10. Electrochemistry

Question 1

What is the oxidation state (oxidation number) of an atom?

Answer 1

The oxidation number explains how many electrons that atom is donating or accepting in the overall bonding of a molecule.

Any reaction in which the oxidation numbers of the reactants change is called a redox reaction.

Question 2

What are the seven oxidation state rules in order?

Answer 2

1. The ON is 0 for any element in its standard state (O2 g = 0, Br2 L = 0, etc.)
2. The ON of the whole molecule must equal the charge on that molecule
3. Group 1 metals = +1, Group 2 metals = +2
4. Fluorine is -1
5. hydrogen is -1 when bound to something less EN than carbon, +1 when bound to something more EN than carbon, and 0 when bound to carbon
6. oxygen is always -2 unless its bound in a peroxide in which case it is -1
7. halogens are usually -1

Question 3

What is a galvanic (voltaic) cell? Explain its general set up.

Answer 3

Redox reactions involve the movement of electrons. The movement of charge is current. A galvanic cell uses spontaneous redox reactions to generate some current..

A galvanic cell is composed of two separate beakers filled with electrolytes. One beaker has a metal cathode and one has a metal anode which are connected by wire to an ammeter. At the anode, the electrodes atoms oxidize and free electrons enter the wire and pass over to the cathode. The electrons jump off at the cathode (reducing the solution).

Thus electrons flow anode to cathode

Question 4

What is the anode and cathode of a galvanic cell?

Answer 4

An Ox, Red Cat.

Thus, oxidation occurs at the anode. Here the electrodes atoms are oxidized and free electrons enter the wire.

Reduction occurs at the cathode. Electrons arrive here and jump off onto electrolytes thereby reducing them.

Question 5

What is the salt bridge in a galvanic cell? Which way does the salt ions move?

Answer 5

The salt bridge is a connecting bridge between the two half cells that contains some salt (e.g. KNO3) which allows the galvanic cell to keep running.

e.g. anode: ZnSO4, cathode: CuSO4

At the anode, metal cations are being produced as electrons leave the metal cathode. The anionic ion in the salt bridge (NO3-) thus moves to the anode to interact with the produced cations (Zn+).

At the cathode, positive cations are being removed from the solution as electrons combine with them to form metal (Cu). This leaves a higher [SO4-] and so the cationic ion of the salt bridge (K+) moves to the cathode to interact with these left over anions.n

Salt-bridge: cations move to the cathode, anions move to the anode.

Question 6

What is a shorthand cell diagram?

Answer 6

Anode | Anode solution || Cathode solution | Cathode

A before C

If concentrations not given, assume 1 M.

Question 7

What would occur if you removed the salt bridge from a galvanic cell?

Answer 7

Anode: atoms continue to oxidize, leaving extra cations in solution
Cathode: electrons continue to reduce cations in solution, leaving extra anions in solution.

With no salt bridge, there is a separation of charge which creates a capacitor (C = QV).

Question 8

What are standard reduction potentials (E°)?

Answer 8

The standard reduction potential is the voltage of any have redox reaction relative to the potential of a hydrogen reduction (2H+ + 2e –> H2) which is E° = 0.

If we need to find the voltage (E°) of an oxidation half reaction, we just find the standard reduction potential and flip the sign.

Question 9

What is ΔG° = - nFE° for galvanic cells?

Answer 9

This is the standard conditions (1 atm, 298K, 1M) free energy of a galvanic cell. We use this equation to determine if a redox reaction will be spontaenous and thus generate current.

What is ΔG° = nFE°

n = moles of electrons transferred 
F = Faradays constant = the amount of charge in a mole of electrons = 96,000 C. 

Since a spontaneous reaction has a negative What is ΔG°, E° must be positive for a redox reaction to be spontaneous.

Question 10

Explain how you find the overall E° of the reaction

Au + Cu(2+) –> Au(3+) + Cu

Au reduction potential E° = 1.5V
Cu reduction potential E° = 0.34V

Answer 10

Au —> Au(3+) + 3e
Cu(2+) + 2e–> Cu

We must first balance the equation (balance the electrons)
2Au –> 2Au(3+) + 6e
3Cu(2+) + 6e –> 3Cu

despite balancing the equation, we do NOT multiply the standard potentials.

