Calculate the equilibrium constant K for the reactions at 298 K.
a) Br2(g) + 2Cl-(aq) ---> 2Br-(aq) + Cl2(g)
b) A Galvanic Cell with the SHE and Fe+2/Fe(s)

Calculate the equilibrium constant K for the reactions at 298 K. a) Br2(g) + 2Cl-(aq) --->...
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.
Exercise 18.67: Problems by Topic - Cell Potential, Free Energy, and the Equilibrium Constant Calculate the equilibrium constant for each of the reactions at 25 ∘C. Part A Pb2+(aq)+Mg(s)→Pb(s)+Mg2+(aq) Express your answer using one significant figure. K = 5×1075 SubmitMy AnswersGive Up All attempts used; correct answer displayed Part B Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g) Express your answer using one significant figure. K = 6•10−10 SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part C MnO2(s)+4H+(aq)+Cu(s)→Mn2+(aq)+2H2O(l)+Cu2+(aq) Express your answer using one significant figure....
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)
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= ?
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 Cu2+(aq)+2e− →Cu(s) 0.16 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Cl2(g)+2e− →2Cl− 1.36 I2(s)+2e− →2I− 0.54 Part B O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq) (Express your answer using two significant figures. ) Part C Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s) (Express your answer using two significant figures.)
Question 5 (1 point) For the cell diagram Pt(s) | Br2(1) | Br"(aq) || C1-(aq)| Cl2(g) | Pt(s) which reaction occurs at the cathode? O 2C1-(aq) --> Cl2(g) + 2e- O 2Br (aq) --> Br2(1) + 2e O Br2(l) +2e --> 2Br (aq) O Cl2(g) + 2Br (aq) --> 2C1-(aq) + Br2(1) O C12(g) + 22 --> 2C1“(aq)
Consider a galvanic cell that uses the half reactions: Fe3+(aq) + 3e → Fe(s) Ered=-0.04V Co2+ (aq) + 2e → Co(s) Ered - -0.28 V • 1) Calculate AGº for this reaction in kJ/mol. • 2) Calculate the equilibrium constant, K, for this reaction at 298 K • 3) What is the cell potential at 25°C when [Fe3+] - 1.09 M and [Co2+] = 0.057 M?
3. Calculate deltaS and deltaG and the equilibrium constant 298
K for each of these reactions, and indicate whether they are
spontaneous under standard conditionsa.
H2 (g) + F2 (g) --> 2HF (g)
b. C2H5OH (g) --> C2H4 (g) + H2O (g) (Calculate deltaH for
this one as well )
c.2HCl (g) + Br2 (g) → 2HBr(g) + Cl2 (g)
d. PCl3 (g) + 3 H2 (g) --> PH3 (g) + 3 HCl (g) (Calculate H
for this one as...
Determine the equilibrium constant (Keq) at 25°C for the reaction?Cl2(g) + 2Br- (aq)? 2Cl- (aq) + Br2(l)A. 1.5 × 10-10 B. 6.3 × 109 C. 1.3 × 1041 D. 8.1 × 104 E. 9.8Please show your work.
2. Calculate the standard cell emf (cell potential) for electrochemical cells having the following overall cell reactions and predict if they are spontaneous: a) Sn(s) + Pb2+ (aq) → Sn?" (aq) + Pb(s) Eºcell = Volts. Spontaneous? b) 2002 (aq) + Zn?"(aq) → 2Co3+ (aq) + Zn(s) Eºcell__volts; Spontaneous? c) Cl2(g) + 2Br (aq) + 2Cl(aq) + Br2(1); E° = volts; Spontaneous?