9 Question 9: A galvanic cell is based on the reaction:8 H+ + 5 Fe²+ + MnO4¯ →5 Fe³+ + Mn²+ + 4 H₂OCalculate the cell potential using the data in table 17.1 and report your result to the correct numberof significant figures:E cell =VCheckReuse<> EmbedH-P

Answered: Question 9: A galvanic cell is based on…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter18: Introduction To Electrochemistry

Section: Chapter Questions

Problem 18.18QAP

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An 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?
Consider a galvanic cell based on the following half-reactions: a. What is the standard potential for this cell? b. A nonstandard cell is set up at 25C with [Mg2+] = 1.00 105 M. The cell potential is observed to be 4.01 V. Calculate [Au3+] in this cell.
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)
Consider a galvanic cell based on the following half-reactions: a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at 25C with [Fe2+] = 2.00 104 M and [La3+] = 3.00 103 M, what is the expected cell potential?
It took 150. s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of 1+.
Given this reaction, its standard potential, and the standard half-cell potential of 0.34 V for the Cu2+ |Cu half-cell, calculate E° for the Fe(s)|Fe2+(aq) half-cell.
Calculate the standard cell potential of the following cell at 25C. Sn(s)Sn2+(aq)I2(aq)I(aq)
For each reaction listed, determine its standard cell potential at 25 C and whether the reaction is spontaneous at standard conditions. (a) Mn(s)+Ni2+(aq)Mn2+(aq)+Ni(s) (b) 3Cu2+(aq)+2Al(s)2Al3+(aq)+3Cu(s) (c) Na(s)+LiNO3(aq)NaNO3(aq)+Li(s) (d) Ca(NO3)2(aq)+Ba(s)Ba(NO3)2(aq)+Ca(s)
Calculate 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)
At 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.
Consider a voltaic cell in which the following reaction occurs. Zn(s)+Sn2+(aq)Zn2+(aq)+Sn(s) (a) Calculate E° for the cell. (b) When the cell operates, what happens to the concentration of Zn2+? The concentration of Sn2+? (c) When the cell voltage drops to zero, what is the ratio of the concentration of Zn2+ to that of Sn2+? (d) If the concentration of both cations is 1.0 M originally, what are the concentrations when the voltage drops to zero?
At 298 K, the solubility product constant for Pb(IO3)2 is 2.6 1013, 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 Pb(IO3)2(s)+2ePb(s)+2IO3(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/Pb(IO3)2 electrode in a 3.5 103 M solution of NaIO3.

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Question 9: A galvanic cell is based on the reaction: 8 H+ + 5 Fe²+ + MnO4¯ →5 Fe³+ + Mn²+ + 4 H₂O Calculate the cell potential using the data in table 17.1 and report your result to the correct number of significant figures: E cell = V Check Reuse <> Embed H-P