Question

Estimate the cell potential of a Daniell cell having $1.0MZ{n}^{2+}$ and originally haiving $1.0MC{u}^{2+}$ after sufficient ammonia has been added to the cathode compartment to make the $N{H}_3$ concentration 2.0 M. Given $E_{Zn/Z{n}^{2+}}^o,E_{Cu/C{u}^{2+}}^o$ are 0.76 and -0.34 V respectively. Also equilibrium constant for the $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ formation is $1\times {10}^{12}$:

• A. 0.71
• B. 1.42
• C. 0.63
• D. 0.86

Answer 1

The correct option is A 0.71

$C{u}^{2+}+4N{H}_3\rightleftharpoons \left[Cu\right(N{H}_3{)}_4{]}^{2+}$

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

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

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, $E_{cell}=E_{O{P}_{Zn/Z{n}^{2+}}}^o+E_{R{P}_{C{u}^{2+}/Cu}}^o+\frac{0.059}{2}lo{g}_{10}\frac{C{u}^{2+}}{Z{n}^{2+}}$

$=0.76+0.34+\frac{0.059}{2}lo{g}_{10}\left[\frac{6.25\times {10}^{-14}}{1}\right]$

$E_{cell}=0.71V$

Hence, option A is correct.