Is it possible to separate Bi, Cu and Ag from a mixture containing:

c(BiO+)=0,08M
c(Cu2+)=0,242M
c(Ag)=0,106M
c(HClO4)=1M

Eo Bi = 0,32 V, Eo Cu = 0,337 V, Eo = 0,792 V
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Using Nernst equation (E=E°-0,0592/n -logQ)
i see it's possible to separate Ag, but what about Bi and Cu? I suppose no.
I got
E(Cu)=0,319V
E(Bi)= 0,298V
1% Cu.... 2,42*10^-3
Inserting this in Nernst equation I get E=0,260V, which is already in a range for depositing Bi.

Is my way of thinking correct?

Thank you for your help.

Yes, your way of thinking is correct. To determine whether it is possible to separate Bi, Cu, and Ag from the given mixture, you can compare their standard reduction potentials (Eo) to the potential at which the species will deposit or react.

First, let's consider Ag. From the given data, we have c(Ag) = 0.106 M. Ag has a standard reduction potential (Eo Ag) of 0.792 V. Using the Nernst equation:

E(Ag) = Eo Ag - (0.0592/n) * log(Q)

Since Ag is being reduced, n = 1. The reaction is Ag+ + e- -> Ag. The concentration quotient Q for this reaction is [Ag+] = 0.106 M.

Plugging in the values, we get:

E(Ag) = 0.792 V - (0.0592/1) * log(0.106) ≈ 0.442 V

Since E(Ag) is positive (greater than zero), Ag will deposit on the electrode. Therefore, it is possible to separate Ag from the mixture by electrodeposition.

Now let's consider Cu. We have c(Cu2+) = 0.242 M and Eo Cu = 0.337 V. Using the same process as above:

E(Cu) = Eo Cu - (0.0592/n) * log(Q)

In this case, Cu is being reduced, so n = 2. The reaction is Cu2+ + 2e- -> Cu. The concentration quotient Q for this reaction is [Cu2+]^2 = (0.242)^2.

Plugging in the values, we get:

E(Cu) = 0.337 V - (0.0592/2) * log((0.242)^2) ≈ 0.269 V

Since E(Cu) is positive, Cu will also deposit on the electrode. Therefore, it is possible to separate Cu from the mixture by electrodeposition.

Now let's consider Bi. We have c(BiO+) = 0.08 M and Eo Bi = 0.32 V. Using the same process as above:

E(Bi) = Eo Bi - (0.0592/n) * log(Q)

In this case, Bi is being reduced, so n = 1. The reaction is BiO+ + e- -> Bi. The concentration quotient Q for this reaction is [BiO+] = 0.08 M.

Plugging in the values, we get:

E(Bi) = 0.32 V - (0.0592/1) * log(0.08) ≈ 0.273 V

Since E(Bi) is positive, Bi will also deposit on the electrode. Therefore, it is possible to separate Bi from the mixture by electrodeposition.

In summary, using the Nernst equation, we can determine that Ag, Cu, and Bi can all be separated from the mixture by electrodeposition.