Question

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:

(a) ${\mathbf{\text{Al(s) + Cr}}}^{\mathbf{\text{3 +}}}{\mathbf{\text{(aq) n Al}}}^{\mathbf{\text{3 +}}}\mathbf{\text{(aq) + Cr(s)}}$

(b) ${\mathbf{\text{Cu}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) + SO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + 2H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l) n}}$$\mathbf{\text{Cu(s) + S}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq) + 4H}}}^{\mathbf{\text{+}}}\mathbf{\text{(aq)}}$

Short answer

a)${\text{Al}}_{\text{(s)}}\left|{\text{Al}}_{\text{(aq)}}^{\text{3 +}}{\text{PCr}}_{\text{(aq)}}^{\text{3 +}}\right|{\text{Cr}}_{\text{(s)}}$

b)${\text{Pt}}_{\text{(s)}}\left|{\text{SO}}_{\text{2(g)}}\right|{\text{SO}}_{\text{4(aq)}}^{\text{2 -}}{\text{PCu}}_{\text{(aq)}}^{\text{2 +}}\mid {\text{Cu}}_{\text{(s)}}$

Step-by-step solution

Definition of redox reaction

Electrons are transferred from one chemical material to another in many chemical processes. The oxidation-reduction or Redox reaction is the name given to these electron transport reactions.

Determining the (a) reaction

Divide states for either side of the reaction must be separated, and the double bar must separate the two electrodes. A comma must be used to separate different ions in the same state.

$$\begin{gathered} Oxidation\text{:Al}\to {\text{Al}}^{\text{3 +}}{\text{+ 3e}}^{\text{-}} \\ {\text{Reduction: Cr}}^{\text{3 +}}{\text{+ 3e}}^{\text{-}}\to \text{Cr} \end{gathered}$$

The anode, which is on the left side of the double bar, undergoes oxidation, whereas the cathode, which is on the right side, undergoes reduction.

Answer: ${\text{Al}}_{\text{(s)}}\left|{\text{Al}}_{\text{(aq)}}^{\text{3 +}}{\text{PCr}}_{\text{(aq)}}^{\text{3 +}}\right|{\text{Cr}}_{\text{(s)}}$

Determining the (b) reaction

$$\begin{gathered} Reduction{\text{: Cu}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}\to \text{Cu} \\ Oxidation:{\text{SO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}\to {\text{SO}}_{\text{4}}^{\text{2 -}}{\text{+ 4H}}^{\text{+}}{\text{+ 2e}}^{\text{-}} \end{gathered}$$

The anode, which is on the left side of the double bar, undergoes oxidation, whereas the cathode, which is on the right side, undergoes reduction. Because none of the species in the anode are solid phase, it must be accompanied by an inactive electrode. Spectator ions in the oxidation half process, such as ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}^{\text{+}}$, are not included in the cell notation.

Answer: ${\text{Pt}}_{\text{(s)}}\left|{\text{SO}}_{\text{2(g)}}\right|{\text{SO}}_{\text{4(aq)}}^{\text{2 -}}{\text{PCu}}_{\text{(aq)}}^{\text{2 +}}\mid {\text{Cu}}_{\text{(s)}}$