Concept explainers
(a)
Interpretation:
The oxidation half-cell reaction for the electrolytic cell that involves the reaction
Concept introduction:
The conversion of chemical energy to electrical energy and vice versa is known as
(b)
Interpretation:
The reduction half-cell reaction for the electrolytic cell that involves the reaction
Concept introduction:
The conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.
(c)
Interpretation:
The anode and cathode for the electrolytic cell that involves the reaction,
Concept introduction:
The conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.
(d)
Interpretation:
The direction of flow of electron for the electrolytic cell that involves the reaction
Concept introduction:
The conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.
(e)
Interpretation:
The direction of flow of
Concept introduction:
The conversion of chemical energy to electrical energy and vice versa is known as electrochemistry. In an electrochemical cell, two electrodes are connected by a wire and electrons are free to move between two compartments. Each compartment is known as a half-cell.
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Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
- Calculate the standard cell potential of the cell corresponding to the oxidation of oxalic acid, H2C2O4, by permanganate ion. MnO4. 5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l) See Appendix C for free energies of formation: Gf for H2C2O4(aq) is 698 kJ.arrow_forwardAn electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, Hg2Cl, containing 1.00 M Cl . The cell potential of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is 0.268 V. What is the solubility product of mercury(I) chloride?arrow_forwardCalculate 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_forward
- Calculate the standard cell potential of the following cell at 25C. Cr(s)Cr3(aq)Hg22(aq)Hg(l)arrow_forwardA voltaic cell is constructed in which one half-cell consists of a silver wire in an aqueous solution of AgNO3.The other half cell consists of an inert platinum wire in an aqueous solution containing Fe2+(aq) and Fe3+(aq). (a) Calculate the cell potential, assuming standard conditions. (b) Write the net ionic equation for the reaction occurring in the cell. (c) Which electrode is the anode and which is the cathode? (d) If [Ag+] is 0.10 M, and [Fe2+] and [Fe3+] are both 1.0 M, what is the cell potential? Is the net cell reaction still that used in part (a)? If not, what is the net reaction under the new conditions?arrow_forwardAn electrolysis experiment is performed to determine the value of the Faraday constant (number of coulombs per mole of electrons). In this experiment, 28.8 g of gold is plated out from a AuCN solution by running an electrolytic cell for two hours with a current of 2.00 A. What is the experimental value obtained for the Faraday Constant?arrow_forward
- You have 1.0 M solutions of Al(NO3)3 and AgNO3 along with Al and Ag electrodes to construct a voltaic cell. The salt bridge contains a saturated solution of KCl. Complete the picture associated with this problem by a writing the symbols of the elements and ions in the appropriate areas (both solutions and electrodes). b identifying the anode and cathode. c indicating the direction of electron flow through the external circuit. d indicating the cell potential (assume standard conditions, with no current flowing). e writing the appropriate half-reaction under each of the containers. f indicating the direction of ion flow in the salt bridge. g identifying the species undergoing oxidation and reduction. h writing the balanced overall reaction for the cell.arrow_forwardConsider the following cell running under standard conditions: Fe(s)Fe2+(aq)Al3+(aq)Al(s) a Is this a voltaic cell? b Which species is being reduced during the chemical reaction? c Which species is the oxidizing agent? d What happens to the concentration of Fe3+(aq) as the reaction proceeds? e How does the mass of Al(s) change as the reaction proceeds?arrow_forwardAt 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction PbC2O4(s)+2ePb(s)+C2O42(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/PbC2O4 electrode in a 0.025 M solution of Na2C2O4.arrow_forward
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