Question

A Cu/Cu2+ concentration cell has a voltage of 0.22 V at 298K. The concentration of Cu2+ in one of the half-cells is 1.6E−3 M. What is the concentration of Cu2+ in the other half-cell?

Answer 1

`[Cu^2+]_("Anode")=5.7xx10^(-11)M`

Explanation:

Let us first describe the half-equations happening in the cell and write the overall equation:

Anode : `underbrace(Cu(s)->Cu^(2+)(aq)+2e^-)_color(blue)("Oxidation")` `" "-E^@=-0.34V`

Cathode : `underbrace(Cu^(2+)(aq)+2e^(-)->Cu(s))_color(blue)("Reduction")` `" "E^@=0.34V`

Overall equation : `Cu(s)+ underbrace(Cu^(2+)(aq))_(Cathode) -> Cu^(2+)(aq) + Cu(s)`

`E_("cell")^@=E_("Anode")^@+E_("Cathode")^@=0V`

For concentration cells, we can calculate the cell potential `E_("cell")` using the Nernst Equation:

`E_("cell")=E_("cell")^@` `-(0.0591)/(n)` `log(([Cu^(2+)]_(Anode))/([Cu^(2+)]_C))`

Replacing the giving terms in this equation assuming that the given concentration is `[Cu^2+]_("Cathode")=1.6xx10^(-3)M`

and that `[Cu^2+]_("Anode")=x`

`0.22=0-(0.0591)/(2)log((x)/(1.6xx10^(-3)))`

Solve for `x=5.7xx10^(-11)M`

Here is a video that explains in details the concentration cell and the Nernst Equation:

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