2Cr(s) +3Fe+2 gives 2cr+3(aq) + 3 Fe(s)
cr oxidaton number increase from zero to +3 so oxidation takes place it is called anode . Fe+2 oxidation number changes from +2 to zero then oxidation number reduced it is reduction called cathode.
cell representation; anode // cathode
cr(s)/cr+3(aq) II Fe+2(aq)IFe(s) so option (3) correct
Question 14 (4 points) The standard cell notation for the following redox reaction is 2 Cr(s)...
Question 5 (Q3) The cell notation for a voltaic cell with the following redox reaction is: Fe2+ (aq) + 2e → Fe(s) (oxidation) » Cr3+ (aq) + 3e" (reduction) Cr(s) Cr3+ (aq) | Cr (s) "Fe (s) | Fe2+ (aq) Cr3+ (aq) | Cr (s) " Fe2+ (aq) | Fe (s) Fe2+ (aq)| Fe (s) "Cr (s) [Cr3+ (aq) II Cr(s) | Cr3+ (aq) Fe2+ (aq)| Fe (s) None of the above
Question 2 (1 point) Given: Cr3+ (aq) + 3e --> Cr(s); E = -0.74 V Fe2+(aq) + 2e- --> Fe(s): E° = -0.41 V What is the standard cell potential for the following reaction? 2Cr(s) + 3Fe2+(aq) --> 3Fe(s) + 2 Cr3+(aq)? +0.33 v -0.33 v +1.15 V O-1.15 V +0.25 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
Choose the overall chemical reaction that takes place in a voltaic cell represented by the the following cell diagram Cr(s)| Cr3+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s). Cr3+(aq) + 3Fe2+(aq) → Cr(s) + 3Fe3+(aq) O Cr(s) + 3Fe3+(aq) → Cr3+(aq) + 3Fe2+(aq) O Cr(s) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq) Cr(s) + 3Fe2+(aq) → Cr3+(aq) + 3Fe3+(aq) O Cr(s) + 2Fe3+(aq) → Cr3+(aq) + 2Fe2+(aq)
26. Determine the redox reaction represented by the following cell notation. Fe(s) I Fe2+(aq) I| Cu2+(aq) I Cu(s) A) Cu(s) + Fe2+(aq) B) Fe(s)+Cu2+(aq) Cu(s)+ Fe2+(aq) C) 2 Fe(s)+ Cu2+(aq)Cu(s) + 2 Fe2+(aq) D) 2 Cu(s) + Fe2+(aq)Fe(s)+ 2 Cu2+(aq) E) 3 Fe(s) + 2 Cu2+(aq) 2 Cu(s)+3 Fe2+(aq) Fe(s) + Cu2+(aq)
Table 20.2 Half-reaction E (V) -0.74 Cr3+ (aq) 3e --Cr (s) Fe2+(aq)+2e-Fe (s) Fe3+ (aq) 0.440 +eFe2+ (s) Sn4+ (aq) 2eSn2 (aq) +0.771 +0.154 8. Based on Table 20.2, which of the following reactions will occur spontaneously as written? A) Sn4+ (aq)+ Fe3+ (aq) Sn2+ (aq) + Fe2+ (aq) B) 3Fe (s)+2Cr3+ (aq)2Cr (s)+3F 2+ (aq) C) Sn4+ (aq)+ Fe2+ (aq) Sn2+ (aq)+ Fe (s) D) 3Sn4+ (aq)+ 2Cr (s)- 2Cr3+ (aq) + 3Sn2+ (aq) E) 3Fe2+ (aq) Fe (s)+2Fe3+...
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-
QUESTION 8 Based on the following cell notation, write a balanced and complete redox reaction: Ca), sat. KCI Fe, Fe2 |Pt Ag) AgCl O A. Ag(s) +CI-(aq)Fe2+ AgCI (s) +Fe3+ = O B. AgCl (s)+ Fe2+ = Ag(s) +CI-(aq) Fe3+ CAg(s) +CI-(aq) + Fe2+ = AgCl(s) + Fe3+ D.Ag(s)CI-(aq) + Fe3+ AgCl(s)+ Fe2+ E. Ag(s)+CI-(aq)+2Fe2+= AgCI(s) +2FE3+
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Use the cell notation to answer the question: Cr(s) | Cr3+(aq) || Br–(aq) | Br2(g) | Pt(s) What is the reducing agent? Group of answer choices Br2(g) Cr3+(aq) Pt(s) Cr(s) Br–(aq)