Question

An electrochemical cell consists of a zinc metal electrode immersed in a solution with (Zn²⁺) = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 M NH; that is also 0,010 M in$\mathbf{Cu}{\mathbf{(}{\mathbf{NH}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{+}}$ , what is the cell potential at 25°C.
${\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{Cu}{\mathbf{(}{\mathbf{NH}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}$

Short answer

Thevalue of cell potential can be found as 0.57 V.

Step-by-step solution

Finding concentration of copper

The expression of equilibrium constant can be denoted as,

$\begin{array}{rcl}\mathrm{K}& =& \frac{\left[\mathrm{Cu}{{\left({\mathrm{NH}}_3\right)}_4}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]{\left[{\mathrm{NH}}_3\right]}^4}\\ \left[{\mathrm{Cu}}^{2+}\right]& =& \frac{\left[\mathrm{Cu}{{\left({\mathrm{NH}}_3\right)}_4}^{2+}\right]}{\mathrm{K}{\left[{\mathrm{NH}}_3\right]}^4}\\ \left[{\mathrm{Cu}}^{2+}\right]& =& \frac{(0.010)}{(1\times {10}^{13}){\left(5\right)}^4}\\ \left[{\mathrm{Cu}}^{2+}\right]& =& 1.6\times {10}^{-18}\mathrm{M}\end{array}$

Finding the value of cell potential

The value can be found as,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \left(\frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\right) \\ {\mathrm{E}}_{\mathrm{cell}}=(1.1 \mathrm{V})-\frac{0.0591}{\mathrm{n}}\log \left(\frac{\left(1 \mathrm{M}\right)}{\left(1.6\times {10}^{-18}\mathrm{M}\right)}\right) \\ {\mathrm{E}}_{\mathrm{cell}}=0.57 \mathrm{V} \end{gathered}$$