Answer:
[tex]1.86*10^{20[/tex]
Explanation:
The equation for the reaction is:
[tex]Zn^{2+}_{(aq)} + 4OH^-_{(aq)}[/tex] ⇄ [tex]Zn(OH)^{2-}_{4(aq)}[/tex]
Oxidation can be defined as the addition of oxygen, removal of hydrogen and/or loss of electron during an electron transfer. Oxidation process occurs at the anode.
On the other hand; reduction is the removal of oxygen, addition of hydrogen and/ or the process of electron gain during an electron transfer. This process occurs at the cathode.
The oxidation-reduction process with its standard reduction potential is as follows:
[tex]Zn(OH)^{2-}_{4(aq)} + 2e^- ----->Zn_{(s)} +4OH^-_{(aq)}[/tex] [tex]E^0_{anode} = -1.36 V[/tex]
At the zinc electrode (cathode); the reduction process of the reaction with its standard reduction potential is :
[tex]Zn^{2+}_{(aq)} +2e^- -----> Zn_{(s)}[/tex] [tex]E^0_{cathode} = -0.76 V[/tex]
The standard cell potential [tex]E^0_{cell}[/tex] is given as:
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
[tex]E^0_{cell}[/tex] = -0.76 V - (- 1.36 V)
[tex]E^0_{cell}[/tex] = -0.76 V + 1.36 V
[tex]E^0_{cell}[/tex] = +0.60 V
Now to determine the formation constant [tex]k_f[/tex] of the [tex]E^0_{cell}[/tex] ; we use the expression:
[tex]E^0_{cell}[/tex] = [tex]\frac{RT}{nF}Ink_f[/tex]
where;
[tex]E^0_{cell}[/tex] = +0.60 V
R = universal gas constant = 8.314 J/mol.K
T = Temperature @ 25° C = (25+273)K = 298 K
n = numbers of moles of electron transfer = 2
F = Faraday's constant = 96500 J/V.mol
[tex]+ 0.60 V = \frac{(8.314)(298)}{n(96500)} Ink_f[/tex]
[tex]+0.60 V = \frac{(0.0257)}{n}Ink_f[/tex]
[tex]+0.60V = \frac{0.0592}{n}log k_f[/tex]
[tex]logk_f = \frac{+0.60V*n}{0.0592}[/tex]
[tex]logk_f = \frac{+0.60V*2}{0.0592}[/tex]
[tex]logk_f = 20.27[/tex]
[tex]k_f= 10^{20.27}[/tex]
[tex]k_f = 1.86*10^{20}[/tex]
Therefore, the formation constant [tex]k_f[/tex] for the reaction is = [tex]1.86*10^{20[/tex]
