contestada

The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn2+ ions. These ions are then oxidized to the deeply colored MnO4- ions by periodate ion (IO4-) in acid solution. Calculate ā„°Ā° and Ī”GĀ° at 25Ā°C for each reaction

Relax

Respuesta :

sahir9397

Answer:

The equations are:

3Mn + 2NO3- + 8H+ -----------> 3Mn2+ + 2NO + 2H2O

2Mn2+ + 5IO4- + 3H2O ---------> 2MnO4- + 5IO3-+ 6H

Ecell = 1.6 V - 1.51 V = 0.09 V

Using the equation of Nernst ,

Ā Ī”G = -nFEcell

where n= number of electrons transferred = 10 mol

F = Farraday's constant = 96485 C/mol

Ecell = 0.09V

Now, Ā  Ī”G = -10 * 96485 C/mol * 0.09 V

Ī”G = -86836.5 J/mol (\because C*V =J)

Ī”G = -86.84 kJ/mol

Now, Ecell = (0.0591/n)*log K

0.09 = 0.0591/10 *logK

0.09 = 0.00591 * log K

log K = 0.09 / 0.00591

log K = 15.23

K = 10-15.23

Ā  K = 5.89*10-16

The combined cell potential, gibbs free energy and value of K are; EĀ°cell = 0.09 V; Ī”G = -86.84 kJ/mol; K = 1.698 Ɨ 10Ā¹ā“

What is the Standard Cell Potential?

A) The balanced equation describing each of the given reactions are;

3Mn + 2NOā‚ƒā» + 8Hāŗ -----------> 3MnĀ²āŗ + 2NO + 2Hā‚‚O

2MnĀ²āŗ + 5IOā“ā» + 3Hā‚‚O ---------> 2MnOā‚„ā» + 5IOā‚ƒā» + 6H

B) The combined cell potential is; EĀ°cell = 1.6 V - 1.51 V

EĀ°cell = 0.09 V

With the aid of the Nernst equation, we can get the gibbs free energy as;

Ī”G = -nFEĀ°cell

where;

n is number of electrons transferred = 10 mol

F is Faraday's constant = 96485 C/mol

EĀ°cell = 0.09V

Thus, Ā 

Ī”G = -10 * 96485 C/mol * 0.09 V

Ī”G = -86836.5 J/mol

Ī”G = -86.84 kJ/mol

Now, EĀ°cell = (0.0591/n)*log K

0.09 = (0.0591/10) Ɨ log K

0.09 = 0.00591 Ɨ log K

log K = 0.09/0.00591

log K = 15.23

K = 1.698 Ɨ 10Ā¹ā“

Read more about Cell potential at; https://brainly.com/question/14115211

Otras preguntas