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GHS Honors Chem

Electro-Electro-ChemistryChemistry

GHS Honors Chem

Electrochemistry

• Electrochemistry is the study of the relationships between electrical energy and chemical reactions

• It’s the study of how chemical energy is changed to electrical energy through the exchange (flow) of electrons

GHS Honors Chem

Oxidation-Reduction Reactions

Oxidation-reduction reactions are chemical reactions involving the exchange of electrons between two substances.

• During an oxidation reaction, there is a loss of electrons.

• For example, the oxidation of Fe(s) to Fe+2(aq) is accompanied by the loss of two electrons. Fe(s) Fe+2(aq) + 2 e -

GHS Honors Chem

Oxidation-Reduction Reactions

• During a reduction reaction, there is a gain of electrons.

• Example: reduction of Cu+2(aq) to Cu(s) is accompanied by the gain of two electrons. Cu+2(aq) + 2 e - Cu(s)

• Short-cut: • Oxidation Is Loss (OIL) of electrons (electrons on the right-hand side of the equation) • Reduction Is Gain (RIG) of electrons(electrons on the left-hand side of the equation) • Remember OIL-RIG.

GHS Honors Chem

Redox Reactions

• Oxidation and reduction occur together. • Hence, they are called redox reactions. • In a redox reaction, one substance

gains electrons (reduction), while the other substance loses electrons (oxidation).Cu+2(aq) + Fe(s) Fe+2(aq) + Cu(s)

• In this reaction, Cu+2(aq) is reduced to Cu (s) and Fe(s) is simultaneously oxidized to Fe+2(aq).

GHS Honors Chem

Redox Reactions

• The substance being reduced is called the oxidizing agent.

• The substance being oxidized is called the reducing agent.

GHS Honors Chem

Oxidation Number

• The oxidation number is a number which tells us how oxidized or reduced a given element of a given substance is.

• The higher the oxidation number is, the more oxidized the element is.

GHS Honors Chem

Assigning Oxidation Numbers

1. Rule # 1: The oxidation number for an element in the elemental state is 0.

– the oxidation numbers for Al in Al(s) is 0.

2. Rule # 2: The oxidation number for oxygen in most oxygen compounds is equal to - 2.

3. Rule # 3: Group IA compounds, have an oxidation number of +1

– Na in NaCl(s) or in Na2SO4(s) has an oxidation number of +1

4. Rule # 4: Group IIA compounds, the oxidation number of the metal is +2.

– Ca in CaCO3 and in Ca(NO3)2 has an oxidation number of +2

GHS Honors Chem


5. Rule # 5: In all HALOGEN-containing compounds, the oxidation number for the halogen is -1.– The oxidation number of F in NaF, CaF2 and AlF3 is always

-1. – In NaCl, Cl has an oxidation number of -1.

6. Rule # 6: In a compound, the oxidation number of hydrogen is +1 if H is bonded to a nonmetal.– H in NH3, in CH4, in H2O and in HCN has the same

oxidation number of +1.

7. Rule # 7: In a compound, the oxidation number of hydrogen is -1 if H is bonded to a metal. (Note: in this case, H behaves like an anion and is called hydride)– H in NaH (sodium hydride), in CaH2 (calcium hydride) has

the oxidation number of -1.

GHS Honors Chem


8. Rule # 8: The sum of the oxidation numbers of all elements in a compound is equal to 0 (the charge of the compound).

– For example in NO2 Ox.# of N + 2 x Ox # O = 0.9. Rule # 9: The sum of the oxidation numbers of all

elements in a polyatomic ion is equal to the charge of the polyatomic ion.

– For example in CO3-2, Ox # C + 3 x Ox # O = -2.

GHS Honors Chem

Assigning Oxidation # Examples

1. Na2SO4: – Oxidation number of Na = “Na” = +1– Oxidation number of O = “O” = -2– Oxidation number of S: = “S” to be

calculated– 2 “Na” + “S” + 4 “O” = 0– 2 x (+1) + “S” + 4 x (-2) = 0– “S” – 6 = 0– “S” = +6

GHS Honors Chem

1. Ox # of Pt in K2PtCl4

2. Ox # of Mn in KMnO4

3. Ox # of Pb in PbSO4

4. Ox # of Pb in PbS2

5. Ox # of each C in C2H3O2-

Assigning Oxidation # Examples

GHS Honors Chem

Redox Reactions

Identifying:Identifying:•What is Oxidized (the What is Oxidized (the Reducing Agent)Reducing Agent)

•What is Reduced (the What is Reduced (the Oxidizing Agent) Oxidizing Agent) … … in a Redox Reactionin a Redox Reaction

How’s It Done?How’s It Done?

