Experiment 34

Electrochemistry

Electric Currents

• Alessandro Volta discovered if you dip two different metals in an acidic bath, a potential will move from one metal to the other or a charge will flow between the two metals.
• These cells are referred to as voltaic cells and can be used as batteries.

Chapter 18

Slide 2

Galvanic Cells

Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy.

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Galvanic (Voltaic) Cell: A spontaneous chemical reaction which generates an electric current.

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Electrolytic Cell: An electric current which drives a nonspontaneous reaction.

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For reversible cells, the galvanic reaction can occur spontaneously and then be reversed electrolytically - rechargeable batteries.

Reduction - Oxidation Reactions

• Combustion
• Corrosion
• Photosynthesis
• Kreb’s Cycle
• Synthesis
• Decomposition
• Single Replacement

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• How do you know you have a redox reaction?

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• How do you balance a redox reaction?

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• How do you know which species is being oxidized and which is being reduced?

Recognizing REDOX reactions

A REDOX reaction occurs if there has been a transfer of electrons.

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One species has increased its oxidation number.

A second species has decreased its oxidation number.

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Zn⁰ + Cu²⁺ → Zn²⁺ + Cu⁰

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The first step of recognizing or balancing a REDOX reaction is to determine the oxidation states of each element present.

Oxidation States

Assigning Oxidation Numbers: All atoms have an “oxidation number” regardless of whether it carries an ionic charge.�

1. An atom or element has an oxidation number of zero.� Examples N₂, Na, P₄, H₂ and O₂

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• An atom written as a monatomic ion has an oxidation number identical to the charge given.

Examples Cu²⁺, Cl^1-

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Oxidation numbers and the periodic table

3. Some observed trends in compounds.

• Oxidation states depend upon the valence electrons present on the atom.
• Elements in the same family, generally have the same oxidation states.
• Metals almost always have positive oxidation numbers.
• Transition metals typically have more than one oxidation number.
• Non metals in binary ionic compounds are negatively charged.
• Non metals in Ternary Ionic, Molecular or Organic compounds may have positive oxidation states.
• No element exists in a compound with an oxidation number greater than +8 or less than -8.

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IA - alkali metals - always +1

IIA - alkali earth metals - always +2

IIIA & IIIB - usually +3

VIA - Oxygen almost always -2; - Sulfur -2 in a binary ionic compound

VIIA – halogens always -1 in a binary compound, F is always -1, � Cl, Br, I can take on other charges with more electronegative atoms.

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The Oxidation Numbers of Elements in their Compounds

Oxidation States

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4) When two non metal elements are bonded together (molecular or covalent compounds), the first element will have a positive oxidation number while the element in the last position will have a negative oxidation number.��When a metal is bonded to a non metal, the non metal element will always have the negative oxidation state.��The oxidation numbers in any chemical species must sum to the overall charge on each element.

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A. H₂SO₄ The over-all net charge is zero

B. ClO₄^– The over-all net charge is -1

Oxidation States

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4) When two non metal elements are bonded together (molecular or covalent compounds), the first element will have a positive oxidation number while the element in the last position will have a negative oxidation number.��When a metal is bonded to a non metal, the non metal element will always have the negative oxidation state.��The oxidation numbers in any chemical species must sum to the overall charge on the species.

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• must sum to zero for any neutral molecule�H₂SO₄ 2 H⁺ + S^x + 4 O^2-

2(+1) + (x) + 4(–2) = 0 net charge

x = 0 – 2(+1) – 4(–2) = +6

Oxidation States

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• must sum to the charge on any polyatomic ion�ClO₄^– Cl^x + 4 O^2-� (x) + 4(–2) = –1 net charge

x = –1 – 4(–2) = +7

Often, it is possible to determine the oxidation number of those elements in a compound by determining the charges on elements that are known.

Follow the previous rules and then assign an oxidation number that insures that the overall compound has no net charge or the charge represented on the polyatomic ion..

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KIO₃

K = +1

O = -2

IF₇

F = -1

x + 7(-1) = 0

I = +7

MnO₄^1-

Mn = +7

What are the oxidation numbers of all the elements in each of these compounds?

