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?

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Answer 1 · 2015-11-13T17:49:18

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

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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