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QUESTION 4 Use the reaction: I2(s) - I2(aq) to calculate the molar solubility of I2(s) in water. 12(aq) + 2 e-2 l'E"-+ 0.615 V 12(s)+2e-2IE 0.5355 V A. -1.1505 B. +1.1505 C.-0.0795 D. +0.0795...
Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V Pb2+(aq) + 2e- Pb(s) -0.126V Cr3+(aq) + 3e- Cr(s) -0.740V The strongest oxidizing agent is: ______enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will I2(s) reduce Cr3+(aq) to Cr(s)? Which species can be reduced by Pb(s)? If none, leave box blank.
Standard Electrode Potentials at 25?C Reduction Half-Reaction E?(V) F2(g)+2e? ?2F?(aq) 2.87 Au3+(aq)+3e? ?Au(s) 1.50 Cl2(g)+2e? ?2Cl?(aq) 1.36 O2(g)+4H+(aq)+4e? ?2H2O(l) 1.23 Br2(l)+2e? ?2Br?(aq) 1.09 NO3?(aq)+4H+(aq)+3e? ?NO(g)+2H2O(l) 0.96 Ag+(aq)+e? ?Ag(s) 0.80 I2(s)+2e? ?2I?(aq) 0.54 Cu2+(aq)+2e? ?Cu(s) 0.16 2H+(aq)+2e? ?H2(g) 0 Cr3+(aq)+3e? ?Cr(s) -0.73 2H2O(l)+2e? ?H2(g)+2OH?(aq) -0.83 Mn2+(aq)+2e? ?Mn(s) -1.18 How can the table be used to predict whether or not a metal will dissolve in HCl? In HNO3? Drag the terms on the left to the appropriate blanks on the right to...
1) Calculate the cell potential, E°cell, for the following electrochemical reaction. Cu(s) + I2(g) → Cu+2(aq) + 2I-1(aq) E = ? a.-0.87 V b.-0.19 V c.0.19 V d.0.87 V 2) Calculate the equilibrium constant (Keq) for the following reaction at 25 °C? (Faraday's Constant = 96,500 J / V · mol) & (R = 8.314 J / mol · K) 2Fe+3(aq) + Sn+2(aq) → 2Fe+2(aq) + Sn+4(aq) K = ? a. 4.04 x 107 b. 3.44 x 1018 c. 4.48...
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
Consider the following half-reactions: Half-reaction E° (V) 12(s) + 2e - →21"(aq) 0.535V 2H+ (aq) + 2e - → H2(g) 0.000V Cr3+(aq) + 3e —— Cr(s) -0.740V The strongest oxidizing agent is: enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will 12(s) reduce Cr3+(aq) to Cr(s)? — Which species can be reduced by H2(g)? If none, leave box blank. Use the References to access important values if needed for this question....
A. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Sn2+(aq) + 2e-Sn(s) -0.140V Cr3+(aq) + 3e-Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Cl2(g) oxidize Cr(s) to Cr3+(aq)? (6) Which species can be oxidized by Sn2+(aq)? If none, leave box blank. B. Half-reaction E° (V) Cl2(g) + 2e-2Cl-(aq) 1.360V Cu2+(aq) + 2e-Cu(s) 0.337V Mn2+(aq) + 2e-Mn(s) -1.180V (1) The...
Table 20.1 Half Reaction E°(V). F2 (g) + 2e →2F (aq) +2.87 Cl2 (g) + 2e → 2CV (aq) +1.359 Br2 (1) + 2e → 2Br (aq) +1.065 O2 (g) + 4H+ (aq) + 4e → 2H20 (1)+1.23 Agt te → Ag (s) +0.799 Fe3+ (aq) + € → Fe2+ (aq) +0.771 12 (s) + 2e → 21+ (aq) +0.536 Cu2+ + 2e → Cu(s) +0.34 2H+ + 2e → H2 (g) Pb2+ + 2e → Pb (s) -0.126 Ni2+...
a. Cl2 b. F2 c. Br2 d. I2 e. All of the halogens have equal strength as oxidizing agents. E°(V +2.87 +1.359 +1.065 +1.23 +0.799 Table 20.1 Half Reaction F2 (8) 2e → 2F- (ag) Cl2(8) 2e → 2Cl- (aq) Br21) 2e → 2Br" (ag) 02 (elut 4H+ (ag) + 4e + 2H20 (1) Agte → Ag (5) Fe3+ (aq) → Fe2+ (aq) 12 (5) + 2e – 21 (ag) Cu2+ + 2e - Cu (s) 2H+ + 2e →...
39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standard potential for the cell reaction for 2 Ag + H2 (g)-2 Ag+2H +0.80 V b. a. +0.40 V +0.12 V d. c. -0.12 v none of these e. 40. Given: Fe (s) + 2 Ag' (aq) Fe (ag)+ 2 Ag (s) with E ell -1.24 V What is the...
Please show all steps taken (prefer typed solution) Half-Reaction E ° (V) Ag+ (aq) + e− → Ag (s) 0.7996 Al3+ (aq) + 3e− → Al (s) −1.676 Au+ (aq) + e− → Au (s) 1.692 Au3+ (aq) + 3e− → Au (s) 1.498 Ba2+ (aq) + 2e− → Ba (s) −2.912 Br2 (l) + 2e− → 2Br− (aq) 1.066 Ca2+ (aq) + 2e− → Ca (s) −2.868 Cl2 (g) + 2e− → 2Cl− (aq) 1.35827 Co2+ (aq) + 2e−...