Answer:
there is an increase by 0.18 V in the cell voltage.
Explanation:
The given equation of the reaction can be well written as
[tex]2AgCl_{(s)} + Zn _{(s)} \to 2Ag_{(s)} + 2 Cl^- _{(aq)}+ Zn^{2+}_{(aq)}[/tex]
By application of Nernst Equation ; we have the expression
[tex]E_{cell} = E^0- \dfrac{0,059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
here in the above equation;
n = number of electrons transferred in the equation of the reaction
n = 2
Also;
[tex]E^0 = E_{cathode} - E_{anode}[/tex]
[tex]E^0 = E_{Ag^+/Ag} - E_{Zn^+/Zn}[/tex]
[tex]E^0 = +(0.80 \ V) - (-0..76 \ V)[/tex]
[tex]E^0 = (0.80 \ V +0..76 \ V)[/tex]
[tex]E^0 = 1.56 \ V[/tex]
If the zinc ion concentration is kept constant at 1 M; we have:
[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log (1)[/tex]
Since log(1) = 0
Therefore;
[tex]E_{cell} = 1.56\ V[/tex]
When the chlorine ion concentration is decreased from 1 M to 0.001 M; we have;
[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]
[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[1*0.001^2}]})[/tex]
[tex]E_{cell} = 1.56 - 0.0295 \ * \ log ({[1*10^{-6}}]})[/tex]
[tex]E_{cell} = + 1.737 \ V[/tex]
The change in voltage = [tex]E_{cell} - E^0[/tex]
=( 1.737 - 1.56 )V
= 0.177 V
≅ 0.18 V
Thus; from the following observation; there is an increase by 0.18 V in the cell voltage.
The voltage of the cell increased by 0.18 V.
The equation of the reaction is; 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+
We know that;
E°cell = 1.36 - (-0.76) = 2.12 V
If the cells are both at 1M concentration the Ecell = E°cell = 2.12 V
When the concentration of Cl- decreased from 1 M to 0.001 M
Ecell = E°cell - 0.0592/n log Q
Substituting values;
Ecell = 2.12 V - 0.0592/2 log (1 × (0.001)^2)
Ecell = 2.298 V
Increase in voltage = 2.298 V - 2.12 V = 0.18 V
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