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)

Explanation:

The equation for the reaction is:

    ⇄    

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:

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

The standard cell potential is given as:

= -0.76 V - (- 1.36 V)

= -0.76 V + 1.36 V

= +0.60 V

Now to determine the formation constant of the ; we use the expression:

=

where;

= +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

Therefore, the formation constant for the reaction is =

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )