ΔG⊖ for the following reaction is: 4A1+3O2+6H2O+4⊖OH→4A1(OH)4⊖ E⊖cell=2.73V ΔfG⊖(⊖OH)=−157kJmol−1 ΔfG⊖(H2O)=−237kJmol−1
A
−3.16×103kJmol−1
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B
−0.79×103kJmol−1
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C
−0.263×103kJmol−1
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D
+0.263×103kJmol−1
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Solution
The correct option is A−3.16×103kJmol−1 As we know, ΔrG⊖cell=−ncellFE⊖cell=−12×96500×2.73 =−3.16×103kJmol−1 For ncell, write the half cell reaction as: Cathode : 2H2O(l)+O2(g)+4e−→4⊖OH(aq) Anode : Al(s)+4⊖OH(aq)→Al(OH)⊖4(aq)+3e−