Verified answer

Galvanic cell.....

*** Note: Your question has some flaws, and they will be shown below.

2(Fe(s) --> Fe3+ + 3e-) .... oxidation half-reaction.... E = +0.04V ..... occurs at anode

3(Cl2(aq) + 2e- --> 2Cl-) .. reduction half-reaction ... E = +1.36V ..... occurs at cathode

----------------- ------------------ -----------------

2Fe(s) + 3Cl2(aq) --> 2Fe3+ + 6Cl- ........ E = +1.40V ..... the reaction is spontaneous

The catch is that there will be additional side-reactions. Fe metal will react with oxygen and water to make Fe(OH)3. Put a piece of iron into water and watch the water slowly become yellowish and cloudy. Cl2 will react with water to make HCl and HOCl. This reaction can be used to make chlorine bleach and pool "chlorine."

4Fe(s) + 3O2(g) + 6H2O(l) --> 4Fe(OH)3(s)

Cl2(aq) + H2O --> HOCl(aq) + HCl(aq) ....... acidic solution.... reacts with Fe(OH)3

*** The preferred reaction is...

Fe(s) --> Fe2+ + 2e- .... oxidation half-reaction.... E = +0.44V ..... occurs at anode

Cl2(aq) + 2e- --> 2Cl- .. reduction half-reaction ... E = +1.36V ..... occurs at cathode

----------------- ------------------ -----------------

Fe(s) + Cl2(aq) --> Fe2+ + 2Cl- ........ E = +1.80V ..... the reaction is spontaneous

*** Since the voltage is greater, the formation of Fe2+ is favored over the formation of Fe3+

Even if Fe3+ was formed, it would react with the iron metal in the electrode to make Fe2+

Fe(s) + 2Fe3+ --> 3Fe2+

*** The bottom line is that the entire premise of your question is flawed. The reaction won't occur as you have indicated. Whoever wrote the original question should have done a bit more homework.