A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ 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 Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) 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: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the 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

Write half-reactions

In this voltaic cell, the two half-reactions are the reduction of Ag+ and the oxidation of Fe: Oxidation half-reaction (anode): \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\) Reduction half-reaction (cathode): \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\)

02

Write overall cell reaction

Now, we will balance the overall cell reaction by combining the two half-reactions, making sure the electrons are balanced. Overall cell reaction: \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\)

03

Calculate the standard cell emf

Using the standard electrode potentials from Appendix E, we can calculate the emf of the cell: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) The values for the given half-reactions are: \(E_{Fe^{2+}/Fe}^{\circ} = -0.44 V\) \(E_{Ag^{+}/Ag}^{\circ} = 0.80 V\) Thus, emf of the cell: \(E_{cell}^{\circ} = 0.80V - (-0.44V) = 1.24V\) (b)

04

Write half-reactions

In this voltaic cell, the two half-reactions are the reduction of H+ and the oxidation of Zn: Oxidation half-reaction (anode): \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\) Reduction half-reaction (cathode): \(2H^{+}(aq) + 2e^- \rightarrow H_{2}(g)\)

05

Write overall cell reaction

Now, combine the two half-reactions, making sure the electrons are balanced: Overall cell reaction: \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_{2}(g)\)

06

Calculate the standard cell emf

Using the standard electrode potentials from Appendix E: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) The values for the given half-reactions are: \(E_{Zn^{2+}/Zn}^{\circ} = -0.76 V\) \(E_{H^{+}/H_{2}}^{\circ} = 0.00 V\) Thus, emf of the cell: \(E_{cell}^{\circ} = 0.00V - (-0.76V) = 0.76V\) (c)

07

Represent the cell

Using the provided reaction, the cell may be represented as: \(Cu|Cu^{2+}||ClO_{3}^{-}, Cl^{-}|Pt\)

08

Write half-reactions

In this voltaic cell, the two half-reactions are: Oxidation half-reaction (anode): \(3Cu(s) \rightarrow 3Cu^{2+}(aq) + 6e^-\) Reduction half-reaction (cathode): \(ClO_{3}^{-}(aq) + 6H^{+}(aq) + 6e^- \rightarrow Cl^{-}(aq) + 3H_{2}O(l)\)

09

Write overall cell reaction

Combining these half-reactions, we obtain the overall cell reaction: \(ClO_{3}^{-}(aq) + 3Cu(s) + 6H^{+}(aq) \rightarrow Cl^{-}(aq) + 3Cu^{2+}(aq) + 3H_{2}O(l)\)

10

Calculate the standard cell emf

Now, we need to calculate the standard emf for the oxidation half-reaction. Given: \(E_{ClO_{3}^{-}/Cl^{-}}^{\circ} = 1.45V\) We need to find the standard emf for the oxidation half-reaction as follows: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) Rearranging, we can find the standard emf for the anode: \(E_{anode}^{\circ} = E_{cathode}^{\circ} - E_{cell}^{\circ}\) Using the given values: \(E_{anode}^{\circ} = 1.45V - E_{Cu^{2+}/Cu}^{\circ}\) From Appendix E, we know: \(E_{Cu^{2+}/Cu}^{\circ} = 0.34 V\) Thus, the standard cell emf: \(E_{cell}^{\circ} = 1.45V - 0.34V = 1.11V\)

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