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
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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.
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