IF₇ MnO₄^1- H₂O₂ KIO₃ FeSO₄

Fe₃O₄

H₂O₂

FeSO₄

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Assign the oxidation numbers to the elements in the compound KMnO₄.

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Step 1: Start with the oxidation numbers you know:

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Step 2: The numbers in the boxes (total contribution to charge) must sum to zero (KMnO₄ is a neutral compound).

K Mn O₄

+1

+1

Oxidation number:

Total contribution to charge:

+7

+7

–2

–8

Oxidation Numbers

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Assign the oxidation numbers to the elements in the compound H₂SO₄.

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Step 1: Start with the oxidation numbers you know:

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Step 2: The numbers in the boxes (total contribution to charge) must sum to zero (the chemical species is neutral).

H₂ S O₄

+1

+2

Oxidation number:

Total contribution to charge:

+6

+6

–2

–8

Oxidation Numbers

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Assign the oxidation numbers to the elements in the ion C₂O₄^2-

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Step 1: Start with the oxidation numbers you know:

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Step 2: The numbers in the boxes (total contribution to charge) must sum to negative two (the chemical species is a –2 anion).

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C₂ O₄ ^2-

+3

+6

Oxidation number:

Total contribution to charge:

–2

–8

^–

Oxidation Numbers

Oxidation Numbers on larger molecules

Oxidation state = Valence Electrons - atoms assigned to the atom in the Lewis structure

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Chapter 4 Unit 2

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OX_a = V_a - Y_a

OX_ca = 4 - 7

OX_ca = -3

OX_b = V_b - Y_b

OX_Cb = 4 - 6

OX_Cb = -2

OX_c = V_c - Y_c

OX_Cc = 4 - 5

OX_Cc = -1

Note the oxidation state of each carbon does not have to be the same value for all three.

Identifying oxidation-reduction reactions.

Oxidation-Reduction - REDOX

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A chemical reaction where there is a net change in the oxidation number of one or more species.

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Both an oxidation and a reduction must occur during the reaction. What are the charges?

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Mg (s) + Cl₂ (g) MgCl₂ (s)

Here the oxidation number of Mg has changed from

zero to +2. Cl has changed from zero to -1.

Is This a REDOX reaction?

2Fe(NO₃)₃ (aq) + Zn(s) → 2Fe(NO₃)₂ (aq) + Zn(NO₃)₂ (aq)

2Fe³⁺(NO₃)₃ + Zn⁰ → 2Fe²⁺(NO₃)₂ + Zn²⁺(NO₃)₂

2Fe³⁺ + Zn⁰ → 2Fe²⁺ + Zn²⁺

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Fe³⁺ is reduced to Fe²⁺, Fe³⁺ is the oxidizing agent

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Zn is oxidized to Zn²⁺, Zn is the reducing agent

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NO₃^- is a spectator ion.

Is it a REDOX reaction?

Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄ + KNO₃

_�oxidation states

Pb²⁺ N⁵⁺ O^2- K⁺ Cr⁶⁺ O^2-

reactants products

K⁺ N⁵⁺ O^2- Pb²⁺ Cr⁶⁺ O^2-

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Is there a difference in oxidation states?

Chapter 4 Unit 2

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Oxidation–Reduction Reactions

Whenever one atom loses electrons and it’s oxidation state increases (more positive), the atom or ion is oxidized, and another atom must gain those electrons, decreasing it’s oxidation state, (more negative) and the atom or ion is reduced.

A substance which loses electrons (oxidized) is called a reducing agent. Its oxidation number increases.

Oxidation Is Loss (OIL) or Loss of Electrons is Oxidation (LEO)

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A substance which gains electrons (reduced) is called the oxidizing agent. Its oxidation number decreases.

• Reduction Is Gain (RIG) or Gain of Electrons is Reduction (GER)

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OIL RIG or LEO GER

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Oxidation-Reduction Reactions

An oxidation-reduction (or redox) reaction is a chemical reaction in which electrons are transferred from one reactant to another.

