analytical chemistry - clutch 1e ch.13 - fundamentals...

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ANALYTICAL CHEMISTRY - CLUTCH 1E

CH.13 - FUNDAMENTALS OF ELECTROCHEMISTRY

CONCEPT: BASIC REDOX CONCEPTS OXIDATION-REDUCTION (REDOX) reactions deal with the transfer of electrons from one reactant to another.

Lose

Electrons

Oxidation

Gain

Electrons

Reduction

Reducing Agent (Reductant)

} }} Element

becomes

more positive

Oxidation

Number

Increases } Element

becomes

more negative

Oxidation

Number

Decreases

Oxidizing Agent (Oxidant)

Li (s) Li+ (aq) + e – Cl2 (g) + 2 e – 2 Cl – (aq)

Li (s) + Cl2 (g) Li+ (aq) + 2 Cl – aq)

Electrical Charge

The units for electrical charge are measured in ________________ (C).

} Charge of 1 electron Faraday Constant

}(1.602×10−19C) ⋅ (6.022×1023mol−1) = 9.647×104C

1mole e−charge mole

e –

q = n ⋅ F Faraday Constant

Electrical Current

The units for electrical current are in __________ (A).

Electrical Voltage

The relationship between work and voltage can be expressed as:

The relationship between Gibbs Free Energy and electric potential can be expressed as:

Ohm's Law

The units for resistance are in __________ (Ω).

Power

Power represents work done per unit of time. The units for power are in __________ (W).

w = E ⋅ qWork Voltage Charge

ΔG = − n ⋅ F ⋅ E GibbsFree Energy

mole e –

Faraday Constant

Voltage

I = ER

Voltage

ResistanceCurrent

P = E ⋅ IPower Voltage Current

Current

Charge

TimeI = q

t



PRACTICE: BASIC REDOX CONCEPTS CALCULATIONS 1 EXAMPLE 1: What happens to the current in a circuit if a 3.0 V battery is removed and replaced by a 1.0 V battery?

EXAMPLE 2: If the voltage of a TE Series Enhanced Balance has a 240 V battery, what is the resistance in the circuit if the current is 0.80 A?

PRACTICE 1: Solve for the missing variable in the following circuit.

R = 320 ohms

I = ?

V = 24 V

PRACTICE 2: Solve for the missing variable in the following circuit.

R = 100 ohms

I = 1.5 A

V = ?



PRACTICE: BASIC REDOX CONCEPTS CALCULATIONS 2 EXAMPLE 1: Calculate ΔG°rxn and E°cell for a redox reaction with n = 4 that has an equilibrium constant of K = 0.130 (at 25 °C).

EXAMPLE 2: Copper can be electroplated at the cathode of an electrolysis cell by the half-reaction:

Cu2+ (aq) + 2 e – → Cu (s)

How much time would it take for 525 mg of copper to be plated at a current of 4.3 A?

PRACTICE: A metal forms the salt MCl3. Electrolysis of the molten salt with a current of 0.700 A for 6.63 h produced 3.00 g of the metal. What is the molar mass of the metal?



PRACTICE: BASIC REDOX CONCEPTS CALCULATIONS 3 EXAMPLE 1: In the following reaction:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

What is the maximum energy produced when 15.0 g of Zn is completely reacted in a Zn-Cu electrochemical cell that has an average cell potential of 1.10 V?

EXAMPLE 2: A chemist weighing 110 lb takes her NMR sample from the first floor to the second floor, which is 12 meters up, in 25 seconds. How much power has she generated?

EXAMPLE 3: Determine the amount of time (in mins) needed to produce 1.7 x 102 watts from 1500 J of work committed.



CONCEPT: BALANCING REDOX REACTIONS Generally, you will need to balance a redox reaction in an acidic or basic solution.

Balancing A Redox Reaction in Acidic Reactions: STEP 1: Write the equation into 2 half-reactions.

STEP 2: Balance elements that are not oxygen or hydrogen.

STEP 3: Balance Oxygens by adding ____________.

STEP 4: Balance Hydrogens by adding ____________.

STEP 5: Balance overall charge by adding electrons (e –) to the more _________________ side. Both half reactions must

have an ___________ number of electrons.

STEP 6: Combine the half-reactions and cross out reaction intermediates.

Balancing A Redox Reaction in Basic Reactions: Follow Steps 1-6 from above.

