Question

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

Answer 1

Step-by-step 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 .