Zn metal loses 2 electrons and is oxidized to Zn²⁺

Zn²⁺ is called the reducing agent

Cu²⁺ gains 2 electrons and is

reduced to Cu metal

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Cu is called the oxidizing agent

Oxidation is the loss of electrons. OIL

Reduction is the gain of electrons. RIG

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Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

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Redox reactions must have both mass balance and charge balance.

Step 1. Determine if the reaction is a redox reaction - � Assign oxidation state to each atom. Determine which elements� are changing oxidation state?

Step 2: Separate the unbalanced reaction into half-reactions.

Step 3: Balance each half-reaction with regard to all atoms

Step 4: Balance each half-reaction with regard to all oxygen

Step 5: Balance each half-reaction with regard to all hydrogen

Step 6: Balance charge by adding electrons.

Step 7: If the number of electrons in the half reactions are not equal, � multiply as necessary to make them equal.

Step 8: Add half-reactions to show the net reaction the electrons must � be equal to add the reactions.

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Half-Reaction method.

With this approach, the reaction is broken into two parts.

Oxidation half-reaction. The portion of the reaction where electrons are lost.

A → Aⁿ⁺ + ne^- note: the electrons are on the product side

Reduction half-reaction. The portion of the reaction where electrons are gained.

Bⁿ⁺ + xe^- → BO₂^m+ note: the electrons are on the reactant side

Balancing Oxidation-Reduction Reactions

Redox reactions must have both mass balance and charge balance.

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Step 1. Assign oxidation state to each atom:

Al³⁺(aq) + Mg (s) → Mg²⁺(aq) + Al(s)

Mg (s) → Mg²⁺(aq)

+ 2e^–

12 e^-

10 e^-

Step 2: Separate the unbalanced reaction into half-reactions.

Step 7: If the number of electrons in the half reactions are not equal, multiply as � necessary to make them equal.� Before adding half-reactions, the electrons must balance.

Oxidation half-reaction:

Reduction half-reaction:

Al³⁺(aq) → Al(s)

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+ 3e^–

10 e^-

13 e^-

Step 3, 4 and 5: Balance each half-reaction with regard to all atoms, O and H.

Step 6: Balance charge by adding electrons.

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Oxidation-Reduction Reactions

Multiply the oxidation half-reaction by 3

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Multiply the reduction half-reaction by 2

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Oxidation half-reaction:

3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

Step 7: If the number of electrons in the half reactions are not

equal, multiply as necessary to make them equal.

Step 8: Add the half-reactions together and cancel electrons

Reduction half-reaction:

( Al³⁺(aq) + 3 e^- → Al(s) )

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Oxidation half-reaction:

3 Mg (s) → 3 Mg²⁺(aq) + 6e^–

2 Al³⁺(aq) + 6 e^- → 2 Al(s)

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Reduction half-reaction:

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Oxidation half reaction

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Reduction half reaction

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Net Reaction

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Which atom is being oxidized? Which atom is being reduced?

�Which species is the reducing agent? Which is the oxidizing agent?

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3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

Balancing REDOX reactions

HCl + KMnO₄ (aq) + H₂O₂ (aq) → MnCl₂ (aq) + O₂ (g) + KCl +H₂O in an acid solution

4 HCl + KMnO₄ (aq) + 2 H₂O₂ (aq) → MnCl₂ (aq) + 2 O₂ (g) + 2 KCl + 4 H₂O

A REDOX reaction is balanced when the mass, the number of atoms and the number of electrons are equal on both sides of the reaction. Spectator ions are not written unless they are part of the species that is being oxidized or reduced.

H₂O₂ + MnO₄^-1 → 3 O₂ + Mn⁺²

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HCl + KMnO₄ (aq) + H₂O₂ (aq) → MnCl₂ (aq) + O₂ (g) + KCl +H₂O

H⁺Cl^1- + K⁺Mn⁷⁺O^2-₄ (aq) + H⁺₂O^1-₂ (aq)

→ Mn²⁺Cl^1-₂ (aq) + O⁰₂ (g) + K⁺Cl^1- +H₂O

Removing spectator species, H⁺ OH^1- and H₂O although spectators are used to balance oxygen in polyatomic ions.