STEP 7: Balance remaining H+ by adding an equal amount ____________ ions to both sides of the chemical reaction.

EXAMPLE: Balance the following redox reaction in acidic solution.

Br – (aq) + MnO4 – (aq) Br2 (l) + Mn2+ (aq)



PRACTICE: BALANCING REDOX REACTIONS CALCULATIONS 1 EXAMPLE: Balance the following redox reaction in acidic solution.

Cr2O72– (aq) + H2O2 (aq) Cr3+ (aq) + H2O (l) + O2 (g) PRACTICE: Balance the following redox reaction in basic solution.

Cr2O72– (aq) + SO32– (aq) Cr3+ (aq) + SO42– (aq)



CONCEPT: GALVANIC CELLS Galvanic/Voltaic Cell: A spontaneous cell that ________________ or ________________ electricity.

Galvanic/Voltaic Cell

__________________ ( – ) __________________ ( + )

Cathode : 3Cu2+(aq) + 6 e− 3Cu (s)Anode : 2 Cr (s) 2 Cr3+(aq) + 6 e−

Reduction Half-Reaction Eo (V)

Ionization Energy __________

Anode ___________

Producing ñ Voltage

[Anode] ______

[Cathode] ______

Electron Affinity __________

Cathode __________



CONCEPT: ELECTROLYTIC CELLS In terms of spontaneity the following correlations between the following variables can be made:

ΔGo

K

Eo

ΔSo

Q vs. K

Reaction Classification

Cell Type

< 0

> 1

> 0

> 0

Q < K

> 0

< 1

< 0

< 0

Q > K

= 0

= 1

= 0

= 0

Q = K

Electrolytic Cell: A non-spontaneous electrochemical cell that _______________ electricity and so requires a battery.

Ionization Energy __________

Electron Affinity __________

Electrolytic Cell

__________________ ( – ) __________________ ( + )



CONCEPT: LINE NOTATION Line notation is a quick, simple method to describe an electrochemical cell without having to draw it out in detail.

| = _____________________ || = _____________________

Lower Oxidation State

Higher Oxidation State

Lower Oxidation State

Higher Oxidation State

Galvanic/Voltaic Cell

__________________ ( – ) __________________ ( + )

EXAMPLE: Write the half reactions as well as the overall net ionic equation for the following line notation:

Cu | Cu2+ (aq, 0.0050 M) || Ag+(aq, 0.50 M) | Ag

Cathode : 3Cu2+(aq) + 6 e− 3Cu (s)Anode : 2 Cr (s) 2 Cr3+(aq) + 6 e−



PRACTICE: LINE NOTATION CALCULATIONS 1 EXAMPLE: Sketch the galvanic cell and determine the cell notation for the following redox reaction:

2 H+ (aq) + Fe (s) H2 (g) + Fe2+ (aq)

PRACTICE: Sketch the galvanic cell and determine the line notation for the following redox reaction:

Ni2+ (aq) + Mg (s) Ni (s) + Mg2+ (aq)



CONCEPT: STANDARD POTENTIALS Voltage (E) represents the amount of work done in an electrochemical cell as electrons travel from one electrode to another.

When combining two half-cell reactions together the cell potential for the total net reaction is given (when the concentrations approach unity) by:

ECell = E+ −E−E+

= Represents the _______________ electrode.

E– = Represents the _______________ electrode.

EXAMPLE: Determine the electric potential that results from the given galvanic cell.

Li Li+ Ag+ AgE = -3.040 V E = 0.799 V

e –

e –

e – e –

e –



PRACTICE: STANDARD POTENTIALS CALCULATIONS 1

EXAMPLE 1: Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. Assume the concentrations have approached unity.

3 Cl2 (g) + 2 Fe (s) 6 Cl – (aq) + 2 Fe3+ (aq)

Cl2 (g) + 2 e – 2 Cl – (aq) E° = + 1.396 V Fe3+ (aq) + 3 e – Fe (s) E° = – 0.040 V

EXAMPLE 2: For the a voltaic cell with the overall reaction:

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) E°cell = 1.10 V

Given that the standard reduction potential of Zn2+ to Zn (s) is – 0.762 V, calculate the standard reduction potential for:

Cu2+ (aq) + 2 e – Cu (s)



CONCEPT: NERNST EQUATION The Nernst Equation reveals the quantitative connection between the concentrations of compounds and cell potential.

