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Chapter 13: Problem 9

Draw a picture of the following cell and write reduction halfreactions for each electrode: \(\operatorname{Pt}(s)\left|\mathrm{Fe}^{3+}(a q), \mathrm{Fe}^{2+}(a q) \| \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q), \mathrm{Cr}^{3+}(a q), \mathrm{HA}(a q)\right| \operatorname{Pt}(s)\)

Short Answer

Expert verified
Reduction half-reactions are: \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) and \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\).

Step by step solution

01

Understanding the Cell Notation

The given cell notation is \(\operatorname{Pt}(s)|\mathrm{Fe}^{3+}(aq), \mathrm{Fe}^{2+}(aq) \| \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq), \mathrm{Cr}^{3+}(aq), \mathrm{HA}(aq)| \operatorname{Pt}(s)\). It represents a galvanic cell with two half-cells. Platinum (Pt) electrodes are used because the components involved are in aqueous phases. The single vertical line (|) separates different phases in the same half-cell, and the double vertical line (||) represents the salt bridge between the two half-cells.
02

Identify the Components of Each Half-Cell

The left half-cell consists of the iron ions \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\), indicating it involves the reduction or oxidation of iron. The right half-cell consists of \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), \(\mathrm{Cr}^{3+}\), and \(\mathrm{HA}\), indicative of chromium redox reaction.
03

Write the Reduction Half-Reaction for the Iron System

In the left half-cell, the reduction reaction converts \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\): \[\mathrm{Fe}^{3+}(aq) + e^- \rightarrow \mathrm{Fe}^{2+}(aq)\]
04

Write the Reduction Half-Reaction for the Chromium System

In the right half-cell, the reduction reaction involves \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) converting to \(\mathrm{Cr}^{3+}\) in an acidic medium: \[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 14\mathrm{H}^+(aq) + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_2\mathrm{O}(l)\] This reaction shows electrons being gained, thus depicting reduction.
05

Drawing the Galvanic Cell Diagram

To draw the galvanic cell diagram, sketch two compartments. Label the left compartment as the iron half-cell with Pt electrode, \(\mathrm{Fe}^{3+}(aq)\), and \(\mathrm{Fe}^{2+}(aq)\). In the left compartment, show a voltmeter connected to the Pt electrode. Label the right compartment as the chromium half-cell with another Pt electrode, showing \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq)\), \(\mathrm{Cr}^{3+}(aq)\), and \(\mathrm{HA}(aq)\). Connect a salt bridge between the two compartments to complete the circuit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduction half-reaction
In electrochemistry, understanding reduction half-reactions is fundamental. A reduction half-reaction illustrates how a substance gains electrons. For the cell in question, the reduction half-reaction simplifies how iron changes state. Iron goes from a higher to a lower oxidation state:
  • The iron ions involved are \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\).
  • The reduction half-reaction is: \[\mathrm{Fe}^{3+}(aq) + e^- \rightarrow \mathrm{Fe}^{2+}(aq)\]
This equation shows that an electron is gained by \(\mathrm{Fe}^{3+}\) to form \(\mathrm{Fe}^{2+}\). Understanding such reactions is crucial because they depict the flow of electrons, a backbone concept in electrochemistry.
Galvanic cell
A galvanic cell, also known as a voltaic cell, is where chemical energy transforms into electrical energy. It consists of two half-cells. Each half-cell contains an electrode and an electrolyte.
  • The two half-cells in our cell are the iron and chromium systems.
  • Electrons flow from one half-cell to the other, generating electricity.
A salt bridge, often a gel containing ions, connects them. This maintains electrical neutrality by allowing ions to flow between half-cells. The galvanic cell diagram drawn in Step 5 visually represents this concept. It highlights how the setup of the cell facilitates the redox reactions and current flow through external circuits.
Cell notation
Cell notation succinctly describes a galvanic cell's composition. It uses certain conventions to specify the reactions happening at electrodes. Our given cell notation is: \(\operatorname{Pt}(s)|\mathrm{Fe}^{3+}(aq), \mathrm{Fe}^{2+}(aq) \| \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq), \mathrm{Cr}^{3+}(aq), \mathrm{HA}(aq)| \operatorname{Pt}(s)\).
  • The single vertical line \((|)\) separates different phases in the same half-cell.
  • The double vertical line \((||)\) represents the salt bridge between the two half-cells.
  • Platinum \((Pt)\) electrodes are inert conductors in electrodes where neither the reactant nor the product is in solid or gaseous form.
This notation encapsulates all essential components, indicating what reactions occur in the galvanic cell.
Iron redox reaction
Redox reactions involve oxidation and reduction processes. In the cell鈥檚 iron half-cell, the redox reaction centers on iron ions. The reduction reaction was previously explained. However, understanding redox requires seeing oxidation too:
  • Oxidation loses electrons, while reduction gains them.
  • In our context, if reduction involves \(\mathrm{Fe}^{3+}\), then oxidation would see \(\mathrm{Fe}^{2+}\) lose electrons, completing the cycle.
Redox reactions are balanced by considering both half-reactions together. Each half-reaction has pivotal roles in the electron transfer process essential for cell operation.
Chromium redox reaction
The chromium system in the galvanic cell involves a more complex reaction. The reduction half-reaction for chromium is:\[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 14\mathrm{H}^+(aq) + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_2\mathrm{O}(l)\].
  • This reaction occurs in an acidic medium, as noted by the presence of \(\mathrm{H}^+\).
  • It involves a reduction from the dichromate ion \((\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-})\) to chromium ions \((\mathrm{Cr}^{3+})\).
Like in the iron reaction, the dichromate ion gains electrons to reduce. In the complete redox balance, observing electron numbers in reduction and oxidation tells how the cell generates power. This balance is key to sustaining electrical flow.

