Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad 8^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\) \(\mathscr{b}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E ^ { \circ }}=1.60 \mathrm{~V}\)

Step by step solution

01

Identify the half-reactions

Since the half-reactions are given, we just need to identify the anode (oxidation) and the cathode (reduction). \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{~V}\) In a galvanic cell, the half-reaction with the higher reduction potential is the one that will undergo reduction. Here, the \(\mathrm{Cl}_2\) reaction has a higher \(\mathscr{E}^{\circ}\), so it will happen at the cathode.

02

Write the overall balanced equation

To obtain the overall balanced equation, we multiply each half-reaction with an appropriate factor so that the number of electrons is the same in both reactions. In this case, we don't need to multiply anything. Overall equation: \(\mathrm{Cl}_{2} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{Cl}^{-} + 2 \mathrm{Br}^{-}\)

03

Determine \(\mathscr{E}^{\circ}\) for the cell

\(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.36 \mathrm{~V} - 1.09 \mathrm{~V} = 0.27 \mathrm{~V}\)

04

Sketch the cell

Anode: Bromine electrode (oxidation) Cathode: Chlorine electrode (reduction) Electron flow: From anode to cathode (Br2 to Cl2) Ion migration: Cl- ions move from cathode to anode, Br- ions move from anode to cathode through the salt bridge. b. Galvanic cell with MnO4- and IO4- reactions:

05

Identify the half-reactions

Since the half-reactions are given, we just need to identify the anode (oxidation) and the cathode (reduction). \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\) Here, the \(\mathrm{IO}_{4}^{-}\) reaction has a higher \(\mathscr{E}^{\circ}\), so it will happen at the cathode.

06

Write the overall balanced equation

To obtain the overall balanced equation, we multiply each half-reaction with an appropriate factor so that the number of electrons is the same in both reactions. We'll need to multiply the first reaction by 2 and the second by 5. Overall equation: \(2 (\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}) + 5(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}) \rightarrow 2(\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}) + 5(\mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O})\)

07

Determine \(\mathscr{E}^{\circ}\) for the cell

\(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.60 \mathrm{~V} - 1.51 \mathrm{~V} = 0.09 \mathrm{~V}\)

08

Sketch the cell

Anode: Manganese electrode (oxidation) Cathode: Iodine electrode (reduction) Electron flow: From anode to cathode (MnO4- to IO4-) Ion migration: Mn2+ ions move from anode to cathode, IO3- ions move from cathode to anode through the salt bridge. H+ ions will also migrate to maintain electric neutrality.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!