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4. Use the table below to provide a redox reaction involving the spontaneous oxidation of Cr...
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
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)
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
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+...
Given the following list of half-reaction reduction potentials, identify the reaction that will occur spontaneously as written: Half-reaction E° (V) -0.74 Cr3+ (aq) + 3 e ---> Cr (s) Sn4+ (aq) + 2 e ---> Sn2+ (aq) +0.154 -0.440 Fe2+ (aq) + 2 e ---> Fe(s) Fe3+ (aq) + e ---> Fe2+ (aq) +0.771 2 Cr (s) + 3 Fe2+ (aq) ---> 3 Fe (s) + 2 Cr3+ (aq) 2 Cr3+ (aq) + 3 Sn2+ (aq) ---> 3 Sn4+ (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...
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)...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2 (0.12 M |Fe2 (0.0012 M) Fe(s) E2 =-0.440 V Efe/Fe = 0.339 V Cu2t/Cu Is the electrochemical cell spontaneous or not spontaneous Ecell V as written at 25 °C? not spontaneous spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2(0.0060 M), Sn4+(0.14 M) Fe3+(0.13 M), Fe2+(0.0056 M) Pt(s)...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...