Chem 1C Final Review Fall 2013

Chem 1C Final ReviewFall 2013

Rimjhim Hemnani Amin Mahmoodi Justin Nguyen

Electrochemistry - Overview• Branch of chemistry that deals with the interconversion of

electrical energy to chemical energy• What to know:• Oxidation number rules• Balancing Redox reactions• Galvanic Cells• Cell Diagram• Standard Reduction Potentials• Thermodynamics of Redox Reactions• Effect of Concentration on Cell EMF• Electrolysis

Oxidation number rules

Posted on Dr. A’s website:https://eee.uci.edu/12s/40200/Oxidation_Numbers.pdf

In a nutshell…1. Natural state element/compounds = 0 (ex. H2)

2. Oxidation # = monatomic ion charge (Na+ = +1)

3. Oxygen is usually -2, except in peroxides (H2O2, Na2O2)

4. Hydrogen is usually +1, except in metal hydrides (LiH, NaH)

5. Sum of oxidation numbers = 0 for electrically neutral compounds. For polyatomic ions, sum of oxidation numbers = charge of compound. (NH4

+ has total oxidation sum of 1)

https://eee.uci.edu/12s/40200/Oxidation_Numbers.pdf



Electrochemistry – Redox Reactions

• Electrons are transferred from one species to another

• When a species loses electron = OXIDATION• Marked by oxidation number going up• A species that is oxidized is also called the reducing agent

• When a species gains electron = REDUCTION• Marked by oxidation number going down• A species that is reduced is also called the oxidizing agent

• “LEO the lion says GER”

Balancing Redox Reactions• Problem: Write a balanced equation to represent the oxidation of

iodide ion (I-) by permanganate ion (MnO4-) in basic solution to yield

molecular iodine (I2) and manganese (IV) oxide (MnO2)

Step 1: Write unbalanced equation:

I- + MnO4- I2 + MnO2




Step 2: Assign oxidation numbers and divide into two half reactions (oxidation and reduction)I- + MnO4

- I2 + MnO2

-1 +7 -2 0 +4 -2





- I2 + MnO2

-1 +7 -2 0 +4 -2

Oxidation: I- I2 (I is going from -1 to 0)





- I2 + MnO2

-1 +7 -2 0 +4 -2

Oxidation: I- I2 (I is going from -1 to 0)Reduction: MnO4

- MnO2 (Mn is going from +7 to +4)




Step 3: Balance each half reaction for number and type of atoms and charges

Oxidation: 2 I- I2 + 2e-

Balancing Redox Reactions

Step 3: Balance each half reaction for number and type of atoms and chargesReduction: MnO4

- MnO2

First add O: MnO4- MnO2 + 2 OH-

Next balance H: 2 H+ + MnO4- MnO2 + 2 OH-

*Neutralize H+: + 2 OH- + 2 H+ + MnO4- MnO2 + 2 OH- + 2 OH-

Combine H and OH: 2 H2O + MnO4- MnO2 + 4 OH-

Balance charge: 2 H2O + MnO4- + 3e- MnO2 + 4 OH-

*Only do this for basic conditions.




Step 4: Add half-reactions together to form overall reaction. Equalize number of electrons so that they cancel out later.Oxidation: 3[ 2 I- I2 + 2e- ]Reduction: 2[2 H2O + MnO4

- + 3e- MnO2 + 4 OH- ]

6 I- 3 I2 + 6e- 4 H2O + 2 MnO4

- + 6e- 2 MnO2 + 8 OH-




Step 4: Add half-reactions together to form overall reaction. Equalize number of electrons so that they cancel out later.Oxidation: 3[ 2 I- I2 + 2e- ]Reduction: 2[2 H2O + MnO4

- + 3e- MnO2 + 4 OH- ]

6 I- 3 I2 + 6e- 4 H2O + 2 MnO4

- + 6e- 2 MnO2 + 8 OH-




Step 6: Final check for balanced charges and atoms

Answer:6 I- + 2 MnO4

- + 4 H2O 3 I2 + 2 MnO2 + 8 OH-

Galvanic (aka voltaic) Cells• Experimental apparatus for generating electricity through the

use of a spontaneous reaction

Galvanic (aka voltaic) Cells• Cathode: where REDUCTION

occurs• Electrons flow to the cathode

• Anode: where OXIDATION occurs• Electrons flow away from the

anode• Salt bridge: an inverted U tube

containing an inert electrolyte solution (i.e. KCl, NH4NO3). • Ions prevent buildup of +

charge on anode and (-) charge on cathode (maintains balance of charge)

• “An ox”, “red cat”

Galvanic (aka voltaic) Cells• Cell voltage – difference in

electrical potential between the anode and cathode• Also called cell potential, or

electromotive force (EMF)• Cell voltage is dependent on:• Nature of electrode and ions• Concentrations of ions• Temperature

Cell DiagramRules:1. Anode written first2. | indicates a phase boundary (i.e. going from solid to aqueous)3. A comma (,) indicates different components in the same phase (i.e.

