Suppose a voltaic cell is made from the following half-reactions: Fe(s)=Fe2+(aq)

Question

Suppose a voltaic cell is made from the following half-reactions:

Fe(s)=Fe2+(aq)+2e-

Pb2+(aq)+2e-=Pb(s)

The short hand notation for this cell is:

i) Fe(s) | Fe2+(1M) | Pb2+(1M) | Pb(s)

ii) Pb(s) | Pb2+(1M) | Fe2+(1M) | Fe(s)

iii) Fe(s) | Fe2+(1M) || Pb2+(1M) | Pb(s)

iv) Pb(s) | Pb2+(1M) || Fe2+(1M) | Fe(s)

b. Calculate the standard cell potential for this voltaic cell.

c. Calculate the cell potential for this voltaic cell at 298K if it is constructed using solutions of .020M Fe(NO3)2 and .50M Pb(NO3)2.

Explanation / Answer

a) The cell notation for the given reaction is written by taking oxidation half cell on left and reduction half cell on right side by seperating with two lines. Therefore the cell notation for the given reaction is option (III)

Fe(s) | Fe2+(1M) || Pb2+(1M) | Pb(s)

b) From the reduction potentials values -

E0(Fe2+/Fe) = -0.44 V, E0(Pb2+/Pb) = -0.13 V

Thus - E0 (cell) = E0(oxi) + E0(red) = 0.44 - 0.13 = 0.31 V

c) Ecell = E0cell - (0.0592/n) log [Fe2+/Pb+2]

           = 0.31 V - (0.0592/2) log (0.020/0.50)

           = 0.351

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