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).
Use the following half-reactions to write 3 spontaneous reactions, calculate E°cell for each reaction, what is...
i just need the last thing using what I've already put in. Here was the question attached to it: Reaction 1 and 2: Oxidizing agents. N20 > Au+ ; reducing agents: Au > N2 Reaction 1 and 3: Oxidizing agents: Au+ A > Cr3+ 4; reducing agents: Cr > Au Reaction 2 and 3: Oxidizing agents: N20 4 > Cr3+ ; reducing agents : Cr > N2 Overall ranking: Oxidizing agents : [Select] > [Select] Select] ; reducing agents :...
3. Calculate the cell potential for the voltaic cell based on the following half reactions at T = 25°C: Cr3+(aq) + 3e Cr(s) E° = -0.74 V TiO2(aq) + 2H(aq) + 1e → Ti+(aq) + H2O(1) E° = + 0.10 V Where, [Cr3+] = 1.0 x 104 M, [TiO2+] = 1.0 x 10-1M, [H+] = 1.0 M, [Ti$+] = 5.0 x 10-2 M.
Consider the following half-reactions.Which of these is the strongest reducing agent listed here? Au (aq) +eAu(s) E 1.69 V N20(g) +2H+(aa) +2e N2/g)+H200 EO - 1.77 V Cr3t(ag) +3 e Cris) Eo0.74 V Selected Answer: 5. Cr3(aa) Correct Answer: 6. Cr(s)
Calculate the Eo cell and indicate whether the reaction shown is spontaneous or nonspontaneous Calculate F cell and indicate whether the overall reaction shown is spontaneous or none 1205) + 2e 21 (aq) Fº=0.53 V Craq) + 3e Cr(s) E=-0.74 V Overall reaction: 20/15) - 312(5) - 2013+ (aq) + (aq) + 61 (aq) Eºcell = -1.27 V. spontaneous Ecell -1.27 V, nonspontancous Eºcell - 1.27 V, spontaneous
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
Consider the following half-reactions: Half-reaction E° (V) F2(g) +2e - →2F (aq) 2.870V 2H*(aq) + 2e - H2(g) 0.000V Cr3+ (aq) + 3e — Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will F2(g) oxidize Cr(s) to Cr3+ (aq)? (6) Which species can be oxidized by H(aq)? If none, leave box blank.
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)
What is the E cell for the cell represented by the combination of the following half- reactions? 2Hg2+(aq) + 2e" =Hg22+(aq); E°= 0.92 V Cr3+ (aq) + 3e Cr(s); E°= -0.74 V O 1) -0.18 V O2) 0.18 V 03) 1.28 V 4) 1.66 V 05) 2.12 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
Given the half reactions below, what would be the standard cell potential (E° cell) in volts for the reaction below? (Report answer with 2 decimal places) 2 Cr3+ (aq)+3 Mn (s) 3 Cr (s) +3 Mn2+ (aq) Cr3+ (aq) +3 e Cr (s) Eored0.50 V Mn2+ (aq) 2 e-Mn (s) E red -1.18 v