First assign oxidation numbers to each atom or ion

H₂O₂ + MnO₄^-1 → O₂ + Mn⁺²

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H₂O₂

H is a +1, O cannot be -2

2 (H⁺¹) + 2 (O^x) = 0

2 + 2 x = 0 2x = -2

x = -1; O^-1

MnO₄^1-

O^-2 , what is charge on Mn?

1(Mn^x) + 4(O^-2) = -1

x -8 = -1; x = +7

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Determine which atom is are changing.

Manganese is decreasing in charge from +7 to +2

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Oxygen is increasing in charge from -1 to 0

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H⁺¹₂O^-2₂ + Mn⁺⁷O^-2₄^-1 → O⁰₂ + Mn⁺²

Balancing Redox Reactions

H⁺₂O^1-₂ + Mn⁷⁺ O^2-₄^-1 → O⁰₂ + Mn⁺²

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Redox reactions can also be balanced using the half-reaction method.

Step 2: Separate the unbalanced reaction into half-reactions.

MnO₄^- → Mn²⁺

H₂O₂ → O₂

Step 3: Balance each half-reaction with regard to atoms � other than O and H.

MnO₄^- → Mn²⁺ note the Mn is balanced

H₂O₄ → O₂ already balance no other � atoms

Balancing Redox Reactions

H⁺₂O^1-₂ + Mn⁷⁺ O^2-₄^-1 → O⁰₂ + Mn⁺²

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Step 4: Balance O by adding H₂O.

Step 5: Balance H by adding H⁺.

MnO₄^- → Mn²⁺ + 4 H₂O

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H₂O₂ → O₂

8 H⁺ + MnO₄^- → Mn²⁺ + 4 H₂O

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H₂O₂ → O₂ + 2 H⁺

Balancing Redox Reactions

H⁺₂O^1-₂ + Mn⁷⁺ O^2-₄^-1 → O⁰₂ + Mn⁺²

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Step 6: Balance charge by adding electrons.

^{25 protons 25 protons}

^{18 electrons 23 electrons}

^{+ 5 electrons}

8 H⁺ + Mn⁷⁺O₄^- → Mn²⁺ + 4 H₂O

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H₂O^1-₂ → O⁰₂ + 2 H⁺

^{2 x 8 protons 16 protons}

^{2 x 9 electrons = 18e 2 x 8 electrons = 16 e}

^{+ 2 electrons}

Reduction:

Gain of electrons

Electrons on reactant side

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Mn in MnO₄^1- is being reduced

Mn is the oxidizing agent.

5e^– +

Oxidation:

Loss of electrons

Electrons on product side

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C in H₂C₂O₄ is being oxidized

C is the reducing agent

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+ 2e^-

Balancing Redox Reactions

H⁺₂O^1-₂ + Mn⁷⁺ O^2-₄^-1 → O⁰₂ + Mn⁺²

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Step 7: If the number of electrons in the half reactions are not equal, multiply as necessary to make them equal.

Reduction:

5e^– + 8 H⁺ + MnO₄^- → Mn²⁺ + 4 H₂O

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Balancing Redox Reactions

H₂O₂ + MnO₄^-1 → O₂ + Mn⁺²

Step 8: Add the half-reactions together and cancel electrons

6 H⁺ + 2 MnO₄^- + 5 H₂O₂ → 5 O₂ + 2 Mn²⁺ + 8 H₂O

Balancing Redox Reactions (7)

Some redox reaction occur in basic solution. In this case, there are two additional steps:

Step 9: For each H⁺ ion, add one OH^– to both sides of the equation. H⁺ + OH^1- forms 1 H₂O molecule.

Step 10: Make any additional cancellations necessary by the new H₂O molecules.

6 H⁺ + 2 MnO₄^- + 5 H₂C₂O₄^2– → 10 CO₂ + 2 Mn²⁺ + 8 H₂O

6 OH^- → 6 OH^-

6 H₂O → 6 OH^-

2 MnO₄^- + 5 H₂C₂O₄^2– → 10 CO₂ + 2 Mn²⁺ + 6 OH^- + 2 H₂O

Balance the following reaction in a base solution.

