Question

Consider the following galvanic cell:

A 15.0-mole sample of NH₃ is added to the Ag compartment

(assume 1.00 L of total solution after the addition).

The silver ion reacts with ammonia to form complex

ions as shown:

$$\begin{gathered} \mathit{A}{\mathit{g}}^{\mathbf{+}}\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\to }\mathit{A}\mathit{g}\mathit{N}{\mathit{H}}_{\mathbf{3}}\left(K_1=2.1\times {10}^3\right) \\ \mathit{A}\mathit{g}\mathit{N}{\mathit{H}}_{\mathbf{3}}^{\mathbf{+}}\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\to }\mathit{A}\mathit{g}{\left(N{H}_3\right)}_{\mathbf{2}}\left(K_2=8.2\times {10}^3\right) \end{gathered}$$

Calculate the cell potential after the addition of

15.0 moles of NH₃.

Short answer

The cell potential after the addition of 15.0 moles of NH₃ will be 0.64 V

Step-by-step solution

analyzing the given data

Given equations are;

$$\begin{gathered} A{g}^{+}+e^{-}\to 2Ag\left(E°=-0.80V\right) \\ Cd\to C{d}^{2+}+2e^{-}\left(E°=0.40V\right) \end{gathered}$$

The final equation is,

$2A{g}^{+}+Cd\to 2Ag+C{d}^{2+}+2e^{-}\left(E_{CELL}^{°}=1.20V\right)$

After reaction of ammonia with silver ion,

$$\begin{gathered} A{g}^{+}+N{H}_3\to AgN{H}_3\left(K_1=2.1\times {10}^3\right) \\ AgN{H}_3^{+}+N{H}_3\to Ag{\left(N{H}_3\right)}_2\left(K_2=8.2\times {10}^3\right) \\ A{g}^{+}+2N{H}_3\to Ag{\left(N{H}_3\right)}_2\left(K=K_1\times K_2=1.7\times {10}^7\right) \end{gathered}$$

calculating the cell potential after the addition of 15 moles of nitric acid

K has a larger value, so the reaction tends to complete and hence, occurs the backward reaction.

$$\begin{gathered} A{g}^{+}+2N{H}_3\to Ag{\left(N{H}_3\right)}_2 \\ K=\frac{\left[Ag{\left(N{H}_3\right)}_2\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2} \\ =1.7\times {10}^7 \end{gathered}$$

Neglecting the change in x, since it is negligible

$$\begin{gathered} 1.7\times {10}^7=\frac{\left[1.00\right]}{\left[x\right]\left[13.0\right]} \\ x=3.5\times {10}^{-10} \\ K=\frac{\left[C{d}^{2+}\right]}{{\left[A{g}^{+}\right]}^2} \\ =\frac{\left[1.0\right]}{{\left[3.5\times {10}^{-10}\right]}^2} \\ =8.16\times {10}^{18} \end{gathered}$$

Using Nernst equation:

$$\begin{gathered} E=E_{CELL}°-\left[\frac{0.059}{n}\right]\log Q \\ =1.20-\left[\frac{0.059}{n}\right]\log \left(8.16\times {10}^{18}\right) \\ =0.64V \end{gathered}$$

Hence, the new cell potential will be 0.64 V.