(a)
Interpretation:
The table for reduction potential of various species is given. The reagent that is capable of the given reduction is to be picked.
Concept introduction:
The species with higher reduction potential undergoes reduction thus, oxidizing the other species while the species with lower reduction potential undergoes oxidation and thus, reduces the other species.
To determine: The species that are capable of the given reductions.
(b)
Interpretation:
The table for reduction potential of various species is given. The reagent that is capable of the given reduction is to be picked.
Concept introduction:
The species with higher reduction potential undergoes reduction thus, oxidizing the other species while the species with lower reduction potential undergoes oxidation and thus, reduces the other species.
To determine: The species that are capable of the given reductions.
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Chapter 17 Solutions
Bundle: Chemistry: An Atoms First Approach, 2nd, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
- A galvanic cell is based on the following half-reactions: In this cell, the copper compartment contains a copper electrode and [Cu2+] = 1.00 M, and the vanadium compartment contains a vanadium electrode and V2+ at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M H2EDTA2, resulting in the reaction H2EDTA2(aq)+V2+(aq)VEDTA2(aq)+2H+(aq)K=? The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 mL H2EDTA2 solution added. At the stoichiometric point, was observed to be 1 .98 V. The solution was buffered at a pH of 10.00. a. Calculate before the titration was carried out. b. Calculate the value of the equilibrium constant, K, for the titration reaction. c. Calculate at the halfway point in the titration.arrow_forwardAn electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode immersed in a solution with [Al3+] = 1.0 M. Sodium hydroxide is added to the aluminum compartment, causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3 has ceased, the concentration of OH is 1.0 104 M and the measured cell potential is 1.82 V. Calculate the Ksp value for Al(OH)3. Al(OH)3(s)Al3+(aq)+3OH(aq)Ksp=?arrow_forwardIf the SHE was assigned a value of 3.00 V rather than 0.00 V, what would happen to all of the values listed in the table of standard reduction potentials?arrow_forward
- Use the table of standard reduction potentials (Table 17-1) to pick a reagent that is capable of each of the following oxidations (under standard conditions in acidic solution). a. oxidize Br to Br2 but not oxidize Cl to Cl2 b. oxidize Mn to Mn2+ but not oxidize Ni to Ni2+arrow_forwardThe overall reaction for the production of Cu(OH)2 from Cu in oxygenated water can be broken into three steps: an oxidation half-reaction, a reduction half-reaction, and a precipitation reaction. a. Complete and balance the two missing half-reactions to give the overall equation for the oxidation of cooper in seawater. Oxidation half-reaction: ? Reduction half-reaction: ? Precipitation:Cu2+(aq)+2OH(aq)Cu(OH)2(s)Overall:Cu(s)+12O2(g)+H2O(l)Cu(OH)2(s) b. Determine the equilibrium constant for the overall reaction at 25 C using standard reduction potentials and the solubility product constant (Ksp) of Cu(OH)2(s).arrow_forwardConsider the following reduction potentials: Co3++ 3e Co = 1.26 V Co2++ 2e Co = 0.28 V a. When cobalt metal dissolves in 1.0 M nitric acid, will Co3+ or Co2+ be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of HNO3 to get a different result in part a? Concentrated HNO3 is about 16 M.arrow_forward
- Calculate the cell potential of a cell operating with the following reaction at 25C, in which [MnO4] = 0.010 M, [Br] = 0.010 M. [Mn2] = 0.15 M, and [H] = 1.0 M. 2MNO4(aq)+10Br(aq)+16H+(aq)2MN2(aq)+5Br2(l)+8H2O(l)arrow_forwardA current of 0.0100 A is passed through a solution of rhodium sulfate, causing reduction of the metal ion to the metal. After 3.00 hours, 0.038 g of Rh has been deposited. What is the charge on the rhodium ion, Rhn+? What is the formula for rhodium sulfate?arrow_forwardCalculate the cell potential of a cell operating with the following reaction at 25C, in which [Cr2O32] = 0.020 M, [I] = 0.015 M, [Cr3+] = 0.40 M, and [H+] = 0.60 M. Cr2O72(aq)+6I(aq)+14H+(aq)2Cr3+(aq)+3I2(s)+7H2O(l)arrow_forward
- Galvanic cells harness spontaneous oxidationreduction reactions to produce work by producing a current. They do so by controlling the flow of electrons from the species oxidized to the species reduced. How is a galvanic cell designed? What is in the cathode compartment? The anode compartment? What purpose do electrodes serve? Which way do electrons always flow in the wire connecting the two electrodes in a galvanic cell? Why is it necessary to use a salt bridge or a porous disk in a galvanic cell? Which way do cations flow in the salt bridge? Which way do the anions flow? What is a cell potential and what is a volt?arrow_forwardAn electrochemical cell consists of a silver metal electrode immersed in a solution with [Ag+] = 1.0 M separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0 M NH3 that is also 0.010 M in Cu(NH3)42+, what is the cell potential at 25C? Cu2+(aq)+4NH3(aq)Cu(NH3)42+(aq)K=1.01013arrow_forwardIdentify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Appendix L are the same as those at each of the melting points. Assume the efficiency is 100%. (a) CaCl2. (b) LiH. (c) AlCl3. (d) CrBr3arrow_forward
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