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

Solution Summary

A few problems based on Nernst equation are solved.

Part 1. Complete problem 1 from the Problems in electrophysiology 2006 ... the "RT/F ln" form for the GHK equation as was ... (Remember that the Nernst equation is the ...

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... Compare the two equations, we have: (RT/nF) lnQ = 0.059/n *logQ (RT/nF ... This solution examines a problem solved using the Nernst equation and assesses its ...

... Calculate the pH of the H+(aq) solution. For this problem you need to use the Nernst equation: ... For this problem you need to use the Nernst equation: ...

... Cell Potentials Under Non-Standard Conditions Using the Nernst Equation. ... hint: you can solve the problem by half ... World file as there are equations involved in ...

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... Banking ethics and the Goldman rule. ... to put much of the blame for our financial problems at the ... but that does not justify the simple equation of derivatives ...

... encounters enter nents of the medical error equation: the amount ... misses, the process errors, and the error-free But ... Fischer G, Fetters MD, Munro AP, Goldman EB. ...