Calculate the equilibrium constant for each of the reactions at 25∘C.
Standard Electrode Potentials at 25 ∘C | |||||||||||||||||||||
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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)
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Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25...
Calculate the equilibrium constant for each of the reactions at 25 ∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Pb2+(aq)+2e− →Pb(s) -0.13 Zn2+(aq)+2e− →Zn(s) -0.76 Br2(l)+2e− →2Br−(aq) 1.09 Cl2(g)+2e− →2Cl−(aq) 1.36 MnO2(s)+4H+(aq)+2e− →Mn2+(aq)+2H2O(l) 1.21 Pb2+(aq)+2e− →Pb(s) -0.13 Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using two significant figures.
Use tabulated half-cell potentials to calculate ΔG∘rxn for each of the following reactions at 25 ∘C. O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq) Br2(l)+2I−(aq)→2Br−(aq)+I2(s)
Calculate the equilibrium constant for each of the reactions at 25 ∘C. a. 2Cr3+(aq)+3Sn(s)→2Cr(s)+3Sn2+(aq) b. O2(g)+2H2O(l)+2Sn2+(aq)→4OH−(aq)+2Sn4+(aq) c. 2Cr3+(aq)+3Ni(s)→2Cr(s)+3Ni2+(aq)
Calculate the equilibrium constant for each of the reactions at 25∘C∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction Cl2(g)+2e ---> 2CI- 1.36 I2(s)+2e --> 2I- 0.54 Part A Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s) K= ?
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...
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the following reaction at 25 °C. 2Fe3+ (aq) + 3Sn(s) → 2Fe(s) + 3Sn2+ (aq) Hint: the standard potential of Fe3+ + 3e + Fe(s) is -0.036 and the standard potential of Sn2+ + 2e → Sn(s) is -0.140. AG° = -5 kJ/mol AG° = -10 kJ/mol AG° = -20 kJ/mol AG° = -30 kJ/mol
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) +...
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.774 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here. Reduction Half-Reaction Standard Potential Ered° (V) F2(g) + 2e– → 2F–(aq) +2.87 O3(g) + 2H3O+(aq) + 2e– → O2(g) + 3H2O(l) +2.076 Co3+(aq) + e– → Co2+(aq) +1.92 H2O2(aq) + 2H3O+(aq) + 2e– → 2H2O(l) +1.776 N2O(g) + 2H3O+(aq) + 2e– → N2(g) + 3H2O(l) +1.766 Ce4+(aq) + e– → Ce3+(aq)...
Using the standard reduction potentials listed, calculate the equilibrium constant for each of the following reactions at 298 K. A) Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) Express your answer using two significant figures. B) Co(s)+2H+(aq)→Co2+(aq)+H2(g) Express your answer using two significant figures. C) 10Br−(aq)+2MnO−4(aq)+16H+(aq)→2Mn2+(aq)+8H2O(l)+5Br2(l) Express your answer using two significant figure. E°(V) -0.83 +0.88 +1.78 +0.79 Half-Reaction E°(V) Half-Reaction Ag+ (aq) + - Ag(s) +0.80 2 H20(1) + 2 e — H2(8) + 2 OH+ (aq) AgBr(s) + - Ag(s) + Br" (aq) +0.10 HO2...
Using standard reduction potential in aqueous solutions at 25c Table, which substance is most likely to be oxidised by O2 (g) in acidic aqueous solution? Select one: a. Br2 (l) b. Br- (aq) c. Ni2+ (aq) d. Ag (s) e. Cu2+ (aq) Cathode (Reduction) Half-Reaction Standard Potential E° (volts) Li+(aq) + e- -> Li(s) -3.04 K+(aq) + e- -> K(s) -2.92 Ca2+(aq) + 2e- -> Ca(s) -2.76 Na+(aq) + e- -> Na(s) -2.71 Mg2+(aq) + 2e- -> Mg(s) -2.38 Al3+(aq)...