6. A half cell is constructed from a 10 mL solution of 0.4 M Cu and a copper wir

Question

6. A half cell is constructed from a 10 mL solution of 0.4 M Cu and a copper wire electrode. A second half cell is constructed from 10 mL of a solution containing 0.1 M SnCl4, 0.2 M SnCl,, and a platinum wire electrode. a) What potential would be measured if these half-cells were operated as a galvanic cell? associated with oxidation? flow for 48 minutes, what cell potential would be measured? b) Diagram the cell using the conventional short-hand notation. Which ½ cell is c) If 100 milliamperes (amperes - coulombs/sec) of current were allowed to Cu* +e Sn E-0.139V

Explanation / Answer

Answer for 1:

Standard reduction potentials for the half reactions of the cell:

Cu+ + e- <===> Cu0    E0 = 0.518 V

Sn4+ + e- <==> Sn2+ E0 = 0.139 V

For calculating standard potential, Eocell, of the overall reaction we need to add standard reduction and oxidation potentials of the half reactions. For obtaining Sn2+ oxidation potential the reduction potential of Sn4+ is taken and the sign is reversed. Thus

Eocell =Eored +Eooxi ………………………………………………….eq.1

Eocell = 0.518 + (-0.139) = 0.379 V

The cell potential of a galvanic cell can be calculated from the Nernst equation :        

Ecell = Eocell – (2.303 RT / n F) log Q

But since we normally work at SATP the above equation appears as:

Ecell = Eocell – (0.0592 / n ) log Q …………………….eq.2

where

Ecell = the actual cell potential

Eocell = the standard cell potential

n = the number of moles of electrons transferred

Q = [product ion]y / [reactant ion]x

Accordingly in the above galvanic cell Sn2+ acts as oxidizing half-cell and Cu+ acts as reducing half-cell. In this galvanic cell 2 moles of electrons are transferred and accordingly

n = 2

product ion concentration, Cu+ = 0.4 M

reactant ion concentration, Sn2+ = 0.2 M

Substituting these values into eq.2 and solving the equation one can obtain the actual cell potential as follows:

Ecell = 0.379– (0.0592 / 2 ) log [0.4]/[0.2]2

Ecell = 0.379– (0.0592 / 2 ) log [0.4]/[0.04]

Ecell = 0.379– (0.0592 / 2 ) log 10

Ecell = 0.379– (0.0592 / 2 ) x 1

Ecell = 0.379– (0.0592 / 2 ) x 1

Ecell = 0.379– 0.0296

Ecell = 0.3494 V

The measured potential of the above galvanic cell is 0.3494 V

Answer for 2:

The standard notation of the above galvanic cell is

Sn2+ I Sn4+ II Cu+ I Cu0  

The Sn2+ I Sn4+ half-cell is associated with oxidation

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