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Answer:
Explanation:
Concentration of Hg⁺² = 7.36 x 10⁻⁴ M
Concentration of Al⁺³ = 1.05 M
2Al + 3Hg⁺² = 2Al⁺³ + 3Hg .
E = E₀ + RT / nF ln [ Al⁺³]² / [ Hg⁺² ]³
E₀ = reduction potential of Hg⁺² minus reduction potential of Al⁺³
= 0.92 V - ( - 1.66 V )
= 2.58 V
E = 2.58 + .059 /n log [ Al⁺³]² / [ Hg⁺² ]³
n = 6 , [Al⁺³] = 1.05 M ; [Hg⁺²] = 7.36 x 10⁻⁴ M
E = 2.58 + .059 /6 log [ 1.05]² / [ 7.36 x 10⁻⁴ ]³
= 2.58 + .059 /6 log 27.65 x 10⁸ .
= 2.58 + .059 /6 [8+ log 27.65 ].
= 2.58 + .059 /6 [8+ log 27.65 ].
= 2.58 + .09
= 2.67 V .