3. Electrochemical cell potential In the voltaic cell working at 25 °C below, th

Question

3. Electrochemical cell potential
In the voltaic cell working at 25 °C below, the two electrodes are made by iron and aluminum. The electrolyte solution contains 100mM Al2(SO4)3 and 10mM FeSO4.

The related standard half cell potential at 25 °C is listed below:

a) Calculate half cell potential. Identify cathode and anode, label electron flow direction and current flow direction. Calculate cell potential.
b) Write half cell equations and overall equation.
c) What will be some practical problems if this system is to be used as a primary battery to provide stable power continuously?

Explanation / Answer

(a) The half cell potential for

Al3+ + 3e- ----------> Al   Eo = -1.662 V

EAl= Eo + (0.0592/3) log ([Al3+]) = Eo + (0.0592/3) log (0.2) = -1.662 + (0.0197 * (-0.699)) = -1.662 - 0.0138 = -1.676 V

The half cell potential for

Fe2+ + 2e- ----------> Fe Eo = -0.447 V

EFe= Eo + (0.0592/2) log ([Fe2+]) = Eo + (0.0592/2) log (0.01) = -0.447 + (0.0296 * (-2)) = -0.447 - 0.0592 = -0.506 V

Since the half cell potential for the reduction of Fe2+ (-0.506 V) is higher than the reduction of Al3+ (-1.676 V), the Fe2+/Fe half cell acts as cathode and the Al/Al3+ half cell acts as anode where oxidation of Al to Al3+ occurs. Hence electrons flow from Al/Al3+ half cell to Fe2+/Fe half cell and current flows in opposite direction.

Cell potential Ecell = EFe - EAl = -0.506 + 1.676 V = 1.170 V

(b) The half cell reactions for

Anode : oxidation : Al (s) ----------> Al3+ (aq) + 3e-  

cathode : reduction : Fe2+(aq) + 2e- ----------> Fe (s)

Overall : 3 Fe2+ (aq) + 2 Al (s) ---------->   3 Fe (s) + 2 Al3+ (aq)

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