Question

The standard reduction potentials of $C{u}^{2+}|Cu$ and $A{g}^{+}|Ag$ electrodes are $0.337$ and $0.799volts$ respectively. Construct a galvanic cell using these electrodes so that its standard emf is positive. For what concentration of $A{g}^{+}$ will the emf of the cell at ${25}^{\circ }C$ be zero if concentration of $C{u}^{2+}$ is $0.01M$?

Answer 1

Given, $E_{C{u}^{2+}/Cu}^{\circ }=0.337volt$ and $E_{A{g}^{+}/Ag}^{\circ }=0.799volt$. The standard emf will be positive if $Cu/C{u}^{2+}$ is anode and $A{g}^{+}/Ag$ is cathode. The cell can be represented as:
$Cu|C{u}^{2+}\parallel A{g}^{+}|Ag$
The cell reaction is,
$Cu+2A{g}^{+}\to C{u}^{2+}+2Ag$
$E_{cell}^{\circ }=$ Oxid. potential of anode $+$ Red. potential of cathode
$=-0.337+0.799$
$=0.462volt$
Applying the Nernst equation,
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}\log \frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}$
When, $E_{cell}=0$
$E_{cell}^{\circ }=\frac{0,0591}{2}\log \frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}$
or $\log \frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}=\frac{0.462\times 2}{0.0591}=15.6345$
$\frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}=4.3102\times {10}^{15}$
$[A{g}^{+}{]}^2=\frac{0.01}{4.3102\times {10}^{15}}$
$=0.2320\times {10}^{-17}$
$=2.320\times {10}^{-18}$
$\left[A{g}^{+}\right]=1.523\times {10}^{-9}M$.