Respuesta :
The question is missing. Here is the complete question.
Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]?
(a) [tex]Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}[/tex]
(b) [tex]2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}[/tex]
(c)[tex]Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}[/tex]
(d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Answer: (d) [tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.
Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.
For the cell notation: [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex]
Aluminum's half-equation is oxidation:
[tex]Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}[/tex]
For Lead, half-equation is reduction:
[tex]Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}[/tex]
Multiply first half-equation for 2 and second half-equation by 3:
[tex]2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}[/tex]
[tex]3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}[/tex]
Adding them:
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
The balanced redox reaction with cell notation [tex]Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}[/tex] is
[tex]2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}[/tex]
