You are given a solution of Fe2+ ions, and asked to determine its concentration.

Question

You are given a solution of Fe2+ ions, and asked to determine its concentration. You make a 1.00 M solution of Fe2+. You place an iron electrode in each solution, and you connect them with a salt bridge. When you measure the voltage across the electrodes the 1.00 M solution is at a potential of +0.020V compared to the unknown. What is the concentration of Fe2+ ions in the unknown solution? 3. Half RXN E rxn Ag+ (aq) + e – Ag(s) 0.7996 Cu2+(aq) + 2e– Cu(s) 0.337 Fe3+(aq) + 3e– Fe(s) 0.04 Pb2+(aq) + 2e– Pb(s) 0.126 Sn2+(aq) + 2e– Sn(s) 0.13 Mn2+(aq) + 2e– Mn(s) 1.185 Al3+(aq) + 3e– Al(s) 1.662 Mg2+(aq) + 2e– Mg(s) 2.372

Explanation / Answer

If two solutions which differ in concentration are connected by a salt bridge, a concentration ell will be generated. The less concentratied side will try to increase the Fe^2+ ion concentration in the solution and hence Fe(s) from the electrode will oxidise and go to the solution. On the other hand, more concentrated side will try to decrease its concentration. So, Fe^2+ will be reduced and will be deposited on the electrode.

Nernst equation for this cell is as follows :

E = E^0 - 0.059/2 log [(Fe^2+) less conentrated/(Fe^2+) more concentrated ]

E^0 of Fe2+/Fe system = -0.44 V

0.02 = -0.44 - log[1/(Fe^2+)]

or, [Fe^2+] more concentrated = 2.88 M

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