Homework Answers
Answer
Option d
Fe2+ (aq) + Cr3+ (aq) ---> Cr (s) + 3 Fe3+ (aq)
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Given the following list of half-reaction reduction potentials, identify the reaction that will occur spontaneously as...
Table 20.2 Half-reaction E (V) -0.74 Cr3+ (aq) 3e --Cr (s) Fe2+(aq)+2e-Fe (s) Fe3+ (aq) 0.440...
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+...
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction...
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)
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq...
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?
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 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
4. Use the table below to provide a redox reaction involving the spontaneous oxidation of Cr...
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-
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Table 20.2 Ha...
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)...
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) 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
Using the standard reduction potentials given below, decide which of the following reactions will occur spontaneously...
Using the standard reduction potentials given below, decide which of the following reactions will occur spontaneously as written. Fe3+(aq) + e + Fe2+(aq) E° = 0.77 V Sn4+ (aq) + 2e → Sn2+(aq) E° = 0.13V Zn2+(aq) + 2e → Zn(s) E° = -0.77 V Lit(aq) + e + Li(s) E° = -3.05 V 2Li+ (aq) + Sn2+(aq) → 2Li(s) + Sn4+(aq) Sn2+(aq) + Zn2+(aq) → Sn4+(aq) + Zn(s) Lit(aq) + Fe2+(aq) - Li(s) + Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) →...
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5)...
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...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (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+...
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)
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?
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
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-
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)...
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
Using the standard reduction potentials given below, decide which of the following reactions will occur spontaneously as written. Fe3+(aq) + e + Fe2+(aq) E° = 0.77 V Sn4+ (aq) + 2e → Sn2+(aq) E° = 0.13V Zn2+(aq) + 2e → Zn(s) E° = -0.77 V Lit(aq) + e + Li(s) E° = -3.05 V 2Li+ (aq) + Sn2+(aq) → 2Li(s) + Sn4+(aq) Sn2+(aq) + Zn2+(aq) → Sn4+(aq) + Zn(s) Lit(aq) + Fe2+(aq) - Li(s) + Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) →...
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...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
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