Fe2+ to Fe3+)4. || denotes a salt bridge, thus separating cathode and anode5. Pt (s), aka platinum, is used when there are no metals present for

electrons to travel, (i.e. Fe2+ to Fe3+, or H+ (aq) to H2 (gas))

Example #1: Cu2+ (aq) + Zn (s) Cu (s) + Zn2+ (aq)

Cell notation:Zn (s) | Zn2+ (1 M) || Cu2+ (1 M)| Cu (s)

Oxi = anode

Red. = cath

Cell DiagramRules:1. Anode written first2. | indicates a phase boundary (i.e. going from solid to aqueous)3. A comma (,) indicates different components in the same phase (i.e.

Fe2+ to Fe3+)4. || denotes a salt bridge, thus separating cathode and anode5. Pt (s), aka platinum, is used when there are no metals present for

electrons to travel, (i.e. Fe2+ to Fe3+, or H+ (aq) to H2 (gas))

• Example #2: Zn (s) + 2 H+ Zn2+ + H2

• Cell notation: Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (g) | Pt (s)

Oxi = anode

Red. = cath

Standard Reduction Potentials• The voltage associated with a reduction reaction at an electrode when

all solutes are 1 M and all gases are at 1 atm .• The reduction of H+ to H2 is arbitarily set at 0.00 V to determine

relative potentials of other substances.• The more positive E° is, the greater the tendency for the substance to

be reduced (aka the strongest oxidizing agent)• E°cell = E°cathode - E°anode

• A positive E°cell means the reaction is favored• Changing the stoichiometric coefficients of a half-cell reaction will NOT

affect the value of E°.• When a reaction is reversed, the sign of E° changes but not its

magnitude.

Standard Reduction Potentials

Chart offered by Dr. A:https://eee.uci.edu/12s/40200/Reduction_Potentials.pdf

More extensive chart:http://highered.mcgraw-hill.com/sites/dl/free/0023654666/650262/Standard_Reduction_Potential_19_01.jpg

https://eee.uci.edu/12s/40200/Reduction_Potentials.pdf

https://eee.uci.edu/12s/40200/Reduction_Potentials.pdf

Standard Reduction Potentials• Example: Can Sn reduce Zn2+ (aq) under standard-state conditions?

• Sn2+ has a greater tendency to be reduced (more positive E value), while Zn2+ has a greater tendency to be oxidized.

• Answer: No.

Standard Reduction Potentials• A galavanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and a Ag electrode in a 1.0 M AgNO3

solution. Calculate the standard emf

• Ag+ has a greater tendency to be reduced (aka cathode)• Mg is therefore oxidized

Half Reactions:Reduction: Ag+ + e- Ag (s)Oxidation: Mg (s) Mg2+ (aq) + 2 e-

Calculation: E°cell = E°cathode - E°anode

= 0.80 – (-2.37) = 3.17 V

Thermodynamics of Redox Reactions• E°cell is related to ΔG° and K.• 1 Joule = 1 Coulomb * 1 Volt• Joule = energy• Coulomb = electrical charge• Volt = voltage

• Faraday’s constant = 9.65 * 104 C/mol e- (or 96,500)• total charge of one mole of electrons

• Total charge can now be expressed as nF (n = # of moles of e- transferred in reaction)

Thermodynamics of Redox Reactions• Useful equations:

- ΔG = -nFEcell ( a negative ΔG = spontaneous)- ΔG = -RT ln K- Ecell = RT ln k

- nF

R = 8.314 J/K*molF = 98,500 J/V*molT = in KELVINS!n = moles of e- from balanced redox reaction

Thermodynamics of Redox Reactions• Example: Calculate standard free-energy change for the

following reaction at 25 °C:

2 Au (s) + 3 Ca2+ (1.0 M) 2 Au3+ (1.0 M) + 3 Ca (s)--------------------------------------------------------------------------Step 1: Determine what equation to use:

In this case, you are looking for ΔG .

