A voltaic cell employs the following redox reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate the cell potential at 25 ∘C under each of the following conditions. A voltaic cell employs the following redox reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate the cell potential at 25 ∘C under each of the following conditions.

1st condition: E = 2.25 V

2nd condition = E = 2.41 V

Explanation:

A voltaic cell employs the following reaction: 2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq) Calculate cell potential at 25 ∘C under each condition?

1st condition: [Fe3+] = 1.6 * 10^-3 M; [Mg2+] = 2.05 M

2nd condition: [Fe3+] = 2.05 M; [Mg2+] = 1.6*10^-3 M

Step 1: Data given

Temperature = 25 °C

The reduction potential for Fe+3 → Fe (s) = - 0.037

The reduction potential for Mg+2 + 2e - → Mg (s) = + 2.356

Step 2: The balanced equation

2Fe3 + (aq) + 3Mg (s) →2Fe (s) + 3Mg2 + (aq)

Step 3:

Fe+3 + 3e-→ Fe (s) - 0.037

Mg (s) → Mg+2 + 2e - + 2.356

We have to multiply the first equation by 2 and thesecond equation by 3

2Fe+3 + 6e-→ 2Fe (s) (-0.037)

3Mg (s) → 3Mg+2 + 6e - (+2.356)

2Fe+3 + 3Mg → 2Fe (s) + 3Mg+2 E° = + 2.319

Step 4: Calculate the cell potential for the first condition

Nernst Equation: E = E° - 0.0592/n * log [Mg+2]³ / [Fe+3]²

⇒with n = the number of electrons = 6

⇒with [Mg+2] = 2.05 M

⇒with [Fe+3] = 1.6 * 10^-3 M

E = + 2.319 - 0.0592/6 * log ((2.05) ³ / (1.6 x 10^-3) ²)

E = + 2.319 - 0.0592/6 * 6.53

E = 2.319 - 0.0099 * 6.53

E = 2.25 V

Step5 : Calculate the cell potential for the second condition

Nernst Equation: E = E° - 0.0592/n * log [Mg+2]³ / [Fe+3]²

⇒with n = the number of electrons = 6

⇒with [Fe^3+] = 2.05 M

⇒with [[Mg+2] = 1.6 * 10^-3 M

E = + 2.319 - 0.0592/6 * log ((1.6 x 10^-3) ³ / (2.05) ²)

E = + 2.319 - 0.0592/6 * (-9.0)

E = 2.41 V