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. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) \(\mathscr{C}^{\circ}=0.68 \mathrm{~V}\) \(\begin{array}{ll}\text { b. } \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} & \mathscr{E}^{\circ}=-1.18 \mathrm{~V} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.036 \mathrm{~V}\end{array}\)

Step by step solution

01

1. Identify the anode and cathode based on the standard potentials

Reduction potentials are given, and the reaction with the highest standard reduction potential will occur at the cathode, while the reaction with the lower reduction potential will happen at the anode. Cathode reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78\mathrm{~V}\) Anode reaction: \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\mathrm{O}_{2}\quad\mathscr{E}^{\circ}=0.68 \mathrm{~V}\)

02

2. Determine the direction of electron flow and ion migration

Electrons flow from the anode (negative side) to the cathode (positive side). For ion migration, anions move from the cathode to the anode, and cations move from the anode to the cathode through the salt bridge.

03

3. Write the overall balanced equation for the reactions

To get the overall balanced equation, we need to combine the anode and cathode reactions: \(\mathrm{H}_{2} \mathrm{O}_{2} + 2 \mathrm{H}^+ + 2 \mathrm{e}^- + \mathrm{O}_2 + 2 \mathrm{H}^+ + 2 \mathrm{e}^- \rightarrow 2 \mathrm{H}_2 \mathrm{O} + \mathrm{H}_{2}\mathrm{O}_{2}\) After simplifying, the overall balanced equation is: \(\mathrm{O}_{2} + 2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O}_{2}\)

04

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

Standard cell potential (\(\mathscr{E}^{\circ}\)) is determined by subtracting the cathode's standard potential from the anode's standard potential: \(\mathscr{E}^{\circ} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.78 \mathrm{~V} - 0.68 \mathrm{~V} = 1.1 \mathrm{~V}\) #Case b: Sketching the galvanic cell for given half-reactions#

05

1. Identify the anode and cathode based on the standard potentials

Cathode reaction: \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036\mathrm{~V}\) Anode reaction: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{~V}\)

06

2. Determine the direction of electron flow and ion migration

Electrons flow from the anode (negative side) to the cathode (positive side). Anions move from the cathode to the anode and cations move from the anode to the cathode through the salt bridge.

07

3. Write the overall balanced equation for the reactions

To get the overall balanced equation, we need to combine the anode and cathode reactions and balance the number of electrons. By multiplying the cathode reaction by 2 and the anode reaction by 3, the electrons will be balanced: \(2(\mathrm{Fe}^{3+} + 3 \mathrm{e}^-) \rightarrow 6 \mathrm{e}^- + 6 \mathrm{Fe}^{3+} \rightarrow 3\mathrm{Fe^{2+}} + \mathrm{Mn} \rightarrow \mathrm{Mn}^{3+} + 2 \mathrm{Fe}\) After simplifying, the overall balanced equation is: \(2 \mathrm{Fe}^{3+} + 3 \mathrm{Mn}^{2+} \rightarrow 2 \mathrm{Mn}^{3+} + 3 \mathrm{Fe}\)

08

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

\(\mathscr{E}^{\circ} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = -0.036 \mathrm{~V} - (-1.18 \mathrm{~V}) = 1.144 \mathrm{~V}\)

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!