Question

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 ?

Answer 1

Answer:

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