Sn²⁺ + IO₄¹ ⇄ Sn⁴⁺ + I^1-

Galvanic Cells

When zinc metal is placed in a copper(II) solution, Zn is oxidized and Cu²⁺ ions are reduced.

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Zn metal in a copper solution before and after the reaction

Reduction half-reaction:

Oxidation half-reaction:

Zn(s) → Zn²⁺(aq) + 2e^-

Cu²⁺(aq) + 2e^- → Cu(s)

Galvanic Cells

Galvanic Cells

Chapter 17/39

Anode:

• The electrode where oxidation occurs.
• The electrode where electrons are produced.
• Is what anions migrate toward.
• Has a negative sign.
• Zn → Zn²⁺ + 2e^–

Galvanic Cells

Chapter 17/40

Cathode:

• The electrode where reduction occurs.
• The electrode where electrons are consumed.
• Is what cations migrate toward.
• Has a positive sign.
• Cu²⁺(aq) + 2e^– → Cu(s)
• The Cu²⁺ gains �two electrons to �form Copper�

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Oxidation half reaction

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Reduction half reaction

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Net Reaction

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Which reaction occurs at the anode, which at the cathode?

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3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

Shorthand Notation for Galvanic Cells

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Phase boundary

Phase boundary

Electron flow

Salt bridge

Cathode half-cell

Electric current flows from anode to cathode because there is a difference in electrical potential energy between the electrodes.

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The electrical potential is measured by a voltmeter and is called the cell potential (E_cell ).

Overall cell reaction:

Anode half-reaction:

Cathode half-reaction:

Zn(s) → Zn²⁺(aq) + 2e^-

Cu²⁺(aq) + 2e^- → Cu(s)

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Anode half-cell

Determination of Standard Reduction Potentials

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The standard hydrogen electrode (SHE):

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2H⁺ (1M) + 2e^- → H₂ (1 atm)

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H₂ (1 atm) → 2H⁺ (1M) + 2e^-

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E^o = 0 V

The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.

Standard Reduction Potentials

Pt(s) │ H₂(g) │H⁺ (1 M) ││Cu²⁺ (1 M) │Cu(s)

Standard Reduction Potentials

Overall cell reaction:

Anode (oxidation):

Cathode (reduction):

Pt | H₂(g) | H¹⁺(aq) || Cu²⁺ | Cu (s)

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The standard potential of a cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode:

E°_cell = E°_red - E°_ox

The measured potential for this cell: E°_cell = 0.34 V

Standard Reduction Potentials

Overall cell reaction:

Anode (oxidation):

Cathode (reduction):

0.34 V = E°_red – 0 V

E°_cell = E°_red - E°_ox

E° = 0.34 V

A standard reduction potential can be defined:

Standard Reduction Potentials (3)

Zn(s) │ Zn²⁺ (1 M) ││ H⁺ (1 M) │ H₂(g) │ Pt(s)

Standard Reduction Potentials

Overall cell reaction:

Anode (oxidation):

Cathode (reduction):

- 0.76 V = 0 V - E°_ox

E°_cell = E°_red - E°_ox

E° = - 0.76 V

As a standard reduction potential:

E° = 0.76 V

Galvanic Cells

Standard Reduction Potentials

Zn(s) │ Zn²⁺ (1 M) ││ Cu²⁺ (1 M) │ Cu(s)

E°_cell = E°_cathode – E°_anode

0.34 V

This Reaction is spontaneous

E°

–(–0.76 V)

E°_cell = 1.10 V

Standard Reduction Potentials

Standard Reduction Potentials at 25°C (See Table 18.1)

Standard Cell potential, E^o

Standard potentials are defined using specific conditions.

• All solutes are at 1 M concentrations
• Slightly soluble species must be at saturation or solubility limit.
• gases at partial pressure of 1 atm
• Solids and liquid are in pure form
• With the cell at 25 ^oC

Chapter 18

Slide 52

Cell Potentials and Free-Energy Changes �for Cell Reactions

Electromotive Force (emf): The force or electrical potential that pushes the negatively charged electrons away from the anode (- electrode) and pulls them toward the cathode (+ electrode).

