# Nernst Equation problems

You are titrating 100.0 mL of 0.0400 M Fe^2+ in 1 M HClO_4 to give Fe^3+ and Ce^3+ using Pt and calomel electrodes to find the endpoint.

a)

Write the balanced titration reaction.

Ce^4++Fe^2+ Ce^3++Fe^3+

Complete the two half reactions for the Pt electrode.

Ce^4+ + e^- Ce^3+ + e^- E^0 = 1.70 V

Fe^2+ + e^- Fe^3+ +e^- E^0 = 0.767 V

b)

From the list select the two correct Nernst equation for the cell

a. E = 0.767- 0.05916log([Fe^2+]/[Fe^3+]-0.241

b. E = 0.767- 0.05916log([Fe^3+]/[Fe^2+]-0.241

c. E = 1.70- 0.05916log([Fe^2+]/[Fe^3+]-0.241

d. E = 1.70- 0.05916log([Fe^3+]/[Fe^2+]-0.241

e. E = 0.767- 0.05916log([Ce^3+]/[Ce^4+]-0.241

f. E = 0.767- 0.05916log([Ce^4+]/[Ce^3+]-0.241

g. E = 1.70- 0.05916log([Ce^3+]/[Ce^4+]-0.241

h. E = 1.70- 0.05916log([Ce^4+]/[Ce^3+]-0.241

c)

Calculate the values of E for the cell when the following volumes of the Ce^4+ titrant have been added. (Activity coefficients may be ignored as they tend to cancel when calculating concentrations of ratios.

a. 1.50 mL

b. 20.0 mL

c. 39.0 mL

d. 40.0 mL

e. 44.0 mL

f. 80 mL

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#### Solution Summary

A few problems based on Nernst equation are solved.

Problems in Electrophysiology

Part 1. Complete problem 1 from the Problems in electrophysiology 2006 handout. You can find it in the electrophysiology handouts document in the course documents folder. Use the "RT/F ln" form for the GHK equation as was shown in class but multiply RT/F by 1000 to convert volts to millivolts. (Remember that the Nernst equation is the statement of electrochemical equilibrium for a permeable ion.)

The Excel program uses "60 log10" so if you use the excel sheet fix the equation first. Note that the units of RT/F will be volts. The "60" in the other form is 1000 x 2.303 x RT/F (the temperature was not 20C). The "2.303" converts log10 to ln; the factor "1000" converts volts to millivolts.

Part 2. Complete problem 2, parts a through d, from the same handout. Q = CV for a parallel plate capacitor. The same equation can be used for a cell membrane because the membrane is so thin that curvature doesn't affect the results. With axon radius given and by assuming that the total membrane surface area is 1 cm2, the length and volume of the piece of axon can be computed. Then you can use those values to find out how much K is in the piece of axon.

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