12.8: Standard Potentials and Equilibrium Constants

We can calculate Keq values from standard potentials if we by relating the previous expression for Gibbs free energy to the one we’ve learned this chapter.

We know from Ch. 9 that:G° =-RTlnKeq

We know from this chapter that:G°=-nFE°

Setting them equal to each other gives us:nFE° =RTlnKeq (which can be rearranged to)

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lnKeq =nFE°

RT

12.8: Standard Potentials and Equilibrium Constants

We can calculate the emf from standard potentials, so for any reaction that can be expressed as 2 half reactions, we can calculate Keq

RT/F = 0.025693 V @ STP, so we ca rewrite the equation above as:

€

lnKeq =nFE°

RT

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lnKeq =nE°

0.025693V Where V is Volts

A note on K values

Keq = ????KA = ????KB = ????KSP = ????

These are all equilibrium constants (hence the capital K) which means they are all equal to [Products]/[Reactants]

DO NO BECOME CONFUSED OR FORGET THIS!!!

The Triangle of Free Energy

Calculating Keq from Electrochemical Data

1. Write the balanced chemical equation for the reaction2. Scan the table of Standard Potentials or Appendix 2B and

find 2 equation that will combine to give you the equation from Step 1.

3. Reverse on the the reactions as needed (Remember: They are all written as Reduction reactions!)

4. Identify the value of n based upon your balanced chemical equation

5. Calculate E°6. Use

€

lnKeq =nE°

0.025693V

The Nernst Equation

As a reaction proceeds, the G approaches zero• At equilibrium, G = 0, right?

As a battery is discharged, the concentrations of reactants and products change until the emf across the electrodes is zero.• A dead battery is one in which G = 0

We can prove this by remembering that G = -nFE• If E is zero, then what must G be?

The enf of the cell varies with the concentration of species in the cell

The Nernst Equation

Recall the concept of the reaction quotient, Q from chapter 9?

• Q=K at equilibrium• Q is [Products]/[Reactants]

Gr = G°r + RT lnQ

Free energy of the actual reaction

Standard free energy of the reaction

Reaction quotient (How much stuff do we have?)

The Nernst Equation

Gr = G°r + RT lnQ

But Gr = -nFE and G°r = -nFE° , so:

-nFE = -nFE° + RTlnQ Divide by -nF

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E = E° - RT

nF

⎛

⎝ ⎜

⎞

⎠ ⎟lnQ

The Nernst Equation

This equation shows us how the concentration of reactants impacts the actual E of the cell at any given point in the reaction.

The Nernst Equation

@ and 1 atm, RT/F = 0.025693V, so:

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E = E° - RT

nF

⎛

⎝ ⎜

⎞

⎠ ⎟lnQ

This equation allows us to estimate the emf of cells under nonstandard conditions

•Like inside a living cell

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E = E° - 0.025693V

n

⎛

⎝ ⎜

⎞

⎠ ⎟lnQ n is unitless and is the whole number of the moles of electrons in the reaction

Since ln x = 2.303 log x, we could rewrite this also as:

€

E = E° - 2.303RT

nF

⎛

⎝ ⎜

⎞

⎠ ⎟logQ

Special Topic: The Nervous System

The nervous system relies on cells called Neurons to relay information

Neurons communicate with each other by releasing neurotransmitters at the axons which are picked up by the dendrites of the neighboring cell

Special Topic: The Nervous System

The axons propagate an electric impulse along their length in response to neurotransmitters hitting receptors in the dendrite

When the electrical impulse reaches the end of the axon, the change in voltage cause membrane proteins to change shape and release neurotransmitters

The electrical impulse is caused by a gradient of ions, specifically: Sodium, potassium, calcium and chloride

Special Topic: The Nervous System

Ion [Intracellular]

(mM)

[Extracellular]

(mM)

Potential (mV)

Na+ 18 150 +56

K+ 135 3 -102

Cl- 7 120 -76

Ca+2 0.0001 1.2 +125

We can use the Nernst Equation to calculate the membrane potential for each ion.

Examples

1) Calculate the equilibrium constant for the reaction:

AgCl (s) --> Ag+ (aq) + Cl- (aq)

i) This reaction is for the small amount of Ag+ and Cl- that form when AgCl dissolves

ii) Using the steps we had a couple of slides ago, the fist thing to do is balance the equation…Done!

iii) Now we need to find 2 reactions that will give us the balanced equation we want. Scanning through Table 12.1, we find:

AgCl (s) + e- --> Ag (s) + Cl- (aq) E°=+0.22V

Ag+ (aq) + e- --> Ag (s) E°=+0.80V

Examples

AgCl (s) + e- --> Ag (s) + Cl- (aq) E°=+0.22V (reduction)

Ag+ (aq) + e- --> Ag (s) E°=+0.80V (oxidation)

• We need to flip the second reaction so that the Ag (s) on both sides of the reaction arrow cancel

• E°= E°R - E°L = 0.22V - 0.80V = -0.58V

• Use the equation:

€

lnKeq =nE°

0.025693V

Examples

2) Calculate the Ksp of cadmium hydroxide, Cd(OH)2

i) Balanced chemical equation:

Cd(OH)2 (s) --> Cd2+ (aq) + 2OH- (aq)

ii) Look in Appendix 2B for suitable half reactions

Cd2+ (aq) + 2e- --> Cd (s) E°=-0.40V

Cd(OH)2 (s) + 2e- --> Cd (s) + 2OH- (aq) E°=-0.81V

iii) Reversing the first reaction will give us the balanced chemical equation

Examples

Cd(OH)2 (s) --> Cd2+ (aq) + 2OH- (aq) (n=2)

E° = E°R - E°L = -0.81V - (-0.40V) = -0.41V

iv) lnK = n E°/0.025693V = (2)(-9.41V)/0.025693V = -31.92

K = e-31.92 = 1.38x10-14

Examples

3) Calculate the emf of the cell:

Zn (s) | Zn2+ (aq, 1.5M) || Fe2+ (aq, 0.1M) | Fe (s)

i) Balanced equation

Zn (s) + Fe2+ (aq) --> Zn2+ (aq) + Fe (s)

Zn2+ + 2e- --> Zn (s) E°=-0.76V Low E°, Oxid

Fe2+ (aq) + 2e- --> Fe (s) E°=-0.44V Higher E°, Red

ii) Q = [Zn2+]/[Fe2+] = 1.5M/0.1M = 15

iii) E°= E°R- E°L = -0.44V - (-0.76V) = 0.336V

• E = E° - (0.025693/2)ln(15)

E = 0.336V - (0.035V) = +0.30V

Examples

Calculate the molar concentration of Y3+ in a saturated solution of YF3 by using a cell constructed with two yttrium electrodes. The electrolyte in one compartment is 1.0M Y(NO3)3. In the other compartment you have prepared a saturated solution of YF3. The measured cell potential is +0.34V at 298K.

i) Let’s figure out what is going on. This is a concentration cell, a special kind of cell that allows uses the same electrode in each side.

Y3+ (aq) + 3e- --> Y (s) Right, nitrate electrode

Y (s) --> Y3+ (aq) + 3e- Left, YF3 electrode

Y3+right --> Y3+

left

Examples

Y3+right --> Y3+

left (n=3)

E = -(RT/nF)lnQ

Q = [Y3+left]/ [Y3+

right]

lnQ = -EnF/RT = -0.34V(3)/0.025693V

lnQ = -39.699

Q = 5.73x10-18 = [Y3+left]/ [Y3+

right] but [Y3+right]=1.0M

[Y3+left]= 5.73x10-18 M