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Mr. Storie 40S Chemistry Final Exam Review

40S CHEMISTRY FINAL EXAM ANSWER KEY:

ATOMIC STRUCTURE

1. Give the electron configuration for a neutral atom of manganese, strontium, and iron.

Mn: 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Sr: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2

Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6

2. Write the short hand notation for the electron configuration of phosphorus, tungsten, and gold.

P: [Ne] 3s2 3p3

W: [Xe] 6s1 4f14 5d5 (electron promotion)

Au: [Xe] 6s1 4f14 5d10 (electron promotion)

3. What is the wavelength of light with a frequency of 5.6 x1020

Hz?

c = λf

λ = c / f = 3.0 x 10 8 m/s /

5.6 x 10

20 Hz = 5.4 x 10

-13m

4. What is the frequency of light with energy 4.3x 10-15

J ?

E = hf

f = E/h (4.3x 10-15

J ) / (6.626 x 10-34

Js/pho) = 6.5 x 1018

Hz

5. What is the energy of blue light that has a wavelength of 650 nm?

Covert nm m: 650 nm = 650 x 10-9

m

c = λf

f = c / λ = 3.0 x 10 8 m/s /

650 x 10

-9 m = 4.62 x 10

14 Hz

E = hf = (6.626 x 10-34

Js/pho) (4.62 x 1014

J ) = 3.1 x 10-19

J

6. Draw an orbital box diagram of chromium. Which electrons would it lose to form a 2+ ion?

The 2 - 4s electrons would be lost.


7. Calculate the number of orbitals in the third energy level. Show your work.

The number of orbitals is calculated by the formula: n2

Where n=3 (3)2 = 9

(3s 3p 3d) 1 3 5 = 9

8. Draw Lewis dot diagrams and use VSEPR to predict the shapes of the following molecules. Include

valence electron count and account for all electrons.

a) AlCl3

Ionic compound – does not follow VSEPR rules.

b) PO4

-3

5 + 4(6) + 3 = 32 – 8 = 24

4 bp / 0 lp Tetrahedral

c) PH3

5 + 3(1) = 8 - 6 = 2

3 bp / 1 lp Trigonal Pyramidal

d) SF6

6 + 6(7) = 48 – 12 = 36

6 bp / 0 lp Octahedral


e) OCl2

6 + 2(7) = 20 – 4 = 16

2bp / 2 lp Bent

9. Identify which molecules in question 9 are polar.

C and E are polar only.

10. What is the general trend observed for ionization energy? Explain this trend for ionization energies;

include reasoning for the trend in a group and the trend in a period.

Increases across and decreases down the Periodic table.

In a group (down) shell number increases, valence electrons are increasingly far away and there is

more shielding. They experience less nuclear force leading to less energy required energy to

remove.

In a period (across) electrons are added to the same shell number, so no appreciable difference in

shielding or distance. However across there is an increase in nuclear charge with additional

protons added. This increased nuclear force increases energy requirements.

11. What are the trends observed for electronegativity, electron affinity, and metallic character?

EN: Increases across, decreases down.

EA: Increases across, decreases down.

Metallic character: Decreases across.

12. By considering the electron arrangement for the atoms oxygen and nitrogen, explain why the first

ionization energy of oxygen is lower than that of nitrogen in spite of the fact that the trend predicts the

opposite. (Hint: draw the orbital box diagram of each)

O: 1s2 2s2 2p4 - unstable

N: 1s2 2s2 2p3 - stable

Half-filled and completely filled orbitals are MORE stable. Nitrogen has a half-filled p-shell

(3/6). Oxygen is one electron “over” stable (4/6). Oxygen benefits in stability by losing an

electron and therefore it requires less energy than the more stable Nitrogen.


13. Draw Lewis dot diagrams for the following molecules. Show the valence electron count and account for

all electrons.

a. H2S

b. XeF4

14. Using Lewis dot diagrams and VSEPR theory predict the shapes of the following molecules. Show the

valence electron count and account for all electrons. Indicate the number of bonding pairs and lone pairs

of electrons. Draw the 3-D representation of the molecules.

a. NF3

b. PI5


15. Explain the difference between polar covalent bonds, non-polar covalent bonds, ionic bonds and

coordinate covalent bonds.

Polar Covalent bonds: Electronegativity differences between 0.4-1.9, leads to “unequal” sharing of

bonding electrons and +/- dipole regions in the molecule.

Non-polar bonds: EN differences between 0-0.4, leads to “equal” sharing of bonding electrons and

a neutral molecule.

