Answer:
The cadmium carbonate (CdCO3) will precipitate first.
We need 6.9 *10^-11 M CO3^2- to initiate precipitation.
Explanation:
<u>Step 1:</u> Given data
The solution contains
⇒ 1.12*10^-2 M Fe2+
⇒ 1.45*10^−2 M Cd2+
The Ksp of CdCO3 is 10^-12
The Ksp of FeCO3 is 3.13 * 10^-11
<u>Step 2:</u> Calculate concentration of CO3^2- in FeCO3
FeCO3(s) ⇔ Fe2+ + CO3^2- Ksp = 3.1*10^-11
Ksp = [Fe2+][CO3^2-]
[CO3^2-] = Ksp / [Fe2+]
[CO3^2-] = 3.13*10^-11 / 1.12*10^-2
[CO3^2-] = 2.795 * 10^-9 M
<u>Step 3:</u> Calculate concentration of CO3^2- in CdCO3
CdCO3(s) ⇔Cd^2+ + CO3^2- Ksp = 10^-12
Ksp = [Cd2][CO3^2-]
[CO3^2-] = Ksp /[Cd^2+]
[CO3^2-] = 10^-12 / 1.45x10^-2
[CO3^2-] =6.9 *10^-11 M
The cadmium carbonate (CdCO3) will precipitate first.
We need 6.9 *10^-11 M CO3^2- to initiate precipitation.
Answer: The half-reactions represents reduction are as follows.
Explanation:
A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.
For example, the oxidation state of Cr in is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.
Similarly, oxidation state of Mn in is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.
Thus, we can conclude that half-reactions represents reduction are as follows.
The number of lone pairs that are most likely found on the central atom is zero. There are no lone pairs found on the central atom.
I think the effect of increasing temperature would be; the equilbrium will shift back wards. Increase in temperature favors backward reaction since the forward reaction is exothermic and the backward reaction is endothermic. Therefore, the equilibrium will shift back wards, and there will be more reactants (H2 and Cl2) compared to the products