Resources:

Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we'll add them to "Resources" part of this page for other students to use.

Question 1

Check those statements that are correct.

Select one or more:  Question 2

What is the voltage for the reaction under standard conditions: 2Al(s) + 3Pb2+(aq) → 2Al3+(aq) + 3Pb(s)? Explanation

2Al(s) +  → 2Al3+(aq)  + 6e-   Ea= -1.66

3Pb2+(aq) + 6e-→  3Pb(s)     Ec=-0.13

Ecell = Ec-Ea

=-0.13-(-1.66)

=1.53

Question 3

Check the statement which are correct.

Select one or more:  Question 4

What is the voltage for the reaction under standard conditions: 3Ag(s) + Cr3+(aq) → 3Ag+(aq) +Cr(s)? (See above for reduction potentials.)

Standard conditions mean any amount of the solids (as long as there is some), solutions are 1 M, and gases are 1 atm. Explanation

3Ag(s) → 3Ag+(aq) + 3e-     Ea= 0.8

3e- + Cr3+(aq) →  Cr(s)        Ec=-0.13

Ecell = Ec - Ea

=-0.73-0.8

=-1.53

Question 5

Check the statement which are correct.

Select one or more:  Question 6

What is the voltage under standard conditons for the reaction: 2Ag(s) + I2(s) → 2Ag+(aq) + 2I-(aq)? Explanation

2Ag(s) → 2Ag+(aq) + 2e-   Ea=0.80

I2(s) + 2e-→  2I-(aq)   Ec=0.54

Ecell= Ec-Ea

=0.54-0.8

=-0.26

Question 7

Which or the following statements are true about a spontaneous oxidation-reduction reaction?

Select one or more:  Question 8

Find ΔGo in J for: Al3+(aq) + 3e- → Al(s) Eo = -1.66 V Explanation

ΔG=-nFE

=-3(96500)(-1.66)

=480,570

Question 9

Find ΔG in J for: X4+(aq) + 4e- → X(s) E = -1.62 V ΔG=-nFE

=-5(96500)(0.46)

=-221,950

Question 10

Which statement is INCORRECT for the following exothermic reaction?

Cu2+(aq) +Zn(s) → Cu(s) + Zn2+(aq)

Select one: Explanation

Cu2+(aq) + 2e-→ Cu(s)   Ec =-0.34

Zn(s) → 2e- + Zn2+(aq)  Ea=-0.76

Ecell= Ec-Ea

-0.34-(-0.76)

=1.1

Question 11

Which of the following statements is NOT correct about the Galvanic cell shown below?

A battery has a voltage and produces a current (flow of electrons) in a circuit. The voltage and current can be made to do useful work such as running an electric motor. A galvanic cell generates electricity like a battery.

Select one: Question 12

In the Zn-Cu galvanic cell shown above, which of the following statements are true?
Select one or more:   Question 13

What is the standard voltage or cell potential of a Cr-Hg cell? (Remember to switch one-half reaction to oxidation and to balance the number of electrons--use paper.)

Cr3+(aq)+ 3e- → Cr(s) Eo = -0.74 V
Hg2+(aq)+ 2e- → Hg(l) Eo = +0.789 V Ecell=Ec-Ea

=0.789-(-0.74)

=1.529

Question 14

For the above galvanic cell which statments are correct?

Select one or more:   Question 15

For the reaction given below the [H+] is increased from 1.00 M to 3.00 M. What happens to the voltage? Write i for increases, n for no change, and d for decreases.

Answer: Explanation

Increasing the reactant concentration above the stantard concentration increases the driving force or voltage to reduce the stress of more reactant.

Question 16

For the above reaction what happens to the voltage if a larger zinc electrode is used? Write i for increases, n for no change, and d for decreases.

Explanation

Recall, that the amount of a solid or pure liquid has no effect on equilibrium. A larger zinc electrode does increase the maximum current because there is more surface area for the oxidation of the zinc.

Question 17

For the above reaction the [Zn2+] is decreased from 1.00 M to 0.30 M. What happens to the voltage?

Question 18

For the above reaction the H2 pressure is increased from 1.00 atm to 4.2 atm. What happens to the voltage?

Question 19

For the above reaction the Cl- concentration is increased from 1.00 M to 4.50 M by the addition of NaCl. What happens to the voltage?

Explanation

The overall reaction has nothing to do with the chloride ion concentration.

Question 20

For the Zn-H2 cell, what is the voltage if [Zn2+] = 6.67 M, [H+] = 0.0194 M, and the hydrogen pressure is 18.0 atm. Explanation

Ecell=Ecell - 0.0592*log Q/n

Q=(6.67)(18)/(0.0194^2)=319003

=0.76-(0.0592)*log(319003)/2

=0.597

Question 21

What is n (the number of moles of electrons) transferred for the following oxidation/ reduction reaction?

2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

Explanation

2Al(s) → 2Al3+(aq) + 6e-

6e- + 6H+(aq) →  3H2(g)

Question 22

2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

For the above Al-H2 cell, what is the voltage if [Al3+] = 5.0561 M, [H+] = 5.3 M, and the hydrogen pressure is 0.0856 atm. Eo = 1.66 V. Explanation

Ecell=Ecell - 0.0592*log Q/n

Q= (0.0856)^3*(5.0561)^2/(5.3)^6= 0.000000723

=1.66- (0.0592)*log(0.000000723)/6

=1.72

Question 23

In the galvanic cell below, where the process is sponataneous, on which electode is the metal deposited and what is the metal?

Select one or more:  Question 24

In electrolysis a power supply is required to drive the reaction in the opposite direction to which it wants to go.

In the above example of electrolysis, on which electode is the metal deposited and what is the metal?

Select one or more:  Question 25

In the reduction of zinc ions to zinc:

Zn2+(aq) + 2e- → Zn(s)

how many moles of electrons are needed to produce 1 mole of zinc?

Question 26

If a current of 4.96 A flows for 15.32 hr, then what quantity of charge (in C) has flowed? Explanation

4.96A = 4.96 C/s * 15.32 hr *3600 sec/hr=273553.92

Question 27

In the reduction of aluminum ions to alumimum: Al3+(aq) + 3e- → Al(s)

how many moles of alumimun are produced by 0.76 moles of electrons? Explanation

Hint, 3 moles of electrons produce 1 mole of aluminum.

.76e-*1 Al/3mol e-= 0.25333

Question 28

If a current of 8.70 A flows for 18.03 hr, then how many moles of electrons have flowed? Explanation

8.70 A = 8.70 C/s * 18.03 hr * 3600 sec/hr =564699.6 * 1 mol/96500C= 5.85

Question 29

In the reduction of aluminum ions to alumimum: Al3+(aq) + 3e- → Al(s)

how many grams of alumimun are produced by a current of 1.81 A for 7.63 hr? Explanation

1.81 C/s * 7.63 hr * 3600 sec/hr=49717.08*1 mol/96500=0.515203* 1mol Al/3e- = 0.1717 mol Al * 26.98 g/ 1mol Al = 4.633 g