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Ch.20 - Electrochemistry
All textbooksBrown 15th EditionCh.20 - ElectrochemistryProblem 100
Chapter 20, Problem 100

A common shorthand way to represent a voltaic cell is anode | anode solution || cathode solution | 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 Fe | Fe2+ || Ag+ | Ag; calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn | Zn2+ || H+ | H2; 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: ClO3^-_(aq) + 3 Cu_(s) + 6 H+_(aq) -> Cl^-_(aq) + 3 Cu2+_(aq) + 3 H2O_(l); Pt is used as an inert electrode in contact with the ClO3^- and Cl^-. Calculate the standard cell emf given: ClO3^-_(aq) + 6 H+_(aq) + 6 e^- -> Cl^-_(aq) + 3 H2O_(l); E° = 1.45 V.

Verified step by step guidance
1
Step 1: Identify the half-reactions for the given voltaic cells. For (a) Fe | Fe2+ || Ag+ | Ag, the half-reactions are: Fe -> Fe2+ + 2e^- (oxidation at the anode) and Ag+ + e^- -> Ag (reduction at the cathode). For (b) Zn | Zn2+ || H+ | H2, the half-reactions are: Zn -> Zn2+ + 2e^- (oxidation at the anode) and 2H+ + 2e^- -> H2 (reduction at the cathode).
Step 2: Write the overall cell reactions by combining the half-reactions. For (a), combine Fe -> Fe2+ + 2e^- and 2Ag+ + 2e^- -> 2Ag to get Fe + 2Ag+ -> Fe2+ + 2Ag. For (b), combine Zn -> Zn2+ + 2e^- and 2H+ + 2e^- -> H2 to get Zn + 2H+ -> Zn2+ + H2.
Step 3: Calculate the standard cell emf (E°cell) using the standard reduction potentials from Appendix E. For (a), E°cell = E°cathode - E°anode = E°(Ag+/Ag) - E°(Fe2+/Fe). For (b), E°cell = E°cathode - E°anode = E°(H+/H2) - E°(Zn2+/Zn).
Step 4: For part (c), represent the cell using the given reaction: ClO3^-_(aq) + 3 Cu_(s) + 6 H+_(aq) -> Cl^-_(aq) + 3 Cu2+_(aq) + 3 H2O_(l). The cell notation is: Pt | ClO3^-_(aq), Cl^-_(aq) || Cu2+_(aq) | Cu_(s).
Step 5: Calculate the standard cell emf for the reaction in part (c) using the given standard reduction potential for ClO3^- reduction: E° = 1.45 V. Use the standard reduction potential for Cu2+/Cu from Appendix E to find E°cell = E°cathode - E°anode.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Voltaic Cells

A voltaic cell, or galvanic cell, is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions. It consists of two electrodes: an anode where oxidation occurs and a cathode where reduction takes place. The flow of electrons from the anode to the cathode generates an electric current, and the cell's voltage can be calculated using standard reduction potentials.
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Half-Reactions

Half-reactions are the individual oxidation and reduction reactions that occur in an electrochemical cell. Each half-reaction shows the transfer of electrons and the change in oxidation states of the reactants. By writing the half-reactions for both the anode and cathode, one can determine the overall cell reaction and calculate the standard cell potential by combining the standard reduction potentials of the half-reactions.
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Standard Cell Potential (E°)

The standard cell potential (E°) is the measure of the voltage produced by a voltaic cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It is calculated by subtracting the standard reduction potential of the anode from that of the cathode. A positive E° indicates a spontaneous reaction, while a negative E° suggests that the reaction is non-spontaneous under standard conditions.
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Related Practice
Textbook Question

(a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5 × 104 A flowing for a period of 24 h. Assume the electrolytic cell is 85% efficient. (b) What is the minimum voltage required to drive the reaction?

Textbook Question

A disproportionation reaction is an oxidation–reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions:

(b) MnO42-(aq) → MnO4-(aq) + MnO2(s) (acidic solution)

Textbook Question

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to Sn2+ by I2 (to form I-), (b) reduction of Ni2+ to Ni by I- (to form I2), (c) reduction of Ce4+ to Ce3+ by H2O2, (d) reduction of Cu2+ to Cu by Sn2+ (to form Sn4+).

Textbook Question

Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are Au+1aq2 + e- ¡ Au1s2 Ered ° = +1.69 V Au3+1aq2 + 3 e- ¡ Au1s2 Ered ° = +1.50 V (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction 4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2 ¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2 What is being oxidized, and what is being reduced in this reaction?