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 thatrole="math" localid="1659592400010" ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}$ >role="math" localid="1659592428249" ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{\mathbf{\text{o}}}$

(a) How dorole="math" localid="1659592855546" $\mathbf{\left[}{\mathbf{\text{A}}}^{\mathbf{\text{+}}}\mathbf{\right]}$ and $\mathbf{\left[}{\mathbf{\text{B}}}^{\mathbf{\text{+}}}\mathbf{\right]}$change as the cell operates?

(b) How doesrole="math" localid="1659592922361" ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}$ change as the cell operates?

(c) What is role="math" localid="1659592941387" $\mathbf{\left[}{\mathbf{\text{A}}}^{\mathbf{\text{+}}}\mathbf{\right]}\mathbf{\text{/}}\mathbf{\left[}{\mathbf{\text{B}}}^{\mathbf{\text{+}}}\mathbf{\right]}$ when${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}$ =role="math" localid="1659592990948" ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{\mathbf{\text{o}}}$ ? Explain.

(d) Is it possible forrole="math" localid="1659593025964" ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{}$ to be less than ${\mathbf{\text{E}}}_{\mathbf{\text{cell}}}^{\mathbf{\text{o}}}$? Explain

Short 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-by-step solution

 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:

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

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.

Changes in Ecell

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.

Ecell=E0cell

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 $\left[{\text{A}}^{\text{+}}\right]\text{=}\left[{\text{B}}^{\text{+}}\right]$.

Ecell<E0cell

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${\text{E}}_{\text{cell}}^{\text{0}}$.