Choose the correct charge balance equation for an aqueous solution of MnCl₂ that ionizes to Mn²+, CI, MnCI*, and MnOH*. [Mn²+] + [MnCI*] + [MnOH*] = [CI] Mn²+] + [MnCl¹] + [MnOH*] = [CI] [Mn²¹] + [MnCI*] + [MnOH*] = 2[CI] [Mn²+] + [MnCI*] + [MnOH*] = [CI] + [OH-] 2[Mn²+] + [MnCl*] + [MnOH*] = [CI] + [OH-] [Mn²¹] + [MnCI¹] + [MnOH*] =2[CI] + [OH-] [H] + [Mn²+] + [MnCI'] + [MnOH*] = [CI] + [OH-] [H] + 2[Mn²+] + [MnCI] + [MnOH*] = [CI] + [OH-] [H*] + [Mn²*] + [MnCI'] + [MnOH'] = 2[CI] + [OH-] Choose the correct mass balance equation for the same solution. 2[Mn²+] = [CI] [Mn²+] = 2[CI] 2[Mn²¹] + [MnCl*] + [MnOH*] = [MnCI*] + [cr] [Mn²+] + [MnCI*] + [MnOH*] = [MnCI] + 2[cr] 2([Mn²+] + [MnCI+] + [MnOH*]) = [MnCI] + [cr] [Mn²+1+[MnCl] + [MBOH)) - com

Choose the correct charge balance equation for an aqueous solution of MnCl₂ that ionizes to Mn²+, CI, MnCI, and MnOH. [Mn²+] + [MnCI] + [MnOH] = [CI] Mn²+] + [MnCl¹] + [MnOH] = [CI] [Mn²¹] + [MnCI] + [MnOH] = 2[CI] [Mn²+] + [MnCI] + [MnOH] = [CI] + [OH-] 2[Mn²+] + [MnCl] + [MnOH] = [CI] + [OH-] [Mn²¹] + [MnCI¹] + [MnOH] =2[CI] + [OH-] [H] + [Mn²+] + [MnCI'] + [MnOH] = [CI] + [OH-] [H] + 2[Mn²+] + [MnCI] + [MnOH] = [CI] + [OH-] [H] + [Mn²] + [MnCI'] + [MnOH'] = 2[CI] + [OH-] Choose the correct mass balance equation for the same solution. 2[Mn²+] = [CI] [Mn²+] = 2[CI] 2[Mn²¹] + [MnCl] + [MnOH] = [MnCI] + [cr] [Mn²+] + [MnCI] + [MnOH] = [MnCI] + 2[cr] 2([Mn²+] + [MnCI+] + [MnOH]) = [MnCI] + [cr] [Mn²+1+[MnCl] + [MBOH)) - com

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter19: Principles Of Chemical Reactivity: Electron Transfer Reactions

Section19.6: Electrochemistry And Thermodynamics

Problem 1RC

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Iron(II) can be oxidized to iron(III) by permanganate ion in acidic solution. The permanganate ion is reduced to manganese(II) ion. (a) Write the oxidation half-reaction, the reduction half-reaction, and the overall redox equation. (b) Calculate E° for the reaction. (c) Calculate the percentage of Fe in an ore if a 0.3500-g sample is dissolved and the Fe2+ formed requires for titration 55.63 mL of a 0.0200 M solution of KMnO4.
The photographs below (a) show what occurs when a solution of iron(III) nitrate is treated with a few drops of aqueous potassium thiocyanate. The nearly colorless iron(III) ion is converted to a red [Fe(H2O)5SCN)2+ ion. (This is a classic test for the presence of iron(III) ions in solution.) [Fe(H2O)6]3+(aq) + SCN(aq) [Fe(H2O)5SCN]2+(aq) + H2O() (a) As more KSCN is added to the solution, the color becomes even more red. Explain this observation. (b) Silver ions form a white precipitate with SCN ions. What would you observe on adding a few drops of aqueous silver nitrate to a red solution of [Fe(H2O)5 SCN]+ ions? Explain your observation.
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Calculate the cell potential, the equilibrium constant, and the free-energy change for:Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values:Ca2+(aq)+2e−→Ca(s) Eo = -2.87 V Mn2+(aq)+2e−→Mn(s) Eo = -1.18 V
This method requires acidic conditions to prevent iron from precipitating out as ferric hydroxide EDTA Fajans Volhard Mohr
Use standard reduction potentials to calculate the equilibrium constant for the reaction:2Ag+(aq) + Fe(s)2Ag(s) + Fe2+(aq)Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.Equilibrium constant: G° for this reaction would be than zero.
Calculate the equilibriumconstant for the disproportionation of the copper(I) ion atroom temperature:2 Cu+(aq)------>Cu2+(aq) + Cu(s).
Calculate the equilibrium constant for the reaction 2 Cu+⇄Cu2++ Cu(s)usingstandard potential data for reactions: Cu+ + e- ----->Cu(s) Eo= + 0.52 V Cu2+ + 2e- ------> Cu (s)Eo= + 0.34 V
Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr2O7 2-1aq2 + I -1aq2 ¡ Cr3+1aq2 + IO3 -1aq2 (acidic solution)
Calculate the reduction potential for a Pt | Cr3+, Cr2+ halfcell in which [Cr3+] is 0.15 M and [Cr2+] is 0.0019 M
place the following metals from most active to least active. Ag, Cu, Mg, Pb, Zn
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