12.8: Standard Potentials and Equilibrium Constants We can calculate K eq 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 ° =-RTlnK eq We know from this chapter that:  G ° =-nFE ° Setting them equal to each other gives us: nFE ° =RTlnK eq (which can be rearranged to)

lnK

eq 

nFE



RT



12.8: Standard Potentials and Equilibrium Constants

lnK

eq 

nFE



RT

We can calculate the emf from standard potentials, so for we can calculate K eq RT/F = 0.025693 V @ STP, so we ca rewrite the equation above as:

lnK

eq 

nE



0.025693V

Where V is Volts 

A note on K values K eq K A = ????

= ????

K B K SP = ????

= ????

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 K eq from Electrochemical Data 1. Write the balanced chemical equation for the reaction 2. 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

lnK

eq 

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]  G r =  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  G r =  G ° r + RT lnQ But  G r = -nFE

and

 G ° r = -nFE ° , so: -nFE = -nFE ° + RTlnQ

E = E



-



RT nF

 

lnQ

The Nernst Equation

Divide by -nF

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

 The Nernst Equation E = E  -  RT nF   lnQ @  and 1 atm, RT/F = 0.025693V, so: 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 This equation allows us to estimate the emf of cells under nonstandard conditions  •Like inside a living cell

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 Na + K + Cl Ca +2 [Intracellular] (mM) 18 135 7 0.0001

[Extracellular] (mM) 150 3 120 1.2

Potential (mV) +56 -102 -76 +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 form when AgCl dissolves that 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 + (aq) + e --> Ag (s) + Cl --> Ag (s) (aq) E ° =+0.22V

E ° =+0.80V

Examples AgCl (s) + e --> Ag (s) + Cl (aq) Ag + (aq) + e --> Ag (s) E ° =+0.22V (reduction) E ° =+0.80V (oxidation) iv. We need to flip the second reaction so that the Ag (s) on both sides of the reaction arrow cancel v. E ° = E ° R - E ° L = 0.22V - 0.80V = -0.58V

 vi. Use the equation:

lnK

eq 

nE 0.025693V



Examples 2) Calculate the K sp of cadmium hydroxide, Cd(OH) 2 i) Balanced chemical equation: Cd(OH) 2 (s) --> Cd 2+ (aq) + 2OH (aq) ii) Look in Appendix 2B for suitable half reactions Cd 2+ (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) --> Cd 2+ (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) | Zn 2+ (aq, 1.5M) || Fe 2+ (aq, 0.1M) | Fe (s) i) Balanced equation Zn (s) + Fe 2+ (aq) --> Zn 2+ (aq) + Fe (s) Zn Fe 2+ 2+ + 2e --> Zn (s) (aq) + 2e --> Fe (s) E ° =-0.76V Low E ° , Oxid E ° =-0.44V Higher E ° , Red ii) Q = [Zn 2+ ]/[Fe 2+ ] = 1.5M/0.1M = 15 iii) E ° = E ° R - E ° L = -0.44V - (-0.76V) = 0.336V

iv) E = E ° - (0.025693/2)ln(15) E = 0.336V - (0.035V) = +0.30V

Examples Calculate the molar concentration of Y 3+ in a saturated solution of YF 3 by using a cell constructed with two yttrium electrodes. The electrolyte in one compartment is 1.0M Y(NO 3 ) 3 . In the other compartment you have prepared a saturated solution of YF 3 . 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.

Y 3+ (aq) + 3e --> Y (s) Y (s) --> Y 3+ (aq) + 3e Y 3+ right --> Y 3+ left Right, nitrate electrode Left, YF 3 electrode

Examples Y 3+ right --> Y 3+ left E = -(RT/nF)lnQ (n=3) Q = [Y 3+ left ]/ [Y 3+ right ] lnQ = -EnF/RT = -0.34V(3)/0.025693V

lnQ = -39.699

Q = 5.73x10

-18 = [Y 3+ left ]/ [Y 3+ right ] [Y 3+ left ]= 5.73x10

-18 M but [Y 3+ right ]=1.0M