7.53 Given the following electrode potentials at 25°C, Fe3+ +e- = Fe2+ E' = 0.771 V Fe2+ + e = Fe(s) E = -0.440 V calcu...
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- Fe(s) Fe2+ + Select one: a. -0.132 b. -0.036 c. 0.081 d.-0.211 e. 0.103
Given the following electrode potentials at 25°C Fe3+ + e- → Fe2+ E° = 0.171 V Fe2+ + 2e-→ Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- → Fe(s) Select one: a.-0.236 b. 0.081 c.-0.036 d. -0.211 e. 0.121
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+...
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...
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?
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)...
Using the following standard reduction potentials Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 V calculate the standard cell potential for the galvanic cell reaction given below, and determine whether or not this reaction is spontaneous under standard conditions. Pb2+(aq) + 2 Fe2+(aq) → 2 Fe3+(aq) + Pb(s) Group of answer choices E° = -0.90 V, spontaneous E° = -0.90 V, nonspontaneous E° = +0.90 V, nonspontaneous E° = +0.90...
A galvanic cell is based on the following half-reactions: Fe2+ + 2e-→ Fe(s) Eº = -0.440 V 2H+ + 2e-→ H2(g) Eº = 0.000 V where the iron compartment contains an iron electrode and [Fe2+] = 1.00 × 10-3 M and the hydrogen compartment contains a platinum electrode, PH2 = 1.00 atm , and a weak acid, HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.320 V at 25ºC, calculate the Ka value for...
Using the following standard reduction potentials: Fe3+ (aq) + e. → Fe2+ (aq) Eo = +0.77 V Pb2+ (aq) + 2 e. → Pb(s) E。--0.13 V Calculate the standard cell potential for the galvanie cell reaction given below, and determine whether or not this reaction is spostaneous under standard conditions. Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb(s) ⓔ A. E.-0.90 V, nonspontaneous OB. E-0.90 V, spontaneous C. Eo +0.90 V, nonspontaneous OD0.90 V, spontaneous