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It is also called the cell potential (E) or the cell voltage.

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Electric Potential is the energy associated with the movement of charge.

Potential (Volts) =

potential energy (joules)

Charge (coulomb)

1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.

Cell Potentials and Free-Energy Changes for Cell Reactions

ΔG° = -nFE°

cell potential

free-energy change

number of moles of electrons transferred in the reaction

faraday or Faraday constant

the electric charge on 1 mol of electrons

96,500 C/mol e^-

ΔG = -nFE

or

E°_cell is related to the thermodynamic quantities ΔG° and K

ΔG < 0 E > 0

ΔG > 0 E < 0

ΔG = 0 or E = 0

at equilibrium

Spontaneity

Oxidation-reduction reactions occur spontaneously when they convert the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent.

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If the reduction cell potential for the half reaction is positive, the reaction is spontaneous in the forward direction.

Chapter 18

Slide 55

Oxidation half reaction

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Reduction half reaction

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Net Reaction

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What is the cell potential?

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Determine the Free Energy, ∆G^o, for the reaction in 4 under standard conditions. �

Is the above reaction spontaneous? Explain.

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3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

Mg(s) → Mg²⁺(aq) + 2 e^-

Al³⁺ (aq) + 3 e^- → Al(s)

Spontaneity of Redox Reactions Under Standard-State Conditions

Calculate ΔG° and K for the following reaction at 25°C:

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3Mg(s) + 2Al³⁺(aq) ⇌ 3Mg²⁺(aq) + 2Al(s)

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Solution

Step 1: Use E° to calculate E°_cell.

E°_cell = E° – E°

E°_cell = –1.66 V – (–2.37 V) = 0.71 V

Al³⁺/Al

Mg²⁺/Mg

Spontaneity of Redox Reactions Under Standard-State Conditions

Calculate ΔG° and K for the following reaction at 25°C:

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3Mg(s) + 2Al³⁺(aq) ⇌ 3Mg²⁺(aq) + 2Al(s)

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Solution

Step 2: Use the equation below to calculate ΔG° :

ΔG° = –nFE°_cell

ΔG° = – (6 e^–)( )(0.71 V)

ΔG° = –411090 J or –411 kJ

96500 J

V·mol e^–

The Nernst Equation at Standard conditions

ΔG = ΔG° + RT ln Q not at equilibrium

At equilibrium: E = 0 ΔG = 0 and Q = K

0 = ΔG° + RT lnK or ΔG° = -RT lnK�ΔG° = –nFE°_cell then –nFE°_cell = -RT lnK

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E°_cell = lnK

RT

nF

Spontaneity of Redox Reactions Under Standard-State Conditions

Calculate ΔG° and K for the following reaction at 25°C:

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3Mg(s) + 2Al³⁺(aq) ⇌ 3Mg²⁺(aq) + 2Al(s)

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Solution

Step 3: Use the equation below to calculate K:

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E°_cell = (RT/nF) lnK

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0.71 V = [(8.314 J/mol•K)(298 K)/(6 mol e^–)(96500 J/V•mol e^–)]lnK

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K = 8 × 10¹⁶⁵

This indicates a reaction that is very product favored

Cell potentials

• You know that both an oxidation and a reduction must occur.
• One of your half reactions must be reversed.
• The spontaneous or galvanic direction for a reaction is the one where E_cell is a positive value.
• The half reaction with the largest positive E value will proceed as a reduction.
• The other half reaction will be reversed and occur as the oxidation.

• Zn with 0.1 M CuSO₄

B. 3 M HCl + Zn

C. 3 M HCl + Copper

D. 0.1 M KI + 0.1 M FeCl₃

E. 0.1 M FeCl₃ + 0.1 M KBr

F. 0.1 M KI + 0.1 M CuSO₄

Using the Nernst Equation, determine the Cell Potential for reaction F. under nonstandard conditions.

