Kaplan MCAT General Chemistry Review

Remember: This last step is extremely important. If mass and charge aren’t balanced, then you made an error in one of the previous steps.

Similar Questions

1) Which atom is being oxidized in the original equation? Which is being reduced? Identify the oxidizing and reducing agents.

2) A disproportionation is a redox reaction in which the same species is both oxidized and reduced during the course of the reaction. One such reaction is shown below. Balance the reaction, assuming that it takes place in acidic solution:

PbSO₄ (s) Pb (s) +

PbO₂ (s) + SO₄^2- (aq)

3) Dentists often use zinc amalgams to make temporary crowns for their patients. It is absolutely vital that they keep the zinc amalgam dry. Any exposure to water would cause pain to the patient and might even crack a tooth. The reaction of zinc metal with water is shown below: Zn (s) + H₂O (l) Zn²⁺ (aq) + H₂ (g)

Balance this reaction, assuming that it takes place in basic solution. Why would exposure to water cause the crown, and perhaps the tooth, to crack?

Electrochemical Cells

Key Concepts

Chapter 11

Oxidation/reduction

Electrochemical cells

Work

Stoichiometry

E°_cell = E°_cathode–E°_anode (V)

G° =–nFE°_cell (kJ/mol)

A galvanic cell is to be constructed using the

MnO₄^- | Mn²⁺ (E°_red = 1.49 V) and Zn²⁺ | Zn

(E°_red = -0.76 V) couples placed in an acidic solution. Assume that all potentials given are measured against the standard hydrogen electrode at 298 K and that all reagents are present in 1 M concentration (their standard states). What is the maximum possible work output of this cell per mole of reactant if it is used to run an electric motor for one hour at room temperature (298 K)? During this amount of time, how much Zn metal would be necessary to run the cell, given a current of 5 A?

1) Determine which half-reaction is occurring at the anode and which is occurring at the cathode of the cell.

MnO₄^- (aq) + 5 e^- Mn²⁺ (aq) E°_red = 1.49 V

Zn²⁺ (aq) + 2 e^- Zn (s) E°_red = -0.76 V

Compare the standard reduction potentials for both reactions. The permanganate reduction potential is greater than the zinc potential, so it would prefer to be reduced and zinc oxidized. Therefore, the zinc is being oxidized at the anode, and the manganese is being reduced at the cathode.

Remember: Oxidation occurs at the anode. (Hint: They both start with a vowel.)

Takeaways

Double-check your work when you balance the cell equation to make sure that you haven’t made any arithmetic errors. One small addition or subtraction mistake can have drastic consequences!

2) Write a balanced reaction for the cell.

Balance the reactions one at a time.

MnO₄^- Mn²⁺

Balance oxygen with water, then hydrogen with acid (H⁺).

8 H⁺ + MnO₄^- Mn²⁺ + 4 H₂O

Balance overall charge with electrons.

8 H⁺ + MnO₄^- + 5 e^- Mn²⁺ + 4 H₂O

This one is easy; all you have to do is balance electrons.

Zn Zn²⁺ + 2 e^-

To combine both equations, we need to multiply each by the appropriate integer to get to the lowest common multiple of 2 and 5, which is 10.

16 H⁺ + 2 MnO₄^- + 10 e^- 2 Mn²⁺ + 8 H₂O

5 Zn 5 Zn²⁺ + 10 e^-

Things to Watch Out For

Remember to balance the electron flow in order to figure out the maximum amount of work that the cell can perform.

Now add the equations up to get the balanced cell equation, and you’re golden.

16 H⁺ (aq) + 2 MnO₄^- (aq) + 5 Zn (s) 2 Mn²⁺ (aq)

+ 5 Zn ²⁺ (aq) + 8 H₂O (l)

3) Calculate the standard potential for the cell as a whole.

E°_cell = 1.49 V - (-0.76 V) = 2.25 V

Use the equation E°_cell = E°_cathode - E°_anode. Because the standard potential for the cell is positive, this confirms that this is a galvanic (or voltaic) cell—once you hook up the electrodes and immerse them in the designated solutions, current will start to flow on its own.

4) ComputeG° for the cell.

G° = -(10 mol e^-)(10⁵ C mol^-1)(2.25 V) = -2.25 × 10⁶ J mol^-1

maximum work output per mole of reactant = 2.25 × 10³ KJ mol^-1

Use the equation G° = -nFE°_cell. The upper limit on the amount of work a reaction can perform is the same thing as G°.

Remember: Power is work over time, and 1 h = 3,600 s 4 × 10³s.

5) Use Faraday’s constant to determine the number of moles of electrons transferred and to do any stoichiometric calculations.