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What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Hg2 concentration is 7.36E-4 M and the Al3 concentration is 1.05 M

Sagot :

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 .

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