Ionic bonds: EN differences > 1.9, leads to “giving and taking” of bonding electrons and positive

and negative ion formation, which stay together due to attractive forces.

Coordinate Covalent: one atom donates both electrons in a single covalent bond.

KINETICS PROBLEMS

1. The reaction, A + 2B Products, was found to have the rate law, Rate = k[A][B]2. While holding the

concentration of A constant, the concentration of B was increased from x to 3x. Predict by what factor

the rate of the reaction will increase.

Rate α [A][B]2

1 α [1][1]2

9 α [1][3]2 A 9x increase in rate.

2. For the hypothetical reaction A + B Products, the following initial rates of reaction have been

measured for the given reactant concentrations.

Experiment [A] (M) [B] (M) Rate (mol/(L∙hr))

1 0.010 0.020 0.020

2 0.015 0.020 0.030

3 0.010 0.010 0.005

a. Calculate the order for reactant A.

Rate α [A]x[B]

y

Trial 2/1: ignore B, no change during these trials

0.030 = 0.015 x

0.020 0.010

1.5 = [1.5] x A is first order (directly proportional)


b. Calculate the order for reactant B.

Rate α [A]x[B]

y

Trial 1/3: ignore A, no change during these these trials

0.020 = 0.020 Y

0.005 0.010

4.0 = [2.0] Y

B is second order (proportional squared)

c. Calculate the overall order for the reaction.

x + y = overall order

first + second = Third order overall

d. Write the rate law expression.

Rate = k[A][B]2

e. Solve for the specific rate constant, k.

k = Rate_ = 0.020 = 5000

[A][B]2

[0.010][0.020]2

3. For a reaction with the rate law - rate = k [A]n [B]

m - Determine the factor by which the rate changes if

both [A] and [B] are doubled given the following values of “n” and “m”. Show all your work.

a. n=0, m=2

rate = k [A]n [B]

m 1 α [1]

2

4 α [2]2 4x increase in rate.

b. n=2, m=2

rate = k [A]n [B]

m 1 α [1]

2[1]

2

16 α [2]2[2]

2 16x increase in rate.

c. n=1, m=3

rate = k [A]n [B]

m 1 α [1]

1[1]

3

16 α [2]1[2]

3 16x increase in rate.


4. For the chemical reaction H2O2 + 2H+

+ 2I

- I2 + 2H2O

the rate law expression is Rate = k[H2O2][I-]. The following mechanism has been suggested.

H2O2 + I- HOI + OH

-

OH- + H

+ H2O

HOI + H+ + I

- I2 + H2O

a. Define the term “intermediate” and identify all intermediates included in this reaction.

Intermediate is a particle that is a product in one step and a reactant in another step of the

mechanism. It does not appear in the overall reaction equation.

Intermediate: OH- and HOI

b. Determine which of the steps is the RDS, and explain how the conclusion was reached.

The RDS is the slowest step. Only changes to particles within the RDS affect the reaction rate.

Only those particles that have an effect on rate appear in the rate law. Therefore the Rate Law

equation indicates which is step is the RDS.

Step 1 has the reactants that appear in the Rate Law, and must be the RDS.

5. Draw a Potential Energy Diagram for the following exothermic reaction:

2 SO2 (g) + O2 (g) 2 SO3 (g) ΔH° = -198 kJ

Pote

ntial E

nerg

y

Reaction Coordinates

ΔH° = -198 kJ

2 SO2 + O

2

2 SO3

A.C.

EA


6. If ammonia gas, NH3, reacts at a rate of 0.090 mol/L s according to the reaction

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

a. at what rate does oxygen gas react under the same conditions? 0.11 mol/L s

b. what is the rate of formation of water? 0.14 mol/L s

c.

d. c. What is the rate of production of nitrogen monoxide? 0.090 mol/L s

7. The reaction A + B + C E + F + G was studied at 0°C and the following data was collected:

Trial [A] [B] [C] Initial Rate

(mol/L) (mol/L) (mol/L) (mol/L•s)

1 1.0 2.0 2.0 1.66

2 2.0 2.0 2.0 3.33

3 1.0 4.0 2.0 1.66

4 1.0 2.0 4.0 6.65

5 1.5 5.0 ? 7.25

a. What is the rate law for this reaction?

