Ans : In the reaction, Chromium (Cr) is oxidised whereas Iodine(I) is reduced.
and Given, E°reduction = 0.53 V
E°oxidation = -0.74 V
We know that,
E°cell = E°reduction - E°oxidation
= 0.53 V - ( - 0.74 V)
= (0.53 + 0.74) V
= 1.27 V
Since, the cell potential is positive, the reaction is spontaneous.
Calculate the Eo cell and indicate whether the reaction shown is spontaneous or nonspontaneous Calculate F...
0 What is the Ecell for the cell represented by the combination of the following 4) half-reactions? 2Hg2+(aq) + 2e-→Hg22+(aq) Cr3t(ag)+3eCr(s) E-0.92 V B) 1.28 V 9) 0.18 V A) 2.12 V D) 1.66 V E)-0.18 V 5) Examine the following half-reactions and select the weakest oxidizing agent among the 5)_ species listed. E。= 0.854 V E1.185 V K+(aq) + e-→ K(s) F20(aq) + 2H+(aq) + 4e-→ 2p-(aq) + H20() E。= 2.153 V A) F2o(ag) B) AuBr4Taq) C) Mn2+(ag) D)...
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
4. Use the table below to provide a redox reaction involving the spontaneous oxidation of Cr (balance your final reaction and provide the Ecell). Half-reaction E (V) Cr3+ (aq) + 3e Cr(s) -0.74 Fe(s) -0.440 Fe3+ (aq) + → Fe2+ (s) +0.771 Sn4+ (aq) + 2e Sn2+ (aq) +0.154 Fe2+ (aq) + 2e-
Use the following half-reactions to write 3 spontaneous reactions, calculate E°cell for each reaction, what is the n number (number of electrons transferred). Au+ (aq) + e− → Au (s) E° = 1.69 V N2O (g) + 2H+ (aq) + 2 e− → N2 (g) + H2O (l) E° = 1.77 V Cr3+ (aq) + 3 e− → Cr (s) E° = -0.74 V Question 7 0.36 pts Use the following half-reactions to write 3 spontaneous reactions, calculate Eºcell for...
4. Write a balanced equation from each cell notation, then calculate thee of the cell, is it spontaneous or nonspontaneous? A. Cr(s) Crot || Cui Cu(s) B. Al(s) Al* | Ce Ce Pt C. Cu (s) Cu? || Al Al (s) Useful information: F= 96485 J/V mol electrons, AG ----F-Ecell AG --RTINK: Ece 0.0592/n*logK; EcellEcett -0.0592/n*logQ Ce (aq) + Ag+ (aq) + e Fet (aq) + Cu" (aq) + Cu (aq) + 2e 2H(aq) + 2 Pb (aq) + 2e...
6. Determine whether the following cell reaction is spontaneous or not at 25°C, and calculate a cell potential. Cu Cu +(0.394 M) || Cu²+(0.258 M) Cu Cu²+(aq) + 2e → Cu(s); <° = 0.34 V
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) + 2e- Fe () -0.440 Fe3+ (aq) + e- → Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. Cr (s) + 3Fe3+ (aq) + 3Fe2+ (aq) + Cr3+ (aq) A) -1.45 B) +2.99 C) +1.51 D) +3.05 E) +1.57
Using the following standard reduction potentials: Fe3+ (aq) + e. → Fe2+ (aq) Eo = +0.77 V Pb2+ (aq) + 2 e. → Pb(s) E。--0.13 V Calculate the standard cell potential for the galvanie cell reaction given below, and determine whether or not this reaction is spostaneous under standard conditions. Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb(s) ⓔ A. E.-0.90 V, nonspontaneous OB. E-0.90 V, spontaneous C. Eo +0.90 V, nonspontaneous OD0.90 V, spontaneous
Name: Chem 1120 Electrochemical Potentials Perform each of the following calculations, showing all work. Use the Electrochemical Potentials table provided on D2L if values are not provided. Standard State Electrochemical Cells 1. In each of the following systems two half-reactions are provided. For each system, a) write the balanced reaction occurring for a spontaneous system, and b) calculate the overall standard cell potential. a. Half Reaction Zn2+(aq) + 2e = Zn(s) Cr3+ (aq) + 38 = Cr(s) Eºred (V) -0.76...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. 35n4+ (aq) + 2Cr (s) → 2Cr3+ (aq) + 3Sn2+ (aq) A) +1.94 B) +0.89 C) +2.53 D) -0.59 E) -1.02