Fe3+ + e- ---> Fe2+ --(i) (E°1 = 0.571 V)
∆G1 = -nFE°1 = -F(0.571) = -0.571F
Fe2+ + 2e- ---> Fe ---(ii) (E°2 = -0.440 V)
∆G2 = -nFE°2 = -2F(-0.440) = 0.880F
(i) + (ii) we get,
Fe3+ + 3e- ---> Fe
∆G3 = -nFE°3 = -3FE°3
=> ∆G1 + ∆G2 = ∆G3
=> 0.309F = -3FE°3
=> E°3 = -0.103 V
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e...
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
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 calculate the electrode potential for [Fe3+ + e = {Fe E; = ? for
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
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?
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)
The following standard reduction potentials have been determined for the aqueous chemistry of iron: Fe3 (a)e>Fe2-(aq 0.770 V Fe2 (aq) 2e>Fe(s E°-0.409 v Calculate the equilibrium constant (K) for the disproportionation of Fe2(aq) at 25 °C. Submit Answer 5 question attempts remaining
Using the standard electrode potentials Co2+(aq) + 2e → Co(s) E° = -0.28 V Fe2+(aq) + 2e → Fe(s) E° = -0.44 V choose the correct statement. Cobalt protects iron against corrosion by forming an insulating layer of oxide. Cobalt will protect iron against corrosion by cathodic protection. Cobalt is easier to oxidize than iron. Cobalt will be used as a sacrificial anode to protect iron against corrosion. Cobalt will serve as the cathode in an electrochemical cell when in...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
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+...