Answer:
The concentration at the cathode is [tex]A_1 = 0.8183M[/tex]
Explanation:
From the question we are told that
The initial concentration is [tex]1.39 M \ for \ Zn^{2+} \ at \ anode \ , 0.312 M \ for \ Zn^{2+} \ at \ cathode[/tex]
The reaction is
At Cathode [tex]Zn^{2+} + 2 e^- ------> Zn[/tex]
At anode [tex]Zn -----> Zn ^{2+} + 2e^{-}[/tex]
The initial potential is [tex]+ 0.037206V \ for \ Cathode \ and \ + 0.037206V \ for \ anode[/tex]
The complete reaction is
[tex]Zn^{2+} + Zn_{(s)} ----> Zn^{2+} + Zn_{(s)}[/tex]
(Cathode) ----------------- (anode)
Looking the reaction above we can see that after the reaction [tex]Zn^{2+}[/tex] at cathode loss some concentration and [tex]Zn^{2+}[/tex] will gain some concentration
Let say the amount of concentration lost/gained is = z M
The after the reaction the concentration of [tex]Zn^{2+}[/tex] at the cathode would be
[tex]A_1 = 1.39 -z[/tex]
While the concentration at the anode would be
[tex]A_2 = 0.312 +z[/tex]
At initial the condition the net potential of the cell is
[tex]E^i_{net} = 0V[/tex]
After time [tex]t = 9.1 hrs = 9.1 *3600 = 32760s[/tex]
The potential of the cell is
[tex]E_{net} = 0.0095376V[/tex]
Generally this potential is defined by Nernst equation as
[tex]E_{cell} = E^i_{cell}- \frac{RT}{nF} ln[\frac{[A_2]}{[A_2]} ][/tex]
Where R is the gas constant with a value of [tex]R = 8.314 J/mol \cdot K[/tex]
T is the temperature given as [tex]T = 289K[/tex]
n is the number of mole of electron which [tex]= 2 mol \ e^-[/tex]
F is the farad constant with a value of [tex]= 96500 C /mol\ e^-[/tex]
Substituting values
[tex]0.0095376 =0.0 -\frac{[8.314][288]}{[2] [9600]} ln [\frac{(0.312 +z)}{(1.39 -z)} ][/tex]
[tex]0.0095376 = 0.12406383 \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ][/tex]
[tex]\frac{0.0095376 }{ 0.12406383} = \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ][/tex]
[tex]0.0768766 = \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ][/tex]
Taking exponent of both sides
[tex]e^{0.0768766 }= [\frac{( 0.312 + z)}{[1.39 -z]} ][/tex]
[tex]1.0799 = [\frac{( 0.312 + z)}{[1.39 -z]} ][/tex]
[tex][1.39 -z ]1.0799 = 0.312 + z \\\\1.501 - 1.0799z = 0.312 +z\\\\1.501-0.312 = 1.0799z + z\\\\1.1891 =2.0799z\\\\ z =\frac{1.1891}{2.0799}\\\\ z =0.5717[/tex]
The concentration at the cathode is [tex]A_1 = 1.39 -0.5717[/tex]
[tex]A_1 = 0.8183M[/tex]