We must reverse the sign of the oxidation reaction (Au reduction: E° = -1.5V).

E° cell = -1.5V + 0.34V = -1.16, which is not spontaneous based on ΔG° = - nFE°

Question 11

Why do we balance the electrons of a redox reaction if it does not effect E° cell?

Answer 11

ΔG° = - nFE°

If we want to find the ΔG° of the reaction, we must know the moles of electrons (n).

Question 12

What does standard reduction potentials tell us about oxidizing and reducing agents?

Answer 12

The more positive the standard reduction potential, the more the atom WANTS to be reduced, making it a stronger oxidizing agent.

The more negative the standard reduction potential, the more the atom WANTS to be oxidized, making it a stronger reducing agent.

Question 13

t or f, coefficients affect standard reduction potentials.

Answer 13

FALSE

Question 14

When we are not at standard conditions (1M, 1 atm, 298K), we cannot use ΔG°.
The Nernst equation ΔG = ΔG° + RTlnQ.

We can then plug in ΔG° = -nFE° and ΔG = -nFE, to find the E of a galvanic cell.

Answer 14

This results in…

E = E° - (RT / nF)lnQ

Question 15

What is true of E = E° - (RT / nF)lnQ when equilibrium is established?

Answer 15

When equilibrium is established, Q = K, ΔG = 0, and E = 0.

Question 16

What is a concentration cell?

Answer 16

A concentration cell is a galvanic cell with two of the same electrodes but each half cell has different ion concentrations. Here E° = 0, but we are not at standard conditions (since concentrations are random).

A higher [] starts at the cathode. As electrons combine with ions, the cathode solution [] goes down. Once the concentrations of anode and cathode are similar / equal, the reaction stops.

Question 17

Similar to acid / base reactions, redox reactions can undergo titrations. Here we add some titrant (e.g. an oxidizing agent) to some unknown solution (something that will be oxidized) until there is no ion left to react with in solution. When this occurs, the solution changes color. What is the equivalence point and half equivalence point of this?

Answer 17

equivalence point - when no more ions exist to react with the titrant being added.

The half equivalence point is the point in which [oxidized] = [un-oxidized] if the titrant is an oxidizing agent.

E = E° - (RT / nF)lnQ 
since ln (1) = 0, E = E°

Question 18

To generate a current, we can apply a voltage (a change in potential) to get charge to move. What is an electrolytic cell?

Answer 18

Unlike a galvanic cell, an electrolytic cell applies an external voltage to drive an unfavorable redox reaction. This process is called electrolysis.

Question 19

Explain the general set up of an electrolytic cell.

Answer 19

unlike a galvanic cell, an electrolytic cell occurs in one large beaker and uses two inert electrodes connected by wire also connected to a voltage source. The solution of electrodes create free ions (e.g. NaCl). The battery forces electrons off of Cl- in solution which then go through an inert electrode and through the wire. Electrons also arrive at the other electrode which react with Na+ creating Na solid.

oxidation still occurs at the anode (Cl- loses electrons) and reduction still occurs at the cathode (Na+ gaining electrons).

Question 20

t or f, electrons always move from anode to cathode.

Answer 20

true

Question 21

what sign is the anode and cathode in a galvanic and electrolytic cell?

Answer 21

galvanic: cathode is positive (it removes cations, leaves anions, thus cations move to the cathode). Anode is negative (it generates cations, thus anions move the the anode).
electrolytic: cathode is negative, anode is positive.

Question 22

t or f, a rechargeable battery is a galvanic cell (discharge) in combination with an electrolytic cell (recharge)

Answer 22

true

Question 23

t or f, Faraday’s law of electrolysis = the amount of chemical change is proportional to the amount of electrons that flow.

I = Q / t
Q = I x t

Answer 23

true

Question 24

Faraday’s law of electrolysis: If 5 amps of current flowed in a NaCl electrolytic cell for 1930 seconds, how much sodium metal is produced?

Answer 24

Q = I x t
Q = 5A x 1930s = 9650C 

convert charge to moles of electrons

9650C / 96,500 C / mole = 0.1 moles of e

Na+ + e –> Na. 1 mole of e gives 1 mole of Na, thus 0.1 moles of Na is produced.

0.1 moles Na (23 g / mole) = 2.3 grams of Na