Kaplan MCAT General Chemistry Review

4 × 10³ s (5 C s^-1) (10^-5 mol e^- C^-1) = 0.2 mol e^-

Remember that current is charge passing though a point per unit of time, and Faraday’s constant tells us how many coulombs of charge make up one mole of electrons.

The balanced half-reaction is used to determine the necessary mole ratio.

Similar Questions

1) How could you alter the cell setup to reverse the direction of current flow?

2) What would the cell potential be if Mn²⁺ and Zn²⁺ were at 2 M concentration and the MnO₄^- concentration remained at 1 M? Would changing the amount of zinc metal present in the cell change this potential? Why or why not?

3) Compute the minimum mass of potassium permanganate (KMnO₄) necessary to run the cell for the same amount of time as specified above.

The Nernst Equation

Key Concepts

Chapter 11

Oxidation and reduction

Electrochemistry

Equilibrium

Solubility equilibria

Nernst equation

E°_cell = E°_cathode - E°_anode (V)

A galvanic cell is created at 298 K using the following net reaction:

2 H⁺ (aq) + Ca (s) Ca²⁺ (aq) + H₂ (g)

Fluoride anions are added to the anode section of the cell only until precipitation is observed. Right at this point, the concentration of fluoride is 1.4 × 10^-2 M, the pH is measured to be 0, the pressure of hydrogen gas is 1 atm, and the measured cell voltage is 2.96 V. Given this information, compute the K_sp of CaF₂ at 298 K.

Additional information:

R = 8.314 J (mol K)^-1

Ca²⁺ (aq) + 2 e^- Ca (s) E°_red = -2.76 V

2 H⁺ (aq) + 2 e^- H₂ (g) E°_red = 0.00 V

F = 96,485 C mol^-1

1) Write down the expression for the K_sp.

The first part of this problem begins as with any other solubility problem: We need to write down the expression for the K_sp.

CaF₂ (s) Ca²⁺ (aq) + 2 F^- (aq)

We’re given the concentration of fluoride right when precipitation begins, so we can plug that right into the K_sp expression above. All we need is the concentration of Ca²⁺ ions, and we’re golden.

K_eq = [Ca²⁺][F^-]² = K_sp

Takeaways

Remember what the superscript ° means: that a reaction is at standard conditions. This means that reagents are at 1.0 M or 1 atm, depending on their phase. If you are working with an electrochemical cell where the concentrations are nonstandard, you must apply the Nernst equation to determine what the effect on the cell voltage will be.

2) Separate the net cell reaction into half-reactions and findE°_cell.

Ca (s) Ca²⁺ (aq) + 2 e^-E ° = 2.76 V

2 H⁺ (aq) + 2 e^- H₂ ( g) E° = 0.00 V

NET: 2 H⁺ (aq) + Ca (s) Ca²⁺ (aq) + H₂ (g)

E°_cell = 0.00 V - (-2.76 V) = 2.76 V

Now we know what the standard potential for the cell is. Our only problem is that, in the situation given in the problem, we are in nonstandard conditions, because the concentration of fluoride is not 1 M.

Remember: E°_cell = E°_cathode- E°_anode.

Things to Watch Out For

Don’t forget to check signs during problems that require computation. When logarithms and exponents are involved, one small sign error can have a massive impact on the answer!

3) Apply the Nernst equation.

The reaction quotient (Q) in this case is of the cell reaction. Recall that the pressure of hydrogen gas is 1 atmosphere. As the pH = 0, [H⁺] = 10^-0 = 1.0 M.

Rearrange the Nernst equation to solve for ln[Ca²⁺].

Start plugging in numbers. Here, n = 2 mol e^-, from the cell equation; F = 96,485 100,000 C mol^-1; T = 298 300 K; R 8 J (mol K)^-1; and 2.96 - 2.76 = 0.2 = 2 × 10^-1 V.

Here, assume–16.7 -20.

–20 = In [Ca²⁺]

–20 = 2.3 log[Ca²⁺]

Recall that ln x = 2.3 log x. Assume that 2.3 2.5 so that - –8.

–8 = log[Ca²⁺ ]

[Ca²⁺] = 10^-8

Similar Questions

1) If the K_sp of copper(I) bromide is 4.2 × 10^-8, compute the concentration of bromide necessary to cause precipitation in an electrochemical cell with the Cu|Cu⁺ and H⁺|H₂ couples. Assume the conditions are as follows: E°_red of Cu⁺ = 0.521 V; pH = 0; P_{H2 (g}) = 1 atm; T = 298 K; E_cell when precipitation begins = 0.82 V.

2) Compute the equilibrium constant at 298 K for the cell comprised of the Zn²⁺ | Zn (E_red° = -0.76 V) and MnO₄^- | Mn²⁺ (E_red° = 1.49 V) couples. Given this number, comment on the oxidizing ability of the permanganate anion.

3) A buffer solution is prepared that is 0.15 M in acetic acid and 0.05 M in sodium acetate. If oxidation is occuring at a platinum wire with 1 atm of H₂ bubbling over it that is submerged in the buffer solution, and the wire is connected to a standard Cu²⁺ | Cu half-cell (E_red° = 0.34 V), the measured cell voltage is 0.592 V. Based on this information, compute the pK_a of acetic acid.

Remember: When you absolutely must do computation, choose numbers that are easy to work with.