# 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?

Nov 13, 2015

#### Answer:

${\left[C {u}^{2} +\right]}_{\text{Anode}} = 5.7 \times {10}^{- 11} M$

#### Explanation:

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

Anode : ${\underbrace{C u \left(s\right) \to C {u}^{2 +} \left(a q\right) + 2 {e}^{-}}}_{\textcolor{b l u e}{\text{Oxidation}}}$$\text{ } - {E}^{\circ} = - 0.34 V$

Cathode : ${\underbrace{C {u}^{2 +} \left(a q\right) + 2 {e}^{-} \to C u \left(s\right)}}_{\textcolor{b l u e}{\text{Reduction}}}$$\text{ } {E}^{\circ} = 0.34 V$

Overall equation : $C u \left(s\right) + {\underbrace{C {u}^{2 +} \left(a q\right)}}_{C a t h o \mathrm{de}} \to C {u}^{2 +} \left(a q\right) + C u \left(s\right)$

${E}_{\text{cell")^@=E_("Anode")^@+E_("Cathode}}^{\circ} = 0 V$

For concentration cells, we can calculate the cell potential ${E}_{\text{cell}}$ using the Nernst Equation:

${E}_{\text{cell")=E_("cell}}^{\circ}$$- \frac{0.0591}{n}$$\log \left(\frac{{\left[C {u}^{2 +}\right]}_{A n o \mathrm{de}}}{{\left[C {u}^{2 +}\right]}_{C}}\right)$

Replacing the giving terms in this equation assuming that the given concentration is ${\left[C {u}^{2} +\right]}_{\text{Cathode}} = 1.6 \times {10}^{- 3} M$

and that ${\left[C {u}^{2} +\right]}_{\text{Anode}} = x$

$0.22 = 0 - \frac{0.0591}{2} \log \left(\frac{x}{1.6 \times {10}^{- 3}}\right)$

Solve for $x = 5.7 \times {10}^{- 11} M$

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

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