FinishRecommended time: 6 minHOThe percentage ionization of a weak acidwith increasing concentration of acid in solution.Under standard conditions, a spontanoeus reaction will occur in a solution containing Ce4+, Ce3+, Fe3+ and Fe2+ jons such thatwill oxidize1Hint:7E°(Ce4+/Ce3+)=1.61 V and E°(Fe3+/Fe2+)=0.77 VReaction is|spontaneous if AH is positive and AS is negative.pką of the weak acid HA is 5.9 and pka of the weak acid HB is 8.1. Hence,|is a stronger acid thannot affectedincreasesdecreasesFe2+Ce3+Ce4+Fe2+at low temperaturesalwaysat high temperaturesneverНА | НВ

The given questions can be answered as follows:

Recommended time: 6 min The percentage ionization of a weak acid with increasing concentration of acid in solution. Under standard conditions, a spontanoeus reaction will occur in a solution containing Ce4+, Ce3+, Fe3+ and Fe2+ jons such that| will oxidize Hint: E°(Ce4+/Ce3+)=1.61 V and E°(Fe3+/Fe2+)=0.77 V Reaction is spontaneous if AH is positive and AS is negative. pka of the weak acid HA is 5.9 and pKa of the weak acid HB is 8.1. Hence, is a stronger acid than not affected increases decreases Fe2+ Ce3+ Ce4+ Fe2+ at low temperatures always at high temperatures never НА | НВ

Recommended time: 6 min The percentage ionization of a weak acid with increasing concentration of acid in solution. Under standard conditions, a spontanoeus reaction will occur in a solution containing Ce4+, Ce3+, Fe3+ and Fe2+ jons such that| will oxidize Hint: E°(Ce4+/Ce3+)=1.61 V and E°(Fe3+/Fe2+)=0.77 V Reaction is spontaneous if AH is positive and AS is negative. pka of the weak acid HA is 5.9 and pKa of the weak acid HB is 8.1. Hence, is a stronger acid than not affected increases decreases Fe2+ Ce3+ Ce4+ Fe2+ at low temperatures always at high temperatures never НА | НВ

