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Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.12 Cu2 and 0.12 MSn2 suppose it is at room temperature.

User Copumpkin
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Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Step-by-step 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.

User Stefano Mondino
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