Question

Consider the reaction:      

2     I^– + Cl₂ " 2 Cl^– + I₂

A.    Name the compounds that are oxidized and reduced on the left hand of the reaction. (2 points)  

B.  Redox reaction involves the transfer of _________________ from one element to another. (2 point)

C. How many electrons does the reducing agent provide in this reaction? Explain. (2 points)

D. In the reaction shown above ______________ is the reducing agent and____________ is the oxidizing agent.

E. Write the equilibrium expression for this reaction.  (4 points)

F. If [Cl₂] = 0.0682M, [I₂] = 0.662 M, [I^-] = 0.0196 and [Cl^-] = 0.956 at equilibrium, what is the value of K?   

G. Are reactants or products favored in this reaction? Explain

H. What is the effect of:

removing some Cl₂ from the equilibrium system? Explain.

adding KI? Explain.

adding I₂?  Explain.

adding a catalyst?  Explain.

Answer 1

2I– + Cl2 -------> 2 Cl– + I2

Oxidation half reaction : 2I- ------> I2 + 2e-

reduction half reaction : Cl2 + 2e- ------> 2Cl-

A) Iodide ion (I-) is oxidised

Chlorine (Cl) is reduced

B) Redox reaction involves the transfer of electrons from one element to other

C) two electrons

D) I- is the reducing agent

Cl2 is the oxidising agent

E) Keq = [Cl-]^2 [I2] / [I-]^2 [Cl2]

F) Keq = (0.956)^2(0.662) / (0.0196)^2(0.0682)

=> Keq = 2.31 x 10^4

G) This reaction is products favoured, since , Keq is greaterthan 1

H) Removing some Cl2 => shifts to reactants

adding some KI => shifts to products side

adding I2 => shifts to reactants side

adding catalyst => has no net effect