What is the cell potential for the reaction mg (s) + fe2 + (aq) →mg2 + (aq) + fe (s) at 71 ∘c when [fe2+] = 3.30 m and [mg2+] = 0.310 m?

The cell potential for this reaction is 1.955 V

Explanation:

Step 1: Data given

Molarity of Fe^2+] = 3.30 M

Molarity of [Mg^2+] = 0.310 M

Temperature = 71°C

E∘ standard potential = 1.92 V

Step 2: The balanced equation

Mg (s) + Fe2 + (aq) → Mg^2 + (aq) + Fe (s)

Step 3: Nernst equation

E = Eo - RT/nF ln Q

⇒with E° = the standard potential = 1.92 V

⇒with R = 8.314 J/K*mol

⇒with T = the temperature = 344 K

⇒with n = the number of electrons transfered = 2

⇒with F = Constant of Faraday F = 96500 C/mol

⇒ with Q = [Mg^2+]/[Fe^2+] = 0.310 / 3.30 = 0.094

E = 1.92 - 8.314*344 / (2*96500) * ln (0.094)

E = 1.92 - 8.314*344 / (2*96500) * (-2.365)

E = 1.955 V

The cell potential for this reaction is 1.955 V