Question

10) . Use the principles that you learned in Part ii to calculate the standard reduction potentials for each half-cell, given that the E° for Cu2+ + 2e-  Cu (s) is 0.34 V.Note that E° for a half-reaction is not dependent on the coefficients, provided of course that the reaction is balanced, i.e., E° for Ag+ + e- Ag (s) is the same as E° for Ag+ + 2e- 2 Ag (s)

Refers to Procedures Part III Step 9

Data Table 1: Data Measurements
	Cu red	Pb red	Sn red	Zn red	Al red	Mg red
Cu black	0	-0.48	-0.49	-0.94	-0.37	-1.75
Pb black	0.49	0	-0.01	-0.49	-0.04	-1.30
Sn black	0.53	0.02	0	-0.48	-0.03	-1.24
Zn black	0.98	0.49	0.48	0	0.19	-0.78
Al black	0.58	0.03	0.04	-0.07	0	-0.94
Mg black	1.76	1.27	1.26	0.77	0.78	0

* Pb results are most unreliable

Answer 1

The Cu half cell is reference and what you measure is how much   the potential of the other half cell is below the reference (because these values are negative)..

       Measured emf = E_M2+/M - E_Cu2+/Cu

Thus

        E_M2+/M = measured emf + E_Cu2+/Cu = measured emf + 0.34 V.

……Now you may calculate,but…………………

I suppose your experimental procedure was:

- each metal M in its half cell had the corresponding oxidised form M²⁺ (except Al/Al³⁺) in solution at a given concentration C

- in all half cells this concentration C was the same.

……………………………….

If C was 1M and T=25^oC, it is OK, all potentials E_M2+/M are in fact standard potentials E^o_M2+/M. But I suppose that C wasn’t 1 M and the temperature was not exactly 25^oC.

In this case we have to justify that what we calculate are standard reduction potentials.

In general

E_cell = E^o_cell – (RT/nF)lnQ          (Q is the rection quotient)

       = ( E^o_M2+/M - E^o_Cu2+/Cu ) - (RT/nF)ln([Cu²⁺]/[M²⁺])

Remember the condition, for all cells [Cu²⁺] = [M²⁺] and ln1=0. Thus

E_cell = E^o_M2+/M - E^o_Cu2+/Cu   and

E^o_M2+/M = E_cell + E^o_Cu2+/Cu

Although you didn’t measure the difference between standard potentials, if you assume that for the Cu²⁺/Cu half cell the potential was standard ( 0.34 V) (in fact it was not), you may calculate standard potentials.

Using the first line of your result table:

Half cell of	Pb	Sn	Zn	Al	Mg
measured emf ,V	-0.48	-0.49	-0.94	-0.37	-1.75
Eo (V) = emf + 0.34	- 0.14	-0.15	- 0.60	- 0.03 !!	- 1. 41!!
Accepted/reference/true value, V	-0.13	-0.14	-0.76	-1.66	-2.37

Your result are very good for Pb,Sn,   good for Zn and poor for Al and Mg (for Al also the reproducibility is poor, if your consider the measurements with reversed wiring).

Explanation for poor results: Zn,Al.Mg are active metals and their surface is covered by a thin layer of oxide. For better results, you have to do a chemical cleaning of the metal surface before experiments. In these special cases you measured a so called mixed potential, due to presence of the oxide.