Electrochemistry Obj B Solutions

P107

1. anode: the electrode which hosts oxidation

cathode: where reduction takes place

2. The cathode is (+) but its (-)'ly charged solution attracts (+) ions known as cations because, as reduction takes place, the positive ions are used up and (-) ions from the solution accumulate.

3. b. Cr = anode

Au = cathode

c. electrons flow from Cr to Au where they are picked up by Au⁺³

d. Au⁺³ + 3e à Au

Crà Cr^{+3 +} 3e

4. a. true

b. false

c. false

d. true

5. a. It would drop to zero.

b. It would also drop to zero.

c. Of course. The Mg would still get oxidized by the Al⁺³.

p 111

1. a. reducing agents are themselves oxidized. From the direction of the reaction, it is Cu that gets oxidized, not Fe⁺². So, yes Cu is a stronger reducing agent than Fe⁺².

b. Most of the time neutral metals are better reducing agents. But not with gold! From the direction of the reaction, it is Fe⁺² that gets oxidized, not Au. So, yes Fe^{+2 is} a stronger reducing agent than Au.

c. The reverse reaction is favoured. That should be obvious from what we argued in (a) and (b). If it's not obvious, you are being distracted by the Bruins' domination of the Habs. You can also check the E values.

d. The strongest oxidizing agent is the one that's best at removing electrons. So the answer is Au⁺³.

Another way of convincing yourself is : Au⁺³ + 3e --> Au E = +1.50 V. This E value is bigger than anything else from your table except fluorine. So from your table, only fluorine is a better oxidizing agent than Au⁺³.

2. a. Li

b. Li. If Cs was stronger the reverse reaction would occur. Also, Li--> Li⁺¹ + 1e ( E = 3.04 V: the largest value)

c. 3Li + Al⁺³ --> 3 Li⁺¹ + Al

d. Al⁺³ is hungrier for electrons than Li⁺¹ is.

3.. When they tell you that A + D⁺² react, then A + D⁺²-->A⁺²+D. This implies that D⁺²is better at getting reduced than A⁺². If you apply this to every reaction, you realize that D⁺²>B⁺²>E⁺²>C⁺²

4. Cr + Al⁺³ à no reaction, so Al is better than Cr at losing electrons

Mg + Al⁺³ à reaction, so Mg is better than Al at losing electrons

Ni⁺² + Al à reaction, so Al is better than Ni at losing electrons

Ni + Cr⁺³ à no reaction, so Cr is better than Ni at losing electrons

So from the best to the worst at losing electrons( best reducing agent) we have

Mg> Al>Cr> Ni

5. To answer the question you need to see from the table of reduction potentials that Cu⁺² is a better oxidizing agent than Cr⁺³. So if you are wondering whether you should store a solution containing Cr⁺³ ( from Cr(NO₃)₃))in copper, you're really wondering whether the following will happen:
Cr⁺³ + Cu --> Cu⁺² + Cr

But it won't because Cu⁺² wins out. So the answer is that, yes, the scientist should store the solution in the copper container, since nothing will happen to it.

6. Ag⁺ + 1e à Ag E = 0.80 V

Zn à Zn⁺² + 2e E = 0.76 V

produce a total of 1.56 V, greater than any other combination. Unfortunately they did not list it as one of the choices. So the answer is B

7. Use Cr⁺² + 2e à Cr E = -0.91 V

Pb⁺² + 2e à Pb E = -0.13 V,

So you need to reverse the first one to obtain a positive overall E value.

Cr à Cr⁺² + 2e E = +0.91 V

Pb⁺² + 2e à Pb E = -0.13 V

Overall: Cr + Pb⁺² à Cr⁺² + Pb E = 0.78 V

1. anode: Cr à Cr⁺² + 2e

2. Pb

3. Pb⁺²

4. Cr + Pb⁺² à Cr⁺² + Pb

5. E = 0.78 V

Red Booklet p 5-52

3. From tables:

Au⁺³ + 3e --> Au E = + 1.50 V

Al⁺³ + 3e --> Al E = - 1.66 V

Reverse 2nd eqn: Al àAl⁺³ + 3e E = 1.66 V

Au⁺³ + 3e àAu E = + 1.50 V

Overall: Al + Au⁺³ à Al⁺³ + Au E = 3.16 V

4. From tables: Cu⁺² + 2eà Cu E = 0.34 V

Cr⁺² + 2e à Cr E = -0.91V

Reverse 2nd eqn: CràCr⁺² + 2e E = 0.91V

Cu⁺² + 2e à Cu E = 0.34 V

Overall: Cr + Cu⁺² à Cr⁺² + Cu

Oxidation reaction: Cr-->Cr⁺² + 2e

Reduction reaction: Cu⁺² + 2e --> Cu

Net reaction: Cr + Cu⁺² --> Cr⁺² + Cu

Increase in mass: Cu

Decrease in mass: Cr ( because it turns into Cr⁺²)

Cell voltage: E = 1.25 V

5. a. E = 0.16 V spontaneous( implies that reaction will proceed as written)

b. E = -0.34 V not spontaneous( implies that reaction will only proceed in the opposite direction)

c. E = -1.24 V not spontaneous

d. E = -0.02 V not spontaneous

6. Pb⁺² + 2e --> Pb E = -0.13 V

X --> X⁺² + 2e E = - x ( negative because I'm reversing the equation to get the electrons to cancel out)

- x + -0.13 = 0.15

x = -0.28 V ( reduction potential)

Look up -0.28 V in table and it corresponds to Co.

But what if you assumed the following?

Pb -->Pb⁺² + 2e E = + 0.13 V

X⁺² + 2e -->X E = x

x + 0.13 = 0.15

x = 0.02 V, but there is nothing in our table with E = 0.02 V, so the first possibility is the one they have in mind.