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

A) 1) Cr202-(aq) + 14 H3O* (aq) + 6 1 2Cr*(aq) + 3 12(s) + 21 H2O(1), 2) 0.697, 3) -403.5, 4) 4.4 x 1070 16) Given the following reduction equations and their associated standard reduction potentials, E, determine the following: 1) The overall redox reaction, 2) Eeu (V), 3) AG" (k) mol-1), and 4) Kc- (170) Cr207-(aq) + 14 H3O*(aq) + 6 e- 2C3+(aq) + 21 H20() E' = 1.232 V 12(s) + 2 e - 21(aq) E' = 0.5355 V ntion is
Using data from the table below, place the following in order of decreasing strength as oxidizing agents (all under standard conditions). Half-reaction E° (V) Fe+ +e + Fe+ Fe?+ + 2e + Fe O2 + 4H+ + 4e + 2H2O Ce+ +e- + Ce+ Cl2 + 2e 2CI Zn+ + 2e Zn 0.77 -0.44 1.23 1.70 1.36 -0.76 Decreasing of strength as oxidizing agents Drag and drop your selection from the following list to complete the answer: O2 Zn2+ Ce+ Fe+ Fe2+ Cl2
For the half-reaction K* (aq) + e --> K (s), Eº = -2.936 V; therefore, %3D O K* (aq) is a good oxidizing agent. K* (aq) is a good reducing agent. OK (s) is a good oxidizing agent. K (s) is a good reducing agent. Previous Page Next Page
(b) Use the listing of standard reduction potentials on page 10 to determine the E° for Ce**(aq) + 4e → Ce(s) Reduction half-reaction E/V Reduction half-reaction E/V Ag Ag Age Ag Agir + Ag+ Br AgCleAg+a 1+ 2e21 15+2e3r In'+eIn In+e In In + 2e In In"+ SeIn In" In K*+ K La+ ela Li Mg+ 2- Mg Mn"+ 2e Mn Mn+e Mn MnO, + 4+ 2e- Ma + 2H,o MnO; + SH + Se Mn+ 4H,0 MnO; +eMno}- Mno + 21,0+ 2eMno, + 4OH Na' +e Na NP+ 2e NI NIOOH + H,0+ Ni(OH), +OH NO;+2H+e NO, + H,0 NO; + H+ - NO+ 2H,0 NO; + H,0+ 2e- NO; + 20H O,+2H,0+ de-→ 4OH 0,+ 4H+ de 2H,0 40.80 +0.54 +1.98 +0.53 +0.0713 -0.14 +0.22 -0.40 AgCro, + 2e 2Ag+ Cro AgF+e Ag +F Agl + Ag+ r Al+ Se Al +0.45 -0.44 +0.78 -0.34 -0.15 -0.49 -1.66 -2.93 Au' +eAu Au+ Se Au Ba+ 2e-Ba Be+ 2eBe +1.69 -2.52 +1.40 -3.05 +2.91 -2.36 -L8S -1.18 +0.20 +1.51 Br, + 2e 2Br Bro +H,0+ 2e Br+ 20H Cal+ 2eCa Ca(OH), + 2e Cd+ 20H Ca + 2eCd Ce+ Ce +1.09 +1.23 +0.76 +1.51 -2.87 +0.56 -0.81 +0.60 -0.40 -2.71 -2.48 -0.23 CeteCe" +1.61 +0.49 a,+ 2e-2 CIO +H,0+ 2→a+20H CO;+ 2H+ 2e…
Lithium metal is a very strong reducing agent and reacts vigorously with 1 M hydrochloric acid to produce hydrogen gas. The relevant half reactions are Li+(ag) + e- = Li(s) E = -3.04 V 2H+(ag) + 2e- = H2g) E = +0.00 V - Lithium also reacts vigorously with pH-neutral iced water. Assuming a hydrogen gas pressure of 0.1 atm and a lithium ion concentration of 0.1 mol/L and assuming that the standard cell potential does not change with temperature, determine the cell potential. - A more accurate calculation for the previous part would need to take into account the change in E° with temperature. What information would be needed to do this, and do you expect the actual cell potential would be higher or lower than that calculated? [Hint: Gibbs Free Energy change.]
Using the provided standard reduction potentials, which of the following cells is an example of a Galvanic cell? Recall the general form of cell notation. Note: The one with smaller value is the anode. -1.5 -1.0 -0.5 0.0 +0.5 +1.0 +1.5 Zn2+/Zn Sn2+/Sn Ag*/Ag Au*/Au Fe2+/Fe Cu2+/Cu
Given the half reactions and En values below, Fe2* + 2e → Fe E° = -0.45 V H2O2 + 2H* + 2e → 2H20 E° = +1.78 V 2H + 2e - H2 E° = 0.00V 4H + NO, + 3e NO + 2H2O E° = +0.96V Determine which species is the strongest reducing agent. O NO O H2 H20 Fe
Refer to the following half-reactions below; MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O E0 = 1.51 V Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O E0 = 1.33 V Calculate the value of equilibrium constant at 25 0C for this reaction?
Determine E°, AG°, and K for the overall reaction from the balanced half-reactions and their standard reduction potentials. 4 Fe3+ + 2 H,0 4 Fe²+ + O, + 4H+ Fe3+ + e-= Fe²+ E° = 0.771 V 0, + 2 H* + 2 e¯2 H,0 E° = 1.229 V E° = V K = Is the overall reaction spontaneous as written? AG° = spontaneous not spontaneous
Please don't provide hand writtin solution....
Given the reduction half-reactions Fe3*(aq) + 3e- Fe(s) E° = -0.036 V Mg2*(aq) + 2e- Mg(s) E° = -2.37 V determine the correct information for this reaction. 2 Fe3+(aq) + 3 Mg(s) 2 Fe(s) + 3 Mg2+*(aq) O Ecell° = +0.085 V and the reaction is nonspontaneous as written O Ecell° = -2.41 V and the reaction is spontaneous as written O Ecello = +2.33 V and the reaction is spontaneous as written O Ecelo = +2.41 V and the reaction is nonspontaneous as written
1. TABLE COMPLETION. Directions: Using the following electrochemical equations, fill in the table completely and correcting. Reactions: (1) KCIO3 + C12H22O11 - KCI + H2O + CO2 Oxidizing Reducing No. Agent Agent Oxidation Reduction (1)