The galvanic cell of the given reaction is represented as
Cr(s) | Cr3+ || Fe2+ | Fe(s)
Given, Cr3+ (aq)+ 3e- -----> Cr(s) Eo = -0.74 V
Fe2+ (aq) + 2e- -----> Fe(s) Eo = -0.41 V
Eocell = E° right – E°left
= (-0.41 V) - (-0.74 V) = -0.41 V + 0.74 V = +0.33 V
Question 2 (1 point) Given: Cr3+ (aq) + 3e --> Cr(s); E = -0.74 V Fe2+(aq)...
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
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+...
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
Question 5 (1 point) Given: Cr3+(aq) + 3e --> Cr(s); E° = -0.74 V Sn2+(aq) + 2e- --> Sn(s); E° = -0.14 V What is the standard cell potential for the following reaction? 2Cr(s) + 3Sn2+(aq) --> 3Sn(s) + 2Cr3+(aq)? 0-0.60 v 0 +0.60 v O +1.06 V 0 +0.88 V 0-0.88 V
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) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
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...
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- Cr(s) -0.74 Fe2+ (aq) + 2e →Fe (s) -0.440 Fe3+ (aq) + e- Fe2+ (8) Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 1) Which of the following reactions will occur spontaneously as written? A) 35n4+ (aq) + 2Cr (s) +2013+ (aq) + 3Sn2+ (aq) B) 3Fe (s) + 2Cr3+ (aq) →2Cr (s) + 3Fe2+ (aq) C)...
Question 2 (1 point) What is Eºcell for the cell reaction: 2Cr(s) + 3Sn4+(aq) --> 3Sn2+(aq) + 2Cr3+ (aq)? Given: Cr3+(aq) + 3e --> Cr(s); E° = -0.74 V Sn4+(aq) + 2e --> Sn2+(aq); E° = +0.15 V 0 +0.45 V O +0.89 V O 1.93 V 0-0.59 V 0 +0.59 V
Given the two following half reactions, Cr(aq)
+ 3e- → Cr(s) E° = -0.74 V Hg2(aq)
+ 2e- → Hg(l) E° = +0.80 V calculate the standard emf for the
following cell: Cr | Cr||
Hg2|
Hg
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)