Rate α [A]x[B]

y[C]

z

Trial 2/1: ignore B and C, no change during these these trials

3.33 = 2.0 x

1.66 1.0

2.0 = [2.0] x A is first order (proportional)

0.090 mol/L·s NH3

4 mol NH3

5 mol O2 = 0.11 mol/L·s O2

0.090 mol/L·s NH3

4 mol NH3

6 mol H2O = 0.14 mol/L·s H2O

0.090 mol/L·s NH3

4 mol NH3

4 mol NO = 0.090 mol/L·s NO


Rate α [A]x[B]

y[C]

z

Trial 3/1: ignore A and C, no change during these these trials

1.66 = 4.0 Y

1.66 2.0

1.0 = [2.0] Y

B is zero order (not included in rate law)

Rate α [A]x[B]

y[C]

z

Trial 4/1: ignore A and B, no change during these these trials

6.65 = 4.0 z

1.66 2.0

4.0 = [2.0] Y

C is second order (proportional squared)

Rate = k[A][C]2

b. What is the value of the rate constant

k = Rate_ = 1.66 = 0.42

[A][C]2

[1.0][2.0]2

c. What is the rate if [A] = 1.5 mol/L, [B] = 2.5 mol/L, [C] = 0.25 mol/L?

Rate = k[A][C]2 = (0.42)(1.5)(0.25)

2 = 0.039 mol/L s

d. What is the [C] in trial 5?

[C] = √ Rate_ = √ 7.25 = 3.4 mol/L

k[A]

[0.42][1.5]

8. Write the rate laws for the following elementary reactions

a. N2 (g) + 3 H2 (g) 2 NH3 (g)

Rate = k[N2][H2]3

b. N2 (g) + 2 O2 (g) 2 NO2 (g)

Rate = k[N2][O2]2


c. H2 (g) + Cl2 (g) 2 HCl (g)

Rate = k[H2][Cl2]

Determine the value of k in 8(a), when [N2] = 0.50 mol/L and

[H2] = 1.5 mol/L and NH3 is produced at a rate of 2.5 mol/L s.

Rate = k[N2][H2]3

k = Rate_ = 2.5 mol/L s = 1.5

[N2][H2]3

[0.50][1.5]3

EQUILIBRIUM PROBLEMS

1. At 773 K, the reaction 2 NO (g) + O2 (g) 2 NO2 (g) produces the following concentrations:

[NO] = 3.49 X 10-4

M; [O2] = 0.80M; [NO2] = 0.25M.

a. What is the equilibrium constant expression for this reaction?

Kc = _[NO]2

_

[NO]2[O2]

b. What is the equilibrium constant for the reaction?

Kc = _[NO]2

_ Kc = _[3.49 X 10-4

M]2_ = 6.4 x 10

5

[NO]2[O2] [0.25M]

2[0.80M]

2. Phosphorous pentachloride decomposes to phosphorous trichloride according to this equation:

PCl5 (g) PCl3 (g) + Cl2 (g) Keq = 1.00 x 10-3

At equilibrium, [PCl5] = 1.00M and [Cl2] = 3.16 x 10-2

M.

a. Calculate the concentration of PCl3.

[I] 0 0

[C] +x +x

[E] 1.00 3.16 x 10-2

3.16 x 10-2

The completed ICE table shows that both products would have the same equilibrium concentration.

BUT, if you plug it in and solve, you get the same answer.


Kc = _[ PCl3][ Cl2]_

[PCl5]

[PCl3] = Kc [PCl5] _ = 1.00 x 10-3

[1.00] _ = 3.16 x 10-2

M

[Cl2]

[3.16 x 10-2

]

b. Chlorine gas, Cl2 (g) is toxic. If we could manipulate the pressure of the container for this

reaction, how can we minimize the formation of this gas? Explain using your knowledge of Le

Chatelier’s principles.

Increase the pressure – increases likelihood of collisions between product side to a greater

advantage than reactant side. Reverse reaction favoured. Shift left.

3. Consider the reaction:

2 HBr (g) H2 (g) + Br2 (g)

Initially, 0.25 mol of hydrogen and 0.25 mol of bromine are placed in 500.0 mL reaction vessel and

heated. The Kc for the reaction is 0.020. Calculate the concentrations at equilibrium.

[I] 0 0.5 0.5

[C] +2x -x -x

[E] 2x 0.5-x 0.5-x

(.78 M) (.11 M) (0.11 M)

Kc = _[ H2][ Br2]_

[HBr]2

√0.020 = √ [ 0.5 - x]2

_ 0.020 = [ 0.5 - x]_

[2x]2 [2x]

0.141 = [ 0. 5 - x]_ 0.283x = 0.5-x 1.283x = 0.5 x = 0.39

[2x]

4. For the given reaction of methanol, the equilibrium concentrations were found to be [CO] = 0.170 M, [H2]

= 0.322, [CH3OH] = 0.0406. Find the value of Kc.