GHS Honors Chem

Redox Reactions

1. Identify the substance oxidized, substance reduced, 1. Identify the substance oxidized, substance reduced, oxidizing agent, reducing agent, and write oxidation oxidizing agent, reducing agent, and write oxidation and reduction half reactionsand reduction half reactions

2H2H22 + O+ O22 →→ 2H2H22OO

Oxidized ReducedOxidized Reduced (RA)(RA) (OA) (OA)

2H2H22 00 → 4H → 4H+1+1 + 4e + 4e-- Oxidation Half Oxidation Half reactionreaction

OO22 0 0 + 4e+ 4e-- → 2O→ 2O-2 -2 Reduction Half Reduction Half ReactionReaction

GHS Honors Chem

Redox Reactions


Mg + ZnMg + Zn+2+2 →→ Mg Mg+2+2 + Zn + ZnOxidized ReducedOxidized Reduced (RA)(RA) (OA) (OA)

Mg → MgMg → Mg+2+2 + 2e + 2e-- Oxidation Half Oxidation Half reactionreaction

ZnZn+2 +2 + 2e+ 2e-- → Zn→ Zn Reduction Half Reduction Half ReactionReaction

GHS Honors Chem

Redox Reactions


SnSn + + SnSnOO2 2 + + 2H2H22SOSO44 → 2SnSO→ 2SnSO44 + 2H + 2H22OO

Oxidized ReducedOxidized Reduced (RA)(RA) (OA) (OA)

SnSn Sn Sn+2+2 + 2e + 2e-- Oxidation Half reactionOxidation Half reaction

SnSn+4+4 + 2e + 2e-- Sn Sn+2+2 Reduction Half ReactionReduction Half Reaction

GHS Honors Chem

Redox Reactions

Redox Reactions Worksheet

GHS Honors Chem

Electrochemical Cells

Let’s Look at the Handouts as an Intro to Electrochemical Cells

GHS Honors Chem

Calculating Standard Cell Potentials

Example: Calculate the standard cell potential for:

Fe (s) Fe+2 (aq) Ag+ (aq) Ag (s)

Fe (s) Fe+2 (aq) + 2 e- Oxidation E°anode = + 0.44 V

Ag+ (aq) + e

- Ag (s) Reduction E°cathode = + 0.80 V

Fe+2 + 2e- Fe(s) Eo = -0.44V

Ag+1 + 1e- Ag(s) Eo = +0.80V

The Ag reduction has a higher EThe Ag reduction has a higher E00 (higher on the list), so it is the (higher on the list), so it is the reduction … the Fe reaction is turned around to an oxidation:reduction … the Fe reaction is turned around to an oxidation:

GHS Honors Chem


• In order to write the overall redox reaction, multiply In order to write the overall redox reaction, multiply the reduction half-reaction by 2, the reduction half-reaction by 2, but NOT the but NOT the potential value.potential value. Thus, Thus,

Hence,Hence,

Overall Redox reaction:

• Fe (s) + 2 Ag+ (aq) Fe+2 (aq) + 2 Ag (s)

Fe (s) Fe+2 (aq) + 2 e-

Oxidation E°anode = + 0.44 V 2 x (Ag+ (aq) + e

- Ag (s)) Reduction E°cathode = + 0.80 V

Fe (s) Fe+2 (aq) + 2 e-

Oxidation E°anode = + 0.44 V 2 Ag+ (aq) + 2 e

- 2 Ag (s) Reduction E°cathode = + 0.80 V

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• Since E°cell is positive, the cell operates spontaneously.

• Reaction will take place only if E°cell is positive

E°cell = E°red + E°ox

E°cell = (+ 0.80 V) + (+0.44 V) E°cell = + 1.24 V

GHS Honors Chem


Example: Calculate the standard cell potential for:

Al (s) Al+3 (aq) Hg+2 (aq) Hg (s)

Al+3 + 3e- Al(s) Eo = -1.66V

Hg+2 + 2e- Hg(s) Eo = +0.85V

The Hg reduction has a higher EThe Hg reduction has a higher E00 (higher on the list), so it is the (higher on the list), so it is the reduction … the Al reaction is turned around to an oxidation:reduction … the Al reaction is turned around to an oxidation:

Al(s) Al+3 + 3e- Eo = +1.66V

Hg+2 + 2e- Hg(s) Eo = +0.85V

GHS Honors Chem


The Hg reduction has a higher EThe Hg reduction has a higher E00 (higher on the list), so it is the (higher on the list), so it is the reduction … the Al reaction is turned around to an oxidation:reduction … the Al reaction is turned around to an oxidation:

(Al(s) Al+3 + 3e-) Eo = +1.66V

(Hg+2 + 2e- Hg(s)) Eo = +0.85V

2 x2 x

3 x3 x

Eo Cell: Eo red + Eo ox

Eo Cell = 0.85V + 1.66V = 2.51V

2Al(s) + 3Hg+2 2Al+3 + 3Hg

ghs honors chem electro- chemistry. ghs honors chem electrochemistry electrochemistry is the study...

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