Question

Calculate the Zn conc. of Zn/Zn++ // Cl/Cl- 0.1M Emf=2.21v

Answer 1

Answer:

Option a. = 0.01 M

Explanation:

To do this, we need to gather the data:

E = 2.21 V

[Cl⁻] = 0.1 M

And the Redox reaction taking place is the following:

Zn(s) + Cl₂(g) <-------> Zn²⁺(aq) + 2Cl⁻(aq)       Q = [Zn] [Cl]²

E° Cl⁻/Cl₂ = 1.36 V

E° Zn/Zn²⁺ = -0.76 V

According to this, the expression to use will be the Nernst equation, and we can assume we are working at 25 °C, therefore, the Nernst equation will be:

E = E° - (0.059/n) logQ

E = E° - (0.059/n) ln([Cl⁻]² * [Zn²⁺])   (1)

From there, we can solve for Zn later.

First, we need to write the semi equation of oxidation and reduction, and get the standard potential of the cell:

Zn(s) --------> Zn²⁺(aq) + 2e⁻       E₁° = 0.76 V

Cl₂(g) + 2e⁻ -----------> 2Cl⁻(aq)   E₂° = 1.36 V

---------------------------------------------------------------

Zn(s) + Cl₂(g) -------> Zn²⁺(aq) + 2Cl⁻(aq)    E° = 0.76 + 1.36 = 2.12 V

Now, let's replace in (1) and then, solve for [Zn]:

2.21 = 2.12 - (0.059/2) log ([0.1]² * [Zn])

2.21 - 2.12 = -0.0295 log (0.01[Zn])

- 0.09 / 0.0295 = log (0.01[Zn])

-3.0508 = log (0.01[Zn])

10^(-3.0508) = 0.01[Zn]

8.8961x10⁻⁴ = 0.01[Zn]

[Zn²⁺] = 0.08896 M

This value can be rounded to 0.1 M. so the correct option will be option A.