Cell Potentials and Free-Energy Changes for Cell Reactions

ΔG° = -nFE°

cell potential

free-energy change

number of moles of electrons transferred in the reaction

faraday or Faraday constant

the electric charge on 1 mol of electrons

96,500 C/mol e^-

ΔG = -nFE

or

E°_cell is related to the thermodynamic quantities ΔG° and K

Cell Potentials and Free-Energy Changes for Cell Reactions

Chapter 17/70

Calculate the standard free-energy change for this reaction at 25 °C.

The standard cell potential at 25 °C is 1.10 V for the reaction:

(1.10 V)

ΔG° = -212 kJ

= -(2 mol e^-)

ΔG° = -nFE°

Concentration dependency of E

The standard cell potential, E^o, are measured under standard conditions.

The cell potential, E, is measured when any of the concentrations vary from standard conditions.

This effect can be experimentally determined by measuring E versus a standard (indicator) electrode, SHE.

Theoretically, the electrode potential can be determined by the Nernst equation.

Chapter 18

Slide 71

Calculation of cell potentials

At nonstandard conditions, we don’t know which will proceed as a reduction until we calculate each E value.

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Steps in determining the spontaneous direction and E of a cell.

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Calculate the E for each half reaction.

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The half reaction with the largest or least negative E value will proceed as a reduction.

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Calculate E_cell

When a reaction is not at equilibrium and not at standard conditions, the Nernst equation can predict spontaneity

The Nernst Equation

ΔG = ΔG° + RT ln Q not at equilibrium

Using: ΔG = -nFE and ΔG° = -nFE°

Nernst Equation:

or

or

at 25^oC

n is the number of moles electrons

F is the Faraday constant (96,500 C/mol e^–)

R is the gas constant (8.314 J/mol K)

When Q > 1, E < E°

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When Q < 1, E > E°

The Nernst Equation at Standard conditions

ΔG = ΔG° + RT ln Q not at equilibrium

Nernst Equation:

All concentrations at standard conditions are 1 atm or 1 M

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Q = 1 log Q = log 1 = 0

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E = E^o or ΔG = ΔG°

The Nernst Equation at Standard conditions

ΔG = ΔG° + RT ln Q not at equilibrium

At equilibrium: E = 0 ΔG = 0 and Q = K

0 = ΔG° + RT lnK or ΔG° = -RT lnK�ΔG° = –nFE°_cell then –nFE°_cell = -RT lnK

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E°_cell = lnK

RT

nF

The Nernst Equation

What is the potential of a cell at 25 °C that has the following ion concentrations?

When a reaction is not at equilibrium and not at standard conditions, the Nernst equation can predict spontaneity.

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Consider a galvanic cell that uses the reaction:

3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

If the concentrations of the ions are not at standard 1 M concentration but we begin the reaction with 0.30 M Mg²⁺ �and 0.20 M Al³⁺ , determine the cell potential using the Nernst equation.

The Nernst Equation

Calculate E°:

E°_cell = –1.66 V – (–2.37 V) = 0.71 V

3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

The Nernst Equation

Calculate E:

log

E = 0.71 V -

log

E =

E = E° -

3 Mg(s) + 2 Al³⁺ (aq) → 2 Al(s) + 3 Mg²⁺(aq)

(0.3 M)³

(0.2 M)²

G. 0.1 M KI + 0.1 M KIO₃

Add 1 mL of 3 M HCl to the solution. What is the experimental observation?

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Using the Nernst Equation, determine the Cell Potential for this reaction under nonstandard conditions.

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12 H⁺(aq) + 10 I^1-(aq) + 2 IO₃^1-(aq) ⇄ 6 I₂(s) + 6 H₂O(l)

at 25^oC

Assume [H⁺] = 1 M

Standard Cell Potentials and �Equilibrium Constants

Three methods to determine equilibrium constants:

3. K from electrochemical data:

1. K from concentration data:

2. K from thermochemical data:

or

Free energy and cell potential

Earlier, we explained that Δ G and the equilibrium constant can be related. Since E_cell is also related to K, we know the following.

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Q ΔG E�

Forward change, spontaneous < K - +

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At equilibrium = K 0 0

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Reverse change, spontaneous > K + -