# An experiment to investigate the change in cell potential with concentration.

8/1/04

Emma Duckworth 7E

An experiment to investigate the change in cell potential with concentration.

Aim:

The purpose of this experiment was to investigate how changing the silver ion concentration in a silver half cell affects the potential of the silver electrode.

Apparatus:

==> Chemicals/ substances:

* Copper (II) sulphate solution

* Copper foil

* Silver nitrate solution (make up to 6 different concentrations)

* Silver wire

* Distilled water

* Saturated potassium nitrate solution

* Safety goggles

* 7 beakers

* 6 pieces of filter paper

* High resistance voltmeter

* 2 connecting leads with crocodile clips

Diagram:

Method:

==> Set up the following cell, using 1 M copper (II) sulphate solution and 0.1 M silver nitrate solution, including a voltmeter and a salt bridge:

Cu(s) Cu 2+ ((aq), 1M) Ag+ ((aq), x M) Ag(s)

==> Measure the potential difference of the cell with the voltmeter and note its polarity. Remove the salt bridge as soon as possible.

==> Dilute the 0.1 M silver nitrate solution to 0.01 M silver nitrate solution, renew the salt bridge and then measure the potential difference of the cell with this concentration (0.01 M) of silver nitrate solution in the silver half cell.

==> Repeat this for each of the listed concentrations (0.1 M, 0.01 M, 0.001M, 0.0001 M, 0.003 M, 0.00033 M) of silver nitrate solutions, by diluting each solution accordingly and replacing the salt bridge each time.

Calculations:

(1) log([Ag+ (aq)]/ M - see table

(2) Electrode potential of the silver half cell for each concentration:

* 0.1 M silver nitrate solution:

Taking E? of copper half cell to be actual value of: + 0.34 V:

E? cell = E?positive - E?negative (or E rgt -E lft)

+ 0.27 = e - (+ 0.34)

? E? of silver electrode = + 0.61 V

* 0.01 M silver nitrate solution:

Taking E? of copper half cell to be actual value of: + 0.34 V:

E? cell = E?positive - E?negative (or E rgt -E lft)

+ 0.26 = e - (+ 0.34)

? E? of silver electrode = + 0.60 V

* 0.0033 M silver nitrate solution:

Taking E? of copper half cell to be actual value of: + 0.34 V:

E? cell = E?positive - E?negative (or E rgt -E lft)

+ 0.25 = e - (+ 0.34)

? E? of silver electrode = +0.59

* 0.001 M silver nitrate solution:

Taking E? of copper half cell to be actual ...