Skip to main content
Pearson+ LogoPearson+ Logo
Ch.20 - Electrochemistry
All textbooksBrown 14th EditionCh.20 - ElectrochemistryProblem 98
Chapter 20, Problem 98

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.
Recommended video:
Guided course
01:21
The Electrolytic Cell

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.
Recommended video:
Guided course
07:19
Redox Half Reactions Example

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.
Recommended video:
Guided course
01:27
Standard Cell Potential
Related Practice
Textbook Question
A mixture of copper and gold metals that is subjected toelectrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowestcommon oxidation state, Te4+, isTe4+1aq2 + 4 e- ¡ Te1s2 E°red = 0.57 VGiven this information, describe the probable fate of tellurium impurities during electrorefining. Do the impuritiesfall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If theygo into solution, do they plate out on the cathode?
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?