Electrochemistry ,497 573u Data Cu/Pb Cu/Pb Cu/Pb Experimental Resultselectrodes

Question

Electrochemistry ,497 573u Data Cu/Pb Cu/Pb Cu/Pb Experimental Resultselectrodes electrodes electrodes electrodes 10M/10M 1.0M/1.0M0.040 M/1.0M 1.0M/0.040 M Avg Cell Potential (V) 1.53Sv |,40/v | ‘497v | ,S7/V Data Analysis A. Compare the average cell potential for the Cu/Pb cell experiment with the E ceil obtained from the standard reduction potentials. Does the experimentally obtained result agree with the accepted voltage? If not, provide some plausible reasons why the values differ B. Compare the average cell potential for the X/Pb cell experiment with the table of Ecell values. If the unknown metal X is either magnesium, silver, or zinc, what is the identity of electrode X from your experimentally obtained voltage?

Explanation / Answer

A. The Measured Average cell potential for the Cu/Pb cell = 0.535V at 1.0M concentrations of Cu and Pb.

The standard redcution potentials which are also calculated at 1.0M conc and 25 deg C can be

Cu2+ + 2e- ----------> Cu (s) E0 = 0.34V

Pb2+ + 2e- ----------> Pb (s) E0 = -0.13V

E0 (cell) = E0 ( red, cathode) - E0 (red, anode) 0.34V -(-0.13V) = 0.47 V

The obtained E0 (0.535 V) is higher the actual value (0.47V)

There can be several reasons for this value to be high or low:

1) The 1 M solutions are not prepared of right concentrations

2) The temperature to be maintained was not accurate (25 deg C)

3) Impure metals for the anode or cathode can cause a voltage change.

4) Previous oxiation of metals from air

5) Every battery has charge/discharge cycle. This also affects the performance of batteries and outputs of a battery keeps changing over time with usage. The battery used in the lab must have been used andhence the difference.

B. The standard potential of the cell with Pb and different metals of X would be:

1)Magnesium:

Mg2+ + 2e- ----------> Mg (s) E0 = 2.37V

Pb2+ + 2e- ----------> Pb (s) E0 = -0.13V

E0 (cell) = E0 ( red, cathode) - E0 (red, anhode) = -0.13V -(-2.37V) = 2.24 V

2) Silver:

Ag+ + e- ----------> Ag (s) E0 = 0.22 V

Pb2+ + 2e- ----------> Pb (s) E0 = -0.13V

E0 (cell) = E0 ( red, cathode) - E0 (red, anhode) = 0.22V -(-0.13V) = 0.35 V

3) Zinc:

Zn2+ + 2e- ----------> Zn (s) E0 = -0.76V

Pb2+ + 2e- ----------> Pb (s) E0 = -0.13V

E0 (cell) = E0 ( red, cathode) - E0 (red, anhode) = -0.13V -(-0.76V) = 0.63 V

hence X would be Ag as its pottential is closest to the measured value of 0.40V.

C.

1. Cu(0.040 M)/Pb(1.0M):

The nernst equation is defined as Ecell = Eocell - (0.0592 /n )* log [Red]/[Ox]

where, Eocell is the standard cell potential

n is no of electrons and [Red] is the concentarion of reductant and [Ox] is the concentration of oxidant in the overall equation: Ox + ne? ? Red

The overall equation in this case would be:

Pb (s) + Cu 2+ (aq) ---> Pb 2+ (aq) + Cu (s)

Ecell = 0.47 - (0.0592 /2 )* log [Pb2+]/[Cu 2+]

Ecell = 0.47 - (0.0592 /2 )* log [1.0]/[0.040]

Ecell = 0.47 - 0.0414 = 0.43V

2. Cu(1.0M)/Pb(0.040 M):

The overall equation in this case would be:

Pb (s) + Cu 2+ (aq) ---> Pb 2+ (aq) + Cu (s)

Ecell = 0.47 - (0.0592 /2 )* log [Pb2+]/[Cu 2+]

Ecell = 0.47 - (0.0592 /2 )* log [0.040]/[1.0]

Ecell = 0.47 - 0.0414 = 0.51V

D. The measured values of the cell potentials are slighlty different from the obtained values.

For Cu(0.040 M)/Pb(1.0M), calculated is 0.43V and the observed is 0.497V

For Cu(1.0 M)/Pb(0.04.0M), calculated is 0.51V and the observed is 0.591 V

The difference in the voltage can be due to the same reasons stated in A. The obtained voltage increased from 0.497 to 0.591V when the concentrations are reversed. Same is the case with calculated ones. Hence the trend we have here is in agreement with the theory.

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