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

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 value of: + 0.34 V:

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

+ 0.21 = e - (+ 0.34)

? E? of silver electrode = + 0.55 V

* 0.0003 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)

+ 0.10 = e - (+ 0.34)

? E? of silver electrode = +0.44 V

* 0.0001 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 V

(3) Graph - see next sheet.

Results:

[Ag+ (aq)] /M

log([Ag+ (aq)]/ M

?E /V (measured)

E(Ag+ Ag) /V

0.10

-1

+ 0.27

+ 0.61 V

0.010

-2

+ 0.26

+ 0.60 V

0.0033

-2.48

+ 0.25

+0.59 V

0.001

-3

+ 0.21

+ 0.55 V

0.0003

-3.52

+ 0.10

+0.44 V

0.0001

-4

+ 0.25

+0.59 V

N.B: These values are probably wrong as we were rushing at the end of the experiment and they do not fit into the general trend of the results.

Questions:

(1) As the concentration of the silver ions increased, the electrode potential of the silver half cell became more positive. This can be explained in terms of Le Chatelier's principle as at the electrode the equilibrium: Ag+ (aq) +e- ? Ag(s) is set up. Applying Le Chatelier's principle, when the concentration of the Ag+ ions is increased the equilibrium shifts more to the right to compensate for the addition of silver ions by using the extra ions up to produce silver metal. When the equilibrium shifts more to the right the electrode potential becomes more positive because this means that the silver will be more readily reduced and will be less likely to be oxidised to form its ions.

(2) The slope on my graph corresponds to the gradient of my line (I drew a line of best fit). Gradient is ?y / ?x = 0.62 -0.52 / -1 - (-3) = 0.09 / 2 = 0. 045 V. In the Nernst equation, the equation that shows the relationship between the potential, E, of a half cell and the standard electrode potential: E = E? + 0.059 V/ z log (Ag+), this gradient corresponds to 0.059 V / z as the Nernst equation fits the general equation for a line y = mx + c. Here, y = E?, x = log (Ag+) and so the gradient must be 0.059 V/ z.

??!

Conclusion:

From my results and graph the following conclusions can be drawn:

==> When the electrode potential of a cell is not measured under standard conditions and the concentration of the ions in a half cell is not the standard 0.1 M, the electrode potential of the half cell and the potential difference of the whole cell is altered.

==> As ions in a half cell become more concentrated the electrode potential of that half cell and the cell potential both become more positive.

==> When the electrode potential of a half cell is graphed against the log of the concentration of ions in that half cell an increasing straight line is produced showing that the electrode potential of the half cell and the log of ions in the half cell increase uniformly to each other in direct proportion.

Evaluation:

From our experiment it is clear that if an electrode potential or cell potential is not measure under standard conditions and the concentration of ions in solution are not 0.1 M then the values are altered. This shows the importance of measuring all standard electrode potentials under standard conditions so that they can be compared to each other.

Our results show clearly that as the concentration of the silver ions in the silver nitrate solution increased the electrode potential became more positive. Le Chatelier's principle can be applied to explain this, as the equilibrium: Ag+ (aq) +e- ? Ag(s) is set up at the silver electrode and the concentration of one of the species in the equilibrium is being altered. According to the principle, when the concentration of silver ions is increased, the position of equilibrium changes so as to minimise this change in concentration, and so in this case the position of equilibrium shifts more to the right to produce more silver metal. If the equilibrium lies more to the right then this means that the electrode potential will become more positive as the silver would be less likely than before to oxidise and form its ions and so at equilibrium there would be less negative charge present causing the electrode potential to be more positive and making the silver an even better oxidising agent.

The change in electrode potential due to the change in concentration was not drastic, however as the potential only changed by a few volts when the solution was diluted by a factor of ten each time. This means that changing the concentration of solutions will not have a large effect on the feasibility of a reaction. Other factors, perhaps, such as pH, would have more of an effect on the potentials as a change in pH causes a quite a much larger change in the concentration of H3O+ ions. For example, changing the pH from 7 to 5 increases the concentration of H3O+ ions by a factor of 100. If the equilibriums involved in the cell reactions contain H3O+ or OH - ions then its cell potential would be likely to depend very much on the pH of the solutions in the half cells. For example, the oxidising abilities of potassium permanganate or the dichromate ion depend very much on the presence of acidic conditions - the decreased concentration of the hydrogen ions causes the equilibrium to shift the left and so the oxidising abilities of the species increases (and the EMF gets more negative) .

The potentials did, however, increase with roughly the same amount each time in proportion which can also be seen on my graph. By calculating the log of the different silver ion concentrations and the resulting electrode potentials of the silver half cell and then plotting these values on a graph, it can be seen that these increase almost uniformly to each other. The slope of this graph corresponds to part of the Nernst equation, an equation that links together the cell potential, electrode potential and concentration of ions and so allows for the fact that concentration of ions effects the electrode potential and cell potential.. The equation is:

E = E? + 0.059 V/ z log (conc. oxidised/ conc. reduced) (z is the number of electrons transferred when the oxidised species changes into the reduced species - in our case the value is 2 as 2 electrons are transferred.) This means that the electrode potential increases as the concentration of the ions increased by a certain relationship.

My results seem to form a sensible pattern so I would assume they are reasonably accurate, but if I were to do this experiment again I would perhaps change the voltmeter and, to investigate the Nernst equation further I could perhaps measure the silver standard electrode potential using a standard hydrogen electrode. (although this may not make any difference) or investigate the effect of pH by changing the pH of solutions in a electrochemical cell that involve hydrogen or hydroxide ions (such as potassium permanganate) . It was very important in this experiment that we kept it a fair test by only changing the concentration of one of the solutions in the cell so that we had the copper cell as a reference point that we could use to calculate the silver electrode potential.