Q21.53P - Chemistry: Molecular Nature Of Matter And Change

Question

A Voltaic cell consists of a metal $\mathbf{A}\mathbf{/}{\mathbf{A}}^{+}$   electrode and a metal  $\mathbf{B}\mathbf{/}{\mathbf{B}}^{+}$ electrode, with the $\mathbf{A}\mathbf{/}{\mathbf{A}}^{+}$  electrode negative. The initial  $\mathbf{\left[}{\mathbf{\text{A}}}^{\mathbf{\text{+}}}\mathbf{\right]}\mathbf{\text{/}}\mathbf{\left[}{\mathbf{\text{B}}}^{\mathbf{\text{+}}}\mathbf{\right]}$  is such that   ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}$ >   ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}^{\mathbf{\text{o}}}$

(a) How do  $\mathbf{\left[}{\mathbf{\text{A}}}^{\mathbf{\text{+}}}\mathbf{\right]}$ and  $\mathbf{\left[}{\mathbf{\text{B}}}^{\mathbf{\text{+}}}\mathbf{\right]}$ change as the cell operates? 

(b) How does  ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}$  change as the cell operates? 

(c) What is   $\mathbf{\left[}{\mathbf{\text{A}}}^{\mathbf{\text{+}}}\mathbf{\right]}\mathbf{\text{/}}\mathbf{\left[}{\mathbf{\text{B}}}^{\mathbf{\text{+}}}\mathbf{\right]}$  when ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}$ = ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}^{\mathbf{\text{o}}}$  ? Explain. 

(d) Is it possible for  ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}^{}$  to be less than  ${\mathbf{\text{E}}}_{\mathbf{\text{cell }}}^{\mathbf{\text{o}}}$? Explain

Step-by-Step Solution

Answer

(a)  $\left[{\text{B}}^{\text{+}}\right]$ decreases while $\left[{\text{B}}^{\text{+}}\right]$ increases.

(b) The ${\text{E}}_{\text{cell }}$ will continue to decrease.

(c)   ${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{0}}$ when  $\left[{\text{A}}^{\text{+}}\right]\text{=}\left[{\text{B}}^{\text{+}}\right]$.

 (d)   ${\text{E}}_{\text{cell}}^{\text{0}}$can be greater than ${\text{E}}_{\text{cell}}$ when    $\left[{\text{A}}^{\text{+}}\right]>\left[{\text{B}}^{\text{+}}\right]$.

Step 1: Cell reaction

The overall cell reaction is represented below.

  

$\begin{array}{l}{{\text{B}}^{\text{+}}}_{\left(\text{aq}\right)}{\text{+\hspace{0.17em}e}}^{\text{-}}\to {\text{B}}_{\left(\text{s}\right)}\\ {\text{A}}_{\left(\text{s}\right)}\to {{\text{A}}^{\text{+}}}_{\left(\text{aq}\right)}{\text{+\hspace{0.17em}e}}^{\text{-}}\\ {\text{B}}_{\text{(aq)}}^{\text{+}}{\text{+A}}_{\text{(s)}}\to {\text{B}}_{\text{(s)}}{\text{+A}}_{\text{(aq)}}^{\text{+}}\end{array}$

 The reaction quotient for the above redox reaction is:

  $\mathrm{Q}=\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$

Step 2: Changes in concentration of [ A + ] and [ B + ]

As it is evident from the cell reaction, Metal A is oxidized to ${\text{A}}^{\text{+}}$  ions whereas ${\text{B}}^{\text{+}}$  ions are reduced to metal B. The atoms of B are deposited on the electrode as a result, the concentration of  ${\text{B}}^{\text{+}}$ ions decrease in the solution. The reverse happens with metal A. The neutral atoms of metal A are oxidized into  ${\text{A}}^{\text{+}}$ and thus the concentration of ${\text{A}}^{\text{+}}$ ions increase in the solution.

Step 3: Changes in E cell

The Nernst equation gives the expression for the cell potential of a given electrolytic cell.

  ${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$

The ${\text{E}}_{\text{cell}}^{\text{0}}$  is the cell potential under standard conditions and is equal to the difference between two half-cell electrode potential. As the cell operates, the concentration of ${\text{A}}^{\text{+}}$ increases and  ${\text{B}}^{\text{+}}$ decreases, the term $\text{ln}\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$ becomes larger. Since ${\text{E}}_{\text{cell}}^{\text{0}}$  is constant,  ${\text{E}}_{\text{cell}}$  decreases with progress of the reaction.

Step 4: E cell = E 0 cell

We will use the Nernst equation to solve the problem. The Nernst equation is:

 ${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{0}}-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$

For  ${\text{E}}_{\text{cell}}{{\text{=E}}^{\text{0}}}_{\text{cell}}$ , the term  $\text{ln}\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$ should be equal to zero.

The above criteria will be fulfilled when $\left[{\text{A}}^{\text{+}}\right]\text{=}\left[{\text{B}}^{\text{+}}\right]$ .

Step 5: E cell < E 0 cell

  Referring to the Nernst equation, it can be said that  ${\text{E}}_{\text{cell}}^{\text{0}}$ can be greater than ${\text{E}}_{\text{cell}}$ when the $\left[{\text{A}}^{\text{+}}\right]>\left[{\text{B}}^{\text{+}}\right]$  . The term  $\text{ln}\frac{\left[{\text{A}}^{\text{+}}\right]}{\left[{\text{B}}^{\text{+}}\right]}$ becomes positive and consequently ${\text{E}}_{\text{cell}}$ becomes less than ${\text{E}}_{\text{cell}}^{\text{0}}$ .