Question

Describe the chemistry that was occurring in the experiment between the time when solutions A (0.3 M KI) and B (0.1 M H2O2) were mixed and STOP TIME

Answer 1

Potassium Iodide (e.g. KI) may result able to react against Hydrogen Peroxide (e.g H2O2).
There are two main chemical reactions :
-) POTASSIUM IODIDE TAKES PART TO REDOX REACTION

2 KI(aq) + H2O2(aq) <---> 2 KOH(aq) + I2(aq)

so Iodine (e.g. I2) darkens the liquid in a REDDISH hue ;

-) POTASSIUM IODIDE ACTS AS CATALYST ON THE HYDROGEN's PEROXIDE's DECOMPOSITION

2 H2O2(aq) <---> 2 H2O(aq) + O2(g)

so Oxygen's bubbles (e.g. O2) leaves up.


ELECTRICAL POTENTIALs
As you know, scientific literature reported Standard Electrical Potentials.
Your trouble involves the following ones :

i) H2O2(aq) + 2 H+(aq) + 2e ---> 2 H2O(aq)
ii) 2 I-(aq) + 2 H2O(aq) ---> 2 IO-(aq) + 4 H+(aq) + 4e
iii) H2O2(aq) ---> 2 H+(aq) + O2(g) + 2e
iv) 2 I-(aq