CH3OH (g) CO (g) + 2 H2 (g)

Kc = _[ CO][H2]2

_ = [0.170][0.322]2 = 0.434

[CH3OH] [0.0406]


5. Hydrogen peroxide can be decomposed as follows:

H2O2 (l) H2 (g) + O2 (g) (endothermic)

Equilibrium is established in a 10 L flask at room temperature. Predict the direction of the

equilibrium shift.

a. Hydrogen gas is added to the flask.

Shift left – increased collisions increases reverse rate b. The temperature is raised to 500 ºC.

Shift right - increased energy favours forward (endothermic) reaction. c. The entire mixture is compressed into a smaller volume.

Shift left – smaller space means more collisions. Reverse reaction rate increases.

ACIDS AND BASES PROBLEMS

1. You have water at 25o to which you add 0.050 M NaOH.

Complete the chart:

[H3O+] [OH

-]

Before adding NaOH 1.0 x 10-7

M 1.0 x 10-7

M

After adding NaOH 2.0 x 10-13

M 0.05 M (+ 1.0 x 10-7 M)

Strong base: all the OH- comes from the base, water contributes little.

Kw = [OH-][H3O

+] 1.0 x 10

-14 = [0.05 M][H3O

+]

[H3O+] = 1.0 x 10

-14 = 2.0 x 10

-13 M

[0.05 M]

2. This time with a weak base. Let’s add 0.0050 M NH4OH. Kb =1.77x 10-5

.

NH4OH(aq) + H2O(l) NH4+

(aq) + OH- (aq)

[I] 0.0050 0 0

[C] -x +x +x

[E] 0.0050-x +x +x

Kb = _[ NH4+][ OH

-]_ 1.77x 10

-5 = __x

2__

[NH4OH] 0.0050 (assume x is small and negligible)

(0.0050)(1.77 x 10-5

) = x2

√(8.9 x 10-8

) = x x = 3.0 x 10-4

M

Kw = [OH-][H3O

+] 1.0 x 10

-14 = [3.0 x 10

-4][H3O

+]

[H3O+] = 1.0 x 10

-14 = 3.3 x 10

-11 M

[3.0 x 10-4

]

[H3O+] [OH

-]

Before adding NH4OH 1.0 x 10-7

M 1.0 x 10-7

M

After adding NH4OH 3.3 x 10-11

M 3.0 x 10-4

M


3. If an acid has a hydronium concentration of 0.000036 M, what is the pOH of the solution?

pH = -log[0.000036]

= 4.4

pOH = 14-4.4 = 9.6

4. What is the hydroxide ion concentration of a solution with a pH of 4.9?

pOH = 14 - 4.9 = 9.1

[OH-] = 10

-9.1 = 8.0 x 10

-10 M

5. What is the % ionization of a 2.5 M solution of acetic acid (Ka = 1.76 x 10-5

).

Weak acid – needs ICE table:

CH3COOH(aq) + H2O(l) CH3COO- (aq) + H3O

+ (aq)

[I] 2.5 M 0 0

[C] -x +x +x

[E] 2.5-x +x +x

Ka = _[ CH3COO- ][ H3O

+ ]_ 1.76x 10

-5 = __x

2__

[CH3COOH] 2.5 (assume x is small and negligible)

(2.5)(1.76 x 10-5

) = x2

√(4.4 x 10-5

) = x x = 0.0066 M

%diss = [ H3O+

]_ x 100 = [0.0066] x 100 = 0.27%

[CH3COOH] [2.5]

6. The pH of 0.50M HNO2 is 4.85. Find Ka.

HNO2 (aq) + H2O(l) NO2- (aq) + H3O

+ (aq)

[I] 0.50 M 0 0

[C] -x +x +x

[E] 0.50-x +x +x

[H+] = 10

-4.85 = 1.4 x 10

-5 M

Ka = _[ NO2- ][ H3O

+ ]_

[HNO2]

(1.4 x 10-5

) (1.4 x 10-5

) = 3.9 x 10-10

(0.50)


7. What is the concentration of phosphoric acid if 15.0 ml of the solution is neutralized by 38.5 mL of a

0.15 M NaOH solution?

3 NaOH + 1 H3PO4 Na3PO4 + 3 H2O

Use the balanced equation to find moles of base, use the molar ratio to convert into moles of acid and

calculate concentration (pay attention to units!):

