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

`1.86*10^{20`

Step-by-step 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)` =
`(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 = ((8.314)(298))/(n(96500)) Ink_f`

`+0.60 V = ((0.0257))/(n)Ink_f`

`+0.60V = (0.0592)/(n)log k_f`

`logk_f = (+0.60V*n)/(0.0592)`

`logk_f = (+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`