Concept explainers
Use the data from the table of standard reduction potentials in Appendix H to calculate the standard potential of the cell based on each of the following reactions. In each case, state whether the reaction proceeds spontaneously as written or spontaneously in the reverse direction under standard-state conditions.
(a)
(b)
(c)
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Chapter 18 Solutions
Chemistry: Principles and Practice
- 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_forwardCalculate the standard cell potential of the following cell at 25C. Sn(s)Sn2+(aq)I2(aq)I(aq)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_forward
- 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.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_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
- a Calculate G for the following cell reaction: Tl(s)Tl+(aq)Pb2+(aq)Pb(s) The Gf for Tl+(aq) is 32.4 kJ/mol. b From G, calculate the standard cell potential for the cell reaction and from this, determine the standard potential for Tl2+(aq)+eTl(s).arrow_forwardIt 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+.arrow_forwardAt 298 K, the solubility product constant for solid Ba(IO3)2 is 1.5 109. Use the standard reduction potential of Ba2+(aq) to find the standard potential for the half-reaction Ba(IO3)2(s)+2eBa(s)+2IO3(aq)arrow_forward
- 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)arrow_forwardGiven 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.arrow_forward
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