(0.15 mol NaOH) x (0.0385 L NaOH) x 1 mole H3PO4 x ___1____ = 0.128 M H3PO4

1L 3 mol NaOH 0.015 L H3PO4

8. Suppose 3.5% of a weak solution of 0.020 M monoprotic acid is ionized. Calculate the Ka.

HA (aq) + H2O(l) A- (aq) + H3O

+ (aq)

[I] 0.020 M 0 0

[C] -x +x +x

[E] 0.020 -x +x +x

%diss = [ H3O+

]_ x 100 (%diss) [HA] = [ H3O+

]

[HA] 100

(3.5%) [0.020 M] = [ H3O+

] [ H3O+

] = 0.0007 M

100

Ka = _[ A- ][ H3O

+ ]_ (concentration of both ions are the same – x)

[HA]

(0.0007) (0.0007) = 2.45 x 10-5

(0.020)

9. Complete and balance the following reactions:

a. Sulphuric acid plus potassium hydroxide

H2SO4(aq) + 2 KOH(aq) K2SO4(aq) + 2 H2O(l)

b. Phosphoric acid + Ca(OH)2

2 H3PO4(aq) + 3 Ca(OH)2(aq) Ca3(PO4)2(aq) + 6 H2O(l)

c. Nitric acid and magnesium hydroxide

2 HNO3(aq) + Mg(OH)2(aq) Mg(NO3)2(aq) + 2 H2O(l)


10. A solution was made by dissolving 28.5 g of KOH in 0.50 L of water. If 0.250 L of this solution was

titrated with 0.136 L of H2SO4, what is the molarity of the acid? 0.938 M

2 KOH + H2SO4 K2SO4 + 2 H2O

Be careful here, you have two volumes of base. You must use the first to find the original concentration

of the base, than use the second to find the number of moles of base you actually used.

First convert grams to moles and divide by volume to get concentration of base:

(28.5 g KOH) x 1 mole KOH x ___1____ = 1.018 M KOH

56 g KOH 0.50 L KOH

Use the balanced equation to find moles of base, use the molar ratio to convert into moles of acid and

calculate concentration (pay attention to units!):

(1.018 mol KOH) x (0.250 L KOH) x 1 mole H2SO4 x ___1____ = 0.936 M H2SO4

1L 2 mol KOH 0.136 L H2SO4

ELECTROCHEMISTRY

1. Consider the particles: Na+, Mg, Cl

-, F2, H2O

a) Which can be oxidized?

Which can lose electrons: Mg, Cl-, H2O

b) Which can be reduced?

Which can gain electrons: Na+, F2, H2O

c) Which is the best reducing agent?

Which one wants to lose electrons and has the lowest potential: Mg

2. For each of the following cells indicate which half-cell is the anode and which is the cathode.

Line notation is written Oxidation cell than Reduction cell. Oxidation occurs at the Anode, Reduction

at the Cathode.

a) Zn│Zn2+

║Cu2+

│Cu Anode: Zn Cathode: Cu

b) Ag + Ni2+

Ag

+ + Ni Anode: Ag Cathode: Ni

c) Sn│Sn2+

║Au3+

│Au Anode: Sn Cathode: Au


3. Sketch a galvanic cell between copper and barium. Show the direction of electron flow and ion

migration, identify the anode and cathode, the species present in each half-cell, indicate the voltage on a

voltmeter, and label the electrodes. Also show the half-reactions, overall cell reaction and the Emf

calculation.

Cathode reaction: Cu2+

+ 2e- Cu(s)

Anode reaction: Ba(s) Ba2+

+ 2e-

Overall reaction: Ba(s) + Cu2+

Ba2+

+ Cu(s)

What is the cell voltage? 2.90 + 0.34 = + 3.24 V

4. For each of the following cells write half-reactions, give the balanced net cell reaction, and calculate

the Emf of the cell.

a) Al3+

(aq) + Mg(s) → Al(s) + Mg2+

(aq)

b) Cr(s)│Cr3+

(aq) (1.0 mol/L)║Ag+

(aq) (1.0 mol/L)│Ag(s)

a) 2 (Al3+

(aq) + 3 e- → Al(s))

3 (Mg(s) → 2 e- + Mg

2+(aq)) Balance half-reactions to make electrons equal.

2 Al3+

(aq) + 3 Mg(s) → 2 Al(s) + 3 Mg2+

(aq)

Ecell = 2.37 + (-1.66) = + 0.71 V

b) 3 (Ag1+

(aq) + 1 e- → Ag(s))

1 (Cr(s) → 3 e- + Cr

3+(aq))

1 Cr(s) + 3 Ag1+

(aq) → 1 Cr3+

(aq) + 3 Ag (s)

Ecell = 0.74 + 0.80 = + 1.54 V

Ba 2+

NO3- Cu

2+

NO3-

Cu

Cathode

+ 3.24 V

K+ NO3-

Anode

Ba


5. Write out anode and cathode reactions for the electrolysis of the following.

a) CaCl2(l) b) CuI2(l)

Electrolysis is the reverse – what you expect to oxidize is reduced and vice versa.

a) Anode (O): 2 Cl- Cl2 (g) + 2 e

-

Cathode (R): Ca2+

+ 2e- Ca(s)

b) Anode (O): 2 I- I2 (g) + 2 e

-

Cathode (R): Cu2+

+ 2e- Cu(s)

6. How many coulombs of charge must pass through a cell in order to deposit 9.45 g of zinc from a solution

of zinc nitrate?

Since we are only considering zinc ion converting to solid, we can ignore the spectator nitrate ion:

Zn2+

+ 2 e- Zn (s)

Use stoich to find the number of moles of electrons:

(9.45 g Zn) x 1 mole Zn x _2 mole-_ = 0.291 mol e

-

65 g Zn 1 mole Zn

Use Faraday’s constant to find coulombs:

(0.291 mol e-) x 96500 C = 28100 C

1 mol e-

7. What happens in an oxidation reaction in terms of electrons?

Oxidation reaction has an atom losing electrons to form an ion.

8. What happens in a reduction reaction in terms of electrons?

Reduction reaction has an ion gaining electrons to form an atom.

9. Assign oxidation numbers to each atom in the following compounds.

+1 +5 -2 +1 +5 -2

a) H3SO4 b. HNO3


10. Which reactant in the following reaction is oxidized and which is the oxidizing agent?

+2 +2 +4 0

CO(g) + PbO(s) → CO2(g) + Pb(s)

Remember name the atom for O/R and the compound as the agents:

O: C

R: Pb

OA: PbO

RA: CO

11. Balance the following equation using the oxidation number method.

Because the O/R is the same atom, you have to massage the coefficient on the product side.

1 e- gained x 5 atoms = 5

-1 +5 0

5 HBr + 1 HBrO3 → 3 Br2 + 3 H2O

5 e- gained x 1 atoms = 5

12. Balance the following equations in acid or base as indicated in brackets.

a. Cu + HNO3 → Cu2+

+ NO2 + H2O (acid)

b. N2O4 + Br- → NO2

- + BrO3

- (base)

a) O: Cu Cu2+

+ 2e-

R: 2x( H+

+ HNO3 + 1e- NO2 + H2O )

O: Cu Cu2+

+ 2e-

R: 2H+

+ 2 HNO3 + 2 e- 2 NO2 + 2 H2O

NET: 2H+

+ 2 HNO3 + Cu 2 NO2 + Cu2+

+ 2 H2O

b) O: 6 OH- + 3 H2O + Br

- BrO3

- + 6e

- + 6 H

+ + 6 OH

-

R: 3x ( N2O4 + 2e- 2 NO2

- )

O: 6 OH- + 3 H2O + Br

- BrO3

- + 6e

- + 6 H

+ + 6 OH

-

R: 3 N2O4 + 6e- 6 NO2

-

NET: 6 OH- + 3 N2O4 + Br

- BrO3

- + 6 NO2

- + 3 H2O


13. Examine the cell below and answer the questions that follow.

a) Cathode reaction: Ag+ + 1e

- Ag(s)

b) Anode reaction: Sn(s) Sn2+

+ 2e-

c) Overall reaction: 2 Ag+ + Sn(s) 2 Ag(s) + Sn

2+

d) What is the cell voltage? Ecell = 0.14 + 0.80 = + 0.94 V

e) Label anode and cathode.

f) Indicate the direction of travel of ions and electrons.

g) While the cell operates which electrode gains in mass? Cathode

h) While the cell operates which electrode decreases in mass? Anode

Ag +

NO3- Sn

2+

NO3-

Ag Sn

Anode

+ 0.94 V

K+ NO3-

Cathode

mr. storie 40s chemistry final exam reviewchem+final+exam...mr. storie 40s chemistry final exam...

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