3 Sn4+ + 2 Cr ------------> 2 Cr3+
+ 3 Sn2+
here the oxidation is (anode)
Cr ------------> Cr3+ + 3e- => (+0.74 )
reduction reaction is (cathode)
Sn4+ + 4e- ------------------> Sn2+ ( +0.154 )
E0 cell = E0 (cathode ) - E0 (anode)
E0 cell = +0.154 - (+0.74)
E0 cell = 0.154 - 0.74 => -0.586 => -0.59
answer => D) = -0.59
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...
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 () -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
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...
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)
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
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
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
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-