E = Eo − RTnFln AB

b

AAa

Eo = Standard Cell Potential

R = Gas Constant = ______________________

n = Number of electrons transferred

F = Faraday's constant = __________________

A = Activity

______ represents the cell potential under non-standard conditions, while ______ represents it under standard conditions.

At 25 oC, RTF

=8.314 J

K ⋅mol⎞

⎠⎟ 298.15 K)(

⎛

⎝⎜

9.649×104 Cmol

⎞

⎠⎟

⎛

⎝⎜

= 0.0257 JC= 0.0257 V

The Nernst Equation then becomes,

ECell = ECello −

0.0257 Vn

lnQ

By multiplying ln by 2.303 we can obtain the log function.

ECell = ECello −

0.05916 Vn

logQ

The cell potential calculated from Nernst equation is the maximum potential at the instant the cell circuit is connected. As the cell discharges and current flows, the electrolyte concentrations will change, Q increases and Ecell deceases.

• Over time the reaction will reach equilibrium at then Q = K and cell potential will equal zero.

ECell = ECello −

RTnF

⎞

⎠⎟

⎛

⎝⎜ lnK = 0 ΔG = ΔGo −RTlnK = 0



PRACTICE: NERNST EQUATION CALCULATIONS 1 EXAMPLE 1: Determine the cell potentials of the following concentration cells:

Ag | Ag+ (aq, 0.0010 M) || Cu2+ (aq, 1.0 M) | Cu E−o = 0.799 V E+

o = 0.339 V

EXAMPLE 2: Consider the following electrochemical cell for the question:

Pt | KBr(aq, 0.01 M), Br2(l) || FeBr2(aq, 1.0M) | Fe E−o =1.078 V E+

o = − 0.440 V

Determine the spontaneity and cell potential based on the given cell notation.



PRACTICE: NERNST EQUATION CALCULATIONS 2

EXAMPLE 1: Consider a standard voltaic cell based on the reaction:

2 H+ (aq) + Sn (s) Sn2+ (aq) + H2 (g)

Which of the following actions would change the emf of the cell?

a) Increasing the pH at the cathode

b) Lowering the pH at the cathode

c) Increasing [Sn2+] at the anode

d) Increasing the hydrogen gas pressure at the cathode

e) All of the above changes will alter the cell potential

EXAMPLE 2: Consider the following half cell reaction at T = 25. °C

Ni2+(aq) + 2 e – Ni(s) E° = – 0.260 V What will be the value for E°, the half cell potential for standard conditions, for the reaction

2 Ni(s) 2 Ni2+ (aq) + 4 e – a) + 0.52 V b) + 0.26 V c) – 0.26 V d) – 0.52 V



CONCEPT: STANDARD CELL POTENTIAL & THE EQUILIBRIUM CONSTANT Galvanic and voltaic cells are able to produce electricity because they are not yet at equilibrium.

• Recall that the chemical reaction will eventually reach equilibrium and then Q = K.

ECell = ECello −

0.05916 Vn

logQ → 0 = ECello −

0.05916 Vn

logK → ECello =

0.05916 Vn

logK

Once we establish the correlation between the cell potential and the equilibrium constant we can reformat the equation:

The relationship between the cell potential, equilibrium constant and Gibbs Free Energy can be seen as:

ECello

K ΔGΔG = −RTlnK

ΔG = −nFECelloECell

o =RTnFlnK

ECello = E+ −E−



PRACTICE: STANDARD CELL POTENTIAL & THE EQUILIBRIUM CONSTANT CALCULATIONS 1 EXAMPLE 1: Determine the equilibrium constant K for the following reaction:

Au+ (aq) + Ce (s) Ce3+ (aq) + Au (s)

The half reactions are determined as:

Au+ (aq) + e – Au (s) Eo =1.690 V

Ce3+ (aq) + 3 e – Ce (s) Eo = − 2.336 V

EXAMPLE 2: From the two half reactions provided the equilibrium constant is calculated as 6.79 x 1030.

ClO (g) + e− ClO− (aq) E+o = ?

Bi (s) Bi3+(aq) + 3 e− E−o = 0.308 V

Determine the standard cell potential for:

ClO (g) + e− ClO− (aq) E+o = ?



analytical chemistry - clutch 1e ch.13 - fundamentals...

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