Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: [tex]2Al-6e^{-}\rightarrow 2Al^{3+}[/tex]
Reduction: [tex]3I_{2}+6e^{-}\rightarrow 6I^{-}[/tex]
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Overall: [tex]2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}[/tex]
Nernst equation for this cell reaction at [tex]25^{0}\textrm{C}[/tex]-
[tex]E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}[/tex]
where n is number of electrons exchanged during cell reaction, [tex]E_{cell}^{0}[/tex] is standard cell emf , [tex]E_{cell}[/tex] is cell emf , [tex][Al^{3+}][/tex] is concentration of [tex]Al^{3+}[/tex] and [tex][Cl^{-}][/tex] is concentration of [tex]Cl^{-}[/tex]
Plug in all the given values in the above equation -
[tex]E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V[/tex]
So, [tex]E_{cell}=2.32V[/tex]
