Question

Calculate the concentration of silver ions in the cell constructed by using $0.1M$ concentration of ${Cu}^{2+}$ and ${Ag}^{+}$ ions. $Cu$ and $Ag$ metals are used as electrodes. The cell potential is $0.422V$.

Answer 1

The electrode potentials of ${Ag}^{+}/Ag$ is positive,so silver silver electrode acts as cathode and copper-copper electrode will acts as anode.

The electrode reaction,

Anode:$Cu\to {Cu}^{2+}+2e^{-}$

Cathode:$Ag\to {Ag}^{+}+e^{-}$

$E_{anode}=+0.34V$

$E_{cathode}=+0.80V$

$E_{cell}^0=+0.80V-0.34V$

$E_{cell}^0=0.46V$

The Emf of the cell is given by the equation,

$E_{cell}=E_{Cathode}-E_{Anode}-\frac{0.0591}{2}log\frac{{Cu}^{2+}}{{Ag}^{+}}0.422V=0.46V-\frac{0.0591}{2}log\frac{0.1}{{\left[A{g}^{+}\right]}^2}$

$log\frac{0.1}{{\left[A{g}^{+}\right]}^2}=\frac{0.46\times 2}{0.0591}=1.286$

$\frac{0.1}{{\left[A{g}^{+}\right]}^2}=antilog\left(1.286\right)=19.32\left[{Ag}^{+}\right]={\left(\frac{0.1}{19.32}\right)}^{\frac{3}{2}}=7.19\times {10}^{-2}mo{L}^{-1}$