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

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Introduction

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 ==> Additional apparatus: * 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)

Middle

(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

Conclusion

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.

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