FRIDAY FRAZZLE #5

FRIDAY FRAZZLE #5 name(s) _______15 points______________

CHEM104 2002

Due: Friday, April 12.

You may work with a partner on this frazzle.

If you have consulted with a partner during the Frazzle, note that under your name above.

Each person's paper will be grading separately.

Use other sheets of paper for working the problem.

Now that we're talking about redox stuff, I can let you in on a little secret:

A pH electrode is really just a self-contained electrochemical cell that is designed to detect and measure hydronium ion concentration. (Some of you may already know this.)

For this frazzle, you should derive an equation that shows the linear relationship between a measured cell potential and pH.

You may assume for the purposes of this problem that the two half reactions that might be combined to make pH electrode are: E^o = + 0.2415V

"Saturated calomel electrode" Hg₂Cl_2(s) + 2K⁺ + 2e^- <==> Hg_(l) + 2Cl^-+ 2 KCl _(s) E^o = + 0.2415V

2H⁺ + 2e^-<==> H_{2 (g)}

Write a mathematical equation showing how the measured E_electrode is linearly related to pH, if all ionic species are kept at a constant concentration except [H⁺].

Identify the overall rxn. 9 points

First identify which are the oxidation and reduction reactions:

H_{2 (g)} <==> 2H⁺ + 2e^- E^o = + 0.0 V (See Table in text)

Hg₂Cl_2(s) + 2K⁺ + 2e^- <==> 2Hg_(l) + 2KCl _(s) E^o = + 0.2415V

H_{2 (g)} Hg₂Cl_2(s) + 2K⁺ Û 2Hg_(l) + 2KCl _(s) +2H⁺

Second use the Nernst equation to derive an equation that relates the cell potential to the pH

E_rxn = E^o – (0.0591/n)logQ*

Q = [H⁺]² where the concentrations of all other ionic species are 1 M

n = 2e^-

E^o = 0.2415 V

E_rxn = 0.2415 V - (0.0591/2)log[H⁺]²

E_rxn = 0.2415 V – 2*(0.0591/2)log[H⁺] where log x² = 2log x

E_rxn = 0.2415 V - (0.0591)log[H⁺]

E_rxn = 0.2415 V + 0.0591*pH where pH = -log[H⁺]

Which is in the form of a line: y = mx + b where y=E_rxn, m=0.0591, x=pH, b=0.2415

*If you started with E_rxn = E^o – (RT/nF)lnQ then use the relationship from GW#14: ln x = 2.3*log x

b) What potential will be read by a voltmeter attached to the pH electrode?

at pH = 7? pH = 1? pH = 13? 3 points

Use the equation that you derived in part a:

E_rxn = 0.2415 V + 0.0591*7 = 0.6550 V

E_rxn = 0.2415 V + 0.0591*1 = 0.3001 V

E_rxn = 0.2415 V + 0.0591*13 = 1.010 V

If you graph the function from part a it looks like this:

It is a line as advertised!

In fact the H⁺ /H_{2 (g)}half reaction is _not_ used in a pH electrode. Can you suggest one reason why this half reaction would be inconvenient in a portable device? 3 points

Designing a non-leaking cell that contains both gas and liquid is potentially tricky for a small portable device. The other problem that H₂ is explosive so any technical glitches in the electrode design could lead to disastrous consequences.