Question

Be sure to answer all parts. Zinc is an amphoteric metal, meaning it reacts with both acids and bases. The standard reduction potential is −1.36 V for the following reaction: (1)Zn(OH)42−(aq) + 2e− → Zn(s) + 4OH−(aq) Calculate the formation constant Kf for the reaction: (2)Zn2+(aq) + 4OH−(aq) ⇌ Zn(OH)42−(aq)

Answer 1

Answer:

$1.86*10^{20$

Explanation:

The equation for the reaction is:

$Zn^{2+}_{(aq)} + 4OH^-_{(aq)}$     ⇄     $Zn(OH)^{2-}_{4(aq)}$

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:

$Zn(OH)^{2-}_{4(aq)} + 2e^- ----->Zn_{(s)} +4OH^-_{(aq)}$            $E^0_{anode} = -1.36 V$

At the zinc electrode (cathode); the reduction process of the reaction with its standard reduction potential is :

$Zn^{2+}_{(aq)} +2e^- -----> Zn_{(s)}$                  $E^0_{cathode} = -0.76 V$

The standard cell potential $E^0_{cell}$ is given as:

$E^0_{cell}=E^0_{cathode}-E^0_{anode}$

$E^0_{cell}$ = -0.76 V - (- 1.36 V)

$E^0_{cell}$ = -0.76 V + 1.36 V

$E^0_{cell}$ = +0.60 V

Now to determine the formation constant $k_f$ of the $E^0_{cell}$ ; we use the expression:

$E^0_{cell}$ = $\frac{RT}{nF}Ink_f$

where;

$E^0_{cell}$ = +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

$+ 0.60 V = \frac{(8.314)(298)}{n(96500)} Ink_f$

$+0.60 V = \frac{(0.0257)}{n}Ink_f$

$+0.60V = \frac{0.0592}{n}log k_f$

$logk_f = \frac{+0.60V*n}{0.0592}$

$logk_f = \frac{+0.60V*2}{0.0592}$

$logk_f = 20.27$

$k_f= 10^{20.27}$

$k_f = 1.86*10^{20}$

Therefore, the formation constant $k_f$ for the reaction is = $1.86*10^{20$