Question

Estimate the cell potential of a Daniel cell having 1.0 M $Z{n}^{++}$ and originally having 1.0 M $C{u}^{++}$ after sufficient $N{H}_3$ has been added to the cathode compartment to make $N{H}_3$ concentration 2.0 M. Given $K_f$ for ${\left[Cu\right(N{H}_3{)}_4]}^{2+}=1\times {10}^{12},$

$E^{\circ }$ for the reaction in V,$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$ is 1.1 V.(write the value to the nearest integer)

Answer 1

$C{u}^{2+}\rightleftharpoons {\left[Cu\right(N{H}_3{)}_4]}^{+2}$

$\therefore K_f=1\times {10}^{12}=\frac{{\left[Cu\right(N{H}_3{)}_4]}^{+2}}{\left[C{u}^{+2}\right][N{H}_3{]}^4}=\frac{1.0}{x(2.0{)}^4}$

$\therefore x=6.25\times {10}^{-14}M$

Note that due to high value of $K_f$ almost all of the $C{u}^{+2}$ ions are converted to $Cu(N{H}_3{)}_4^{2+}$ ion

Now $E_{cell}=E_{cell}^{\circ }+\frac{0.059}{2}{\log }_{10}\frac{C{u}^{2+}}{Z{n}^{+2}}$

$=1.1+\frac{0.059}{2}{\log }_{10}\left[\frac{6.25\times {10}^{-14}}{1}\right]$

$E_{cell}=0.71V$

The nearest integer of 0.71 is $1$