Question

Given the standard electrode potential values 

E°_(Al³⁺/Al) = -1 -66 v E°_(Zn²⁺/Zn) = -0.764 V.

1. Draw a neat diagram of galvanic cell using the above electrodes and write the cell reaction.

2. Calculate the emf of the above cell.

Answer 1

1.

 

2. Cell emf, E_cell = E°_(Zn²⁺/Zn) – E°_(Al³⁺/Al)

= – 0.764 – (-1.66) = + 0.896 V

Comments

Why Al is taken as Anode and Zinc as Cathode?

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At Anode - always oxidation take place.
∵ Reduction potential of E° (Al^+3/Al) = -1.66 V
it means, oxidation potential of E° (Al/Al^+3) = 1.66 V
and similarly oxidation potential of E°(Zn/Zn^+2) = 0.764 V
∵ oxidation potential of Al is greater than the oxidation potential of Zn
because of that Al act as an Anode and Zn as a Cathode.