Answer:
B) increase by 0.18 V
Explanation:
The given chemical spontaneous reaction is :
[tex]2 AgCl_{(s)} + Zn _{(s)} \to 2Ag (s) + 2Cl^- _{(aq)} + Zn ^{2+} _{(aq)}[/tex]
By applying Nernst Equation:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
here ;
n = number of electrons transferred in the reaction
n =2
[tex]E^0 = E^0_{cathode } - E^0_{anode}[/tex]
[tex]E^0 = E^0_{Ag^+/Ag } - E^0_{Zn^{2+}/Zn}[/tex]
[tex]E^0 =+0.80 \ V -(-0.76 \ V)[/tex]
[tex]E^0 =1.56 \ V[/tex]
When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;
[tex]E^0_{cell} = E^0[/tex] is as follows:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)[/tex]
[tex]E_{cell} = 1.56 \ V[/tex]
Now; the [tex]E_{cell}[/tex] value in the decreased concentration of chlorine (aq) ion is calculated as:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1*0.001^2)[/tex]
[tex]E_{cell} = +1.737 \ V[/tex]
Hence; the change in voltage = [tex]E_{cell} - E^0[/tex]
= 1.737 - 1.56
= 0.177 V
≅ 0.18 V
We therefore conclude that: since the [tex]E^0_{cell}[/tex] value after the decreased concentration of Chlorine is greater than the [tex]E^0[/tex] before the change; then there is increase in the value by 0.18 V
