Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Carboxylic acid...........
I think the subatomic particles that are paired with each of their corresponding name would be :
1. X , proton and z, electron
hope this helps
The lysis reaction with the formation of two different gases, and the synthesis reaction with the reaction of hydrogen and oxygen support that water is a compound and not an element.
<h3>What is electrolysis?</h3>
The electrolysis is the analytical process for the separation or the deposition of the element in the presence of the electric field.
The passing of electric current through water results in the release of two types of gases. Thus, it can be said that water lysis or breakdown forms the release of two elements hydrogen and oxygen supporting the fact that water is a compound and not an element.
The production of water is found to be formed with the reaction of hydrogen with oxygen. Thus it supports that water is a compound and, not an element.
Learn more about electrolysis, here:
brainly.com/question/12054569
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