was titrated with acidic mol/L K2Cr2O7 solution. The endpoint was reached when mL of K2Cr2O7 solution had been added. What was the molar concentration of Fe2+ Cr2O Fe2+ → Cr3+ + Fe3+ (balanced??)

Unit 4: Electrochemistry

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