A common shorthand way to represent a voltaic cell is anode \(\mid\) anode solution || cathode solution \(\mid\) cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by $\mathrm{Fe}\left|\mathrm{Fe}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag} ;$ calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{H}^{+}\right| \mathrm{H}_{2} ;$ calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s) &+6 \mathrm{H}^{+}(a q) \\ & \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) $6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) ; E^{\circ}=1.45 \mathrm{~V}$

Step by step solution

01

1a. Identify half-reactions of Fe | Fe²⁺ || Ag⁺ | Ag cell

The anode is Fe, and the cathode is Ag. The anode half-reaction is the oxidation of Fe, and the cathode half-reaction is the reduction of Ag⁺: Anode half-reaction: \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\) Cathode half-reaction: \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\)

02

2a. Identify overall cell reaction

Combine the half-reactions to get the overall cell reaction: Overall reaction: \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\)

03

3a. Calculate the standard cell emf

Look up the standard reduction potentials in Appendix E (not provided here, but values can be found in any standard chemistry textbook) and use them to calculate the standard cell emf: \(E^∘_{Fe^{2+}/Fe} = -0.44 V\) \(E^∘_{Ag^{+}/Ag} = 0.8 V\) Standard cell emf is the difference between the cathode and anode potentials: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0.8 V - (-0.44 V) = 1.24 V\)

04

1b. Identify half-reactions of Zn | Zn²⁺ || H⁺ | H₂ cell

The anode is Zn, and the cathode is H₂. The anode half-reaction is the oxidation of Zn, and the cathode half-reaction is the reduction of H⁺ using Pt as the hydrogen electrode: Anode half-reaction: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\) Cathode half-reaction: \(2H^{+}(aq) + 2e^- \rightarrow H_2(g)\)

05

2b. Identify overall cell reaction

Combine the half-reactions to get the overall cell reaction: Overall reaction: \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_2(g)\)

06

3b. Calculate the standard cell emf

Look up the standard reduction potentials in Appendix E and use them to calculate the standard cell emf: \(E^∘_{Zn^{2+}/Zn} = -0.76 V\) \(E^∘_{H^{+}/H_2} = 0 V\) Standard cell emf is the difference between the cathode and anode potentials: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0 V - (-0.76 V) = 0.76 V\)

07

1c. Represent the cell and calculate the standard cell emf for the given reaction

For the given reaction, chlorine is the anode and the copper is the cathode: Cell representation: \(Pt | ClO_3^-(aq), Cl^-(aq) || Cu^{2+}(aq) | Cu(s)\) Anode half-reaction: \(ClO_3^-(aq) + 6H^{+}(aq) + 6e^- \rightarrow Cl^-(aq) + 3H_2O(l)\) with \(E^∘_{anode}=1.45 V\) Cathode half-reaction: \(3Cu^{2+}(aq) + 6e^- \rightarrow 3Cu(s)\) Look up the standard reduction potential of the Cu²⁺ | Cu half-reaction in Appendix E to calculate the standard cell emf: \(E^∘_{Cu^{2+}/Cu} = 0.34 V\) Standard cell emf is the difference between the cathode and anode potentials, considering the standard reduction potential provided is for the anode half-reaction: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0.34 V - 1.45 V = -1.11 V\)

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