Equation: ΔG = -nFEcell



2 Au (s) + 3 Ca2+ (1.0 M) 2 Au3+ (1.0 M) + 3 Ca (s)--------------------------------------------------------------------------Step 2: Determine what is being reduced and oxidized. Then find Ecell and n.Half Reactions: Reduction: 3 Ca2+ + 6e- 3 Ca (s) (-2.87 V) Oxidation: 2 Au (s) 2 Au3+ (1.0 M) + 6e- (1.50 V)

Ecell = -2.87 – (1.50) = -4.37 Vn = 6



2 Au (s) + 3 Ca2+ (1.0 M) 2 Au3+ (1.0 M) + 3 Ca (s)--------------------------------------------------------------------------Step 3: Plug and chug ΔG = -nFEcell

= -(6)(96500)(-4.37) = 2.53*106 J/mol

Or 2.53*103 kJ/mol



2 Au (s) + 3 Ca2+ (1.0 M) 2 Au3+ (1.0 M) + 3 Ca (s)--------------------------------------------------------------------------Notes: the negative Ecell value that you calculated indicates an unfavorable reaction, which explains why ΔG turned out to be positive (unspontaneous)

The effect of concentration on Cell EMF• Until now, we’ve dealt with STANDARD conditions (1 M)• But what if the concentrations of species aren’t 1 M?

• Use Nernst equation:• E = E° - RT ln Q nF

• E = new EMF that you want to solve• E° = standard EMF• Q = reaction quotient (product / reactant)

The effect of concentration on Cell EMF• Example: Calculate the emf for the following reaction and

determine if it will occur spontaneously, given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M:

Co (s) + Fe2+ (aq) Co2+ + Fe (s)------------------------------------------------------------------------------------Step 1: Determine what is being reduced and oxidized to find E° and n.Half Reactions: Reduction: Fe2+ + 2 e- Fe (s) (-0.44) Oxidation: Co (s) Co2+ + 2e- (-0.28)

E° = (-0.44) – (-0.28) = -0.16 Vn = 2



Co (s) + Fe2+ (aq) Co2+ + Fe (s)------------------------------------------------------------------------------------Step 2: Determine Q

Q = [product]/[reactant] = [0.015]/[0.68] = .221



Co (s) + Fe2+ (aq) Co2+ + Fe (s)------------------------------------------------------------------------------------Step 3: Plug and chug• E = E° - (RT/nF ln Q)

E° R T Q n F -0.16 – (8.314)(298)(ln(.221))/(2*96500) = -0.14

Emf = negative = NOT spontaneous

Electrolysis• The mass of product formed (or reactant consumed) at an

electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question.

• Basically a lot of conversions to get from current to mass (or liters) of substance.

• Current * time Coulombs• Coulombs / Faraday’s constant number of moles of e- • Use mole ratios to find moles of substance being

oxidized/reduced• Moles * molar mass = mass of substance

Electrolysis• Example: A current of 1.26 A is passed through an electrolytic

cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases (H2 and O2) generated at STP.

-----------------------------------------------------------------------------------Step 1: Determine half-cell reactions.- You know that sulfuric acid will give you H+ ions in water.- You also know that sulfuric acid is electrolytic and will conduct

electricity Reactions:Oxidation: 2 H2O (l) O2 (g) + 4 H+ (aq) + 4e-

Reduction: 2 H+ + 2 e- 1 H2 (g)

We see that the gases we need to calculate for are O2 and H2



-----------------------------------------------------------------------------------Step 2: Find the volume of O2 gas generated. convert current to charge

1.26 A * 7.44 hr * (3600 s / 1 hr) = 3.37*104 C

Step 3: convert charge to moles of electrons

3.37*104 C * ( 1 mol e- / 96,500 C) = 0.349 mole e-



-----------------------------------------------------------------------------------Step 4: Use mole ratio of oxygen and electrons.2 H2O (l) O2 (g) + 4 H+ (aq) + 4e-

For every mole of oxygen formed, 4 moles of e- are transferred. 0.349 mol e- * (1 mol O2 / 4 mol e-) = 0.0873 mol O2



-----------------------------------------------------------------------------------Step 5: PV = nRT!

V = nRT/P = (0.0873 mol)(0.0821)(273 K)/(1atm) = 1.96 L



-----------------------------------------------------------------------------------Now for H2 gas, use same methods.2 H+ + 2 e- 1 H2 (g)

3.37*104 C * (1 mole e- / 96500 C) * (1 mol H2 / 2 mol e-) = 0.175 mol H2

V = nRT/P = (.175 mol)(.0821)(273 K)/(1 atm) = 3.92 L

Thank you and good luck!!• TIPS ON STUDYING:• DO NOT PROCRASTINATE• PRACTICE PROBLEMS!• WHEN IN DOUBT, LOOK AT UNITS!• SLEEP WELL• EAT WELL

• AND….DON’T FORGET EVALUATIONS PLZ

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Chem 1C Final Review Fall 2013