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Most popular questions from this chapter

Consider the rechargeable battery: \(\mathrm{Zn}(s)\left|\mathrm{ZnCl}_{2}(a q) \| \mathrm{Cl}^{-}(a q)\right| \mathrm{Cl}_{2}(l) \mid \mathrm{C}(s)\) (a) Write reduction half-reactions for each electrode. From which electrode will electrons flow from the battery into a circuit if the electrode potentials are not too different from \(E^{\circ}\) values? (b) If the battery delivers a constant current of \(1.00 \times 10^{3} \mathrm{~A}\) for \(1.00 \mathrm{~h}\), how many kilograms of \(\mathrm{Cl}_{2}\) will be consumed?

Calculate \(E^{\circ}\) for the half-reaction \(\mathrm{Pd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \rightleftharpoons\) \(\mathrm{Pd}(s)+2 \mathrm{OH}^{-}\)given that \(K_{\mathrm{sp}}\) for \(\mathrm{Pd}(\mathrm{OH})_{2}\) is \(3 \times 10^{-28}\) and \(E^{\circ}=\) \(0.915 \mathrm{~V}\) for the reaction \(\mathrm{Pd}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(s)\).

This problem requires Beer's law from Chapter 17. The oxidized form \((\mathrm{Ox})\) of a flavoprotein that functions as a one-electron reducing agent has a molar absorptivity \((\varepsilon)\) of \(1.12 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(457 \mathrm{~nm}\) at \(\mathrm{pH} 7.00\). For the reduced form (Red), \(\varepsilon=3.82 \times\) \(10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(457 \mathrm{~nm}\) at \(\mathrm{pH} 7.00\). $$ \mathrm{Ox}+\mathrm{e}^{-} \rightleftharpoons \operatorname{Red} \quad E^{0 \prime}=-0.128 \mathrm{~V} $$ The substrate \((S)\) is the molecule reduced by the protein. $$ \text { Red }+S \rightleftharpoons O x+S^{-} $$ Both \(\mathrm{S}\) and \(\mathrm{S}^{-}\)are colorless. A solution at \(\mathrm{pH} 7.00\) was prepared by mixing enough protein plus substrate (Red \(+\mathrm{S}\) ) to produce initial concentrations \([\operatorname{Red}]=[\mathrm{S}]=5.70 \times 10^{-5} \mathrm{M}\). The absorbance at \(457 \mathrm{~mm}\) was \(0.500\) in a \(1.00-\mathrm{cm}\) cell. (a) Calculate the concentrations of \(\mathrm{Ox}\) and Red from the absorbance data. (b) Calculate the concentrations of \(\mathrm{S}\) and \(\mathrm{S}^{-}\). (c) Calculate the value of \(E^{\circ}\) for the reaction \(\mathrm{S}+\mathrm{e}^{-} \rightleftharpoons \mathrm{S}^{-}\).

For the cell \(\mathrm{Pt}(s) \mid \mathrm{VO}^{2+}(0.116 \mathrm{M}), \mathrm{V}^{3+}(0.116 \mathrm{M})\), \(\mathrm{H}^{+}(1.57 \mathrm{M}) \| \mathrm{Sn}^{2+}(0.0318 \mathrm{M}), \mathrm{Sn}^{4+}(0.0318 \mathrm{M}) \mid \mathrm{Pt}(s), E\left(\right.\) not \(\left.E^{\circ}\right)\) \(=-0.289 \mathrm{~V}\). Write the net cell reaction and calculate its equilibrium constant. Do not use \(E^{\circ}\) values from Appendix \(\mathrm{H}\) to answer this question.

Write the Nernst equation for the following half-reaction and find \(E\) when \(\mathrm{pH}=3.00\) and \(P_{\mathrm{A}_{s} \mathrm{H}_{3}}=1.0\) mbar. $$ \operatorname{As}(s)+3 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightleftharpoons \underset{\text { Arsine }}{\mathrm{AsH}_{3}(g)} \quad E^{\circ}=-0.238 \mathrm{~V} $$
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