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)
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:
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!!!
Calculating Keq from Electrochemical Data Write the balanced chemical equation for the reaction Scan the table of Standard Potentials or Appendix 2B and find 2 equation that will combine to give you the equation from Step 1. Reverse on the the reactions as needed (Remember: They are all written as Reduction reactions!) Identify the value of n based upon your balanced chemical equation Calculate E° Use
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 The Nernst Equation Gr = G°r + RT lnQ But Gr = -nFE andG°r = -nFE° , so: -nFE = -nFE° + RTlnQ 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 reaction.
The Nernst Equation @ and 1 atm, RT/F = 0.025693V, so: Since ln x = 2.303 log x, we could rewrite this also as: • 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 We can use the Nernst Equation to calculate the membrane potential for each ion.
Examples Calculate the equilibrium constant for the reaction: AgCl (s) --> Ag+ (aq) + Cl- (aq) This reaction is for the small amount of Ag+ and Cl- that form when AgCl dissolves Using the steps we had a couple of slides ago, the fist thing to do is balance the equation…Done! 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:
Examples 2) Calculate the Ksp of cadmium hydroxide, Cd(OH)2 Balanced chemical equation: Cd(OH)2 (s) --> Cd2+ (aq) + 2OH- (aq) 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 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) 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. 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