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

The cell potential for the given reaction is approximately 0.68 V.

Step-by-step explanation:

To calculate the cell potential for the given reaction, we use the Nernst equation:

E = E° - (0.0592/n) * log(Q)

Where E is the cell potential, E° is the standard cell potential, n is the number of electrons transferred in the reaction, and Q is the reaction quotient.

In this case, the standard cell potential, E°, is given as 0.74 V. The reaction involves the transfer of 2 electrons, so n = 2. The reaction quotient, Q, can be calculated using the concentrations of the reactants and products:

Q = [Mg2+]/[Fe2+]

Plugging in the given concentrations, [Fe2+] = 3.30 M and [Mg2+] = 0.310 M, we can calculate the cell potential:

E = 0.74 V - (0.0592/2) * log(0.310/3.30)

Simplifying the expression, we find that the cell potential for the reaction is approximately 0.68 V.

Answer 2

The cell potential for this reaction is 1.955 V

Step-by-step 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