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An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) ||...

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An electrochemical cell has the following standard cell notation:

Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)

a. Write the balanced oxidation and reduction half-reactions, labeling the oxidation reaction and reduction reaction.


b. What is the standard potential for the cell?

c. Is this a galvanic cell or an electrolytic cell? Explain your answer.

d. Write a balanced redox equation for the cell using the oxidation and reduction half-reactions.

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The cell notation is:

Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)

Electrolytic cells use an external electrical source to push a reaction in a non spontaneous direction. In electrochemical cells, on the other hand, electrical energy is generated in lieu of a spontaneous chemical reaction taking place at the electrodes.

A glimpse at the standard electrode potentials of Al/Al3+(aq.) and Ma/Mg2+(aq.) systems  galvanic cell, reveal their values as – 1.66 V and – 2.37 V respectively. This implies that, Al/Al3+(aq.) electrode, with higher algebraic value of E°(red), should undergo reduction in a spontaneous reaction. Consequently it should constitute the positive electrode (right hand side electrode) of a galvanic cell. But here, it is placed in the left hand side of the cell, i.e. where oxidation takes place. So the resulting overall reaction is non spontaneous, driven by external electrical intervention. Hence this is an electrolytic cell.

By convention, oxidation occurs at the electrode placed at the left hand side. And reduction at the right hand side electrode.

As the system is aqueous, dissociation of water has to be considered. So, there will be more than one cation to be discharged at the cathode and similarly, more than one anion for the anode.

Possible anode (oxidation) reactions are: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−    Eᵒ(red) = +1.23 V

Al → Al˄3+ + 3e Eᵒ(red) =  – 1.66 V

Possible cathode (reduction) reactions are : 2 H+(aq) + 2e− → H2(g)    Eᵒ(red) = 0.00 V

Mg˄2+ + 2e → Mg (s), Eᵒ(red) = – 2.37 V

Preferential electrolysis shall be taking place and the one with higher algebraic value of discharge potential shall get precedence. Clearly, discharge of oxygen at the anode and of hydrogen at the cathode shall get preference, for obvious reasons.

Actual anode (oxidation half-reaction) reaction will thus be: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−    Eᵒ(red) = +1.23 V

And actual cathode (reduction half-reaction) reaction will be:  

2 H+(aq) + 2e− → H2(g) Eᵒ(red) = 0.00V

Overall redox reaction will be:

 2 H2O(l) + 4 H+(aq) + 4e-→  O2(g) + 4 H+(aq) + 4e−  + H2(g)

Or, 2 H2O(l) →  O2(g) + H2(g)

In essence, electrolysis of water will take place, overall.

There will be no spontaneous resaction taking place over there, so, no question of standard potential of the cell. It will be force-driven by external voltage source.


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