Northern Arizona University Quiz #11 CHM 152, General Chemistry II Dr. Brandon Cruickshank Sections 2 & 3, Spring 2004 April 26, 2004 Due by 5:00 p.m. on Friday, April 30. Name _______________________ 1. In the following nuclear equation, identify the missing product.

HXCa 11

4320 +→α+

a) Ti46

22 b) 46 21Sc c) Ti4422 d) 42 Ar18

e) none of these

Remember that an alpha particle has an atomic number of 2 and a mass number of 4. 2. is a radioactive nuclide. When one of these atoms decays, a series of α and βTh232

90− particle emissions occur,

taking the atom through many transformations to end up as an atom of 208 . How many α particles are

emitted in converting into 208 ?

Pb82

Th23290 Pb82

a) 6 b) 8 c) 2 d) 24 e) 4

A beta particle will not change the mass number. The difference in mass numbers is 232 − 208 = 24. An

alpha particle has a mass number of 4, so we need 6 alpha particles to equal a mass of 24. 3. A sample of wood from an Egyptian mummy case gives a 14C count of 9.4 cpm/g C (counts per minute per

gram of carbon). How old is the wood? (The initial decay rate of 14C is 15.3 cpm/g C, and its half-life is 5730 yrs.)

a) 6400 yr b) 4570 yr c) 4030 yr d) 3420 yr e) none of these

12

0.693=t

k

k = 1.21 × 10−4 1/yr

0

[A]ln

[A]

= −

t kt

49.4ln (1.21 10 1/yr)15.3

− = − ×

t

t = 4030 yr 4. If a tree dies and the trunk remains undisturbed for 13,750 years, what percentage of original 14C is still

present? (Half-life of 14C = 5730 yrs)

a) 5.20% b) 18.9 % c) 2.20% d) 45.0% e) none of these

12

0.693=t

k

k = 1.21 × 10−4 1/yr ln[A]t = −kt + ln[A]0 ln[A]t = −(1.21 × 10−4 1/yr)(13,750 yr) + ln(100) ln[A]t = 2.94 [A]t = e2.94 = 18.9% 5. How many electrons are transferred in the following reaction?

2 ClO3− + 12 H+ + 10 I− → 5 I2 + Cl2 + 6 H2O

a) 12 b) 5 c) 2 d) 30 e) 10

The above equation is balanced. Choose either half-reaction to determine the number of electrons

transferred. The iodide/iodine half-reaction is the easier one, so let's use that one. 10 I− → 5 I2 + 10 e− 6. When the equation for the following reaction in basic solution is balanced, what is the sum of the coefficients?

MnO4− (aq) + CN− (aq) → MnO2 (s) + CNO− (aq)

a) 13 b) 8 c) 10 d) 20 e) 11

3(CN− + H2O → CNO− + 2H+ + 2e−)

2(3e− + 4H+ + MnO4− → MnO2 + 2H2O)

3CN− + 2H+ + 2MnO4− → 3CNO− + 2MnO2 + H2O

This is the balanced equation in acidic solution. Add 2OH− to each side of the equation to balance in basic

solution.

3CN− + 2H2O + 2MnO4− → 3CNO− + 2MnO2 + H2O + 2OH−

Canceling one water molecule from each side of the equation gives:

3CN− + H2O + 2MnO4− → 3CNO− + 2MnO2 + 2OH−

The coefficients in the balanced equation add to 13. 7. What is ∆G° at 25°C for the reaction below?

2 Au3+ + 3 Ni → 3 Ni2+ + 2 Au

a) −5.07 × 102 kJ b) +5.07 × 102 kJ c) −7.24 × 102 kJ d) +1.01 × 103 kJ e) −1.01 × 103 kJ

∆G° = −nF cellE

First, calculate E , then calculate ∆G°. cell Au3+ + 3e− → Au red 1.50 V=E

Ni → Ni2+ + 2e− oxi 0.25 V=E cell oxi red 1.50 V 0.25 V 1.75 V= + = + =E E E ∆G° = −nF cellE ∆G° = −(6 mol)(96,500 J/V⋅mol)(1.75 V) ∆G° = −1.01 × 106 J = −1.01 × 103 kJ 8. For a reaction in a galvanic cell both ∆H° and ∆S° are positive. Which of the following is true?

a) E will increase with an increase in temperature. ocell

b) E will decrease with an increase in temperature. cell

c) E will not change when the temperature increases. cell d) ∆G° > 0 for all temperatures. e) None of the above statements is true. This reaction has a favorable ∆S and an unfavorable ∆H. It will be temperature dependent.

Endothermic reactions are favored by an increase in temperature. This should make sense because endothermic reactions absorb heat.

9. Given the following standard reduction potentials in acidic solution, the strongest reducing agent would be?

E°(V) Al3+ + 3 e− ↔ Al (s) −1.66

AgBr (s) + e− ↔ Ag (s) + Br− +0.07

Sn4+ + 2 e− ↔ Sn2+ +0.13

Fe3+ + e− ↔ Fe2+ +0.77

a) Fe3+ b) Fe2+ c) Br− d) Al3+ e) Al The strongest reducing agent will be the substance that is most easily oxidized. Al(s) is the most easily

oxidized with an oxidation potential of +1.66 V. 10. Copper will spontaneously reduce which of the following?

a) Fe2+ and Ag+ b) Fe2+ c) Ag+ d) Al3+ e) Fe2+ and Al3+ Looking at Table 19.1 of the text, you can use the diagonal rule. Cu is diagonally across and below Ag+.

You could also calculate the cell potential.

Ag+ + e− → Ag red 0.80 V=E

Cu → Cu2+ + 2e− oxi 0.34 V= −E

cell oxi red 0.80 V ( 0.34 V) 0.46 V= + = + − =E E E

A positive cell potential indicates that the reaction is spontaneous. 11. Calculate the equilibrium constant at 25°C for the reaction:

Sn2+ + Ni → Sn + Ni2+

a) 1.9 × 10−4 b) 2.1 c) 8.6 d) 1.5 × 1013 e) 5.2 × 103

cell0.0257 V ln=E K

n

First, calculate the cell potential, then calculate the equilibrium constant.

Sn2+ + 2e− → Sn red 0.14 V= −E

Ni → Ni2+ + 2e− oxi 0.25 V=E cell oxi red 0.14 V 0.25 V 0.11 V= + = − + =E E E

cell0.0257 V ln=E K

n

0.0257 V0.11 V ln2

= K

lnK = 8.56 K = e8.56 = 5.2 × 103 12. Consider an electrochemical cell based on the spontaneous reaction

2 AgCl (s) + Zn (s) → 2 Ag (s) + 2 Cl− + Zn2+

If the zinc ion concentration is kept constant at 1 M, and the chloride ion concentration is decreased from 1 M to 0.001 M, the expected change in cell voltage is:

a) increase by 0.06 V b) increase by 0.18 V c) decrease 0.06 V d) decrease by 0.18 V e) increase by 0.35 V

0.0257 V ln= −E E Qn

. This is the Nernst equation which will allow you to calculate a non-standard cell

potential. Since the problem is only asking for the change in cell voltage, we only need to calculate 0.0257 V ln− Q

n

change in cell voltage = 0.0257 V ln− Qn

= ( )2 20.0257 V ln [Cl ] [Zn ]2

− +−

change in cell voltage = ( )20.0257 V ln [0.001 ] [1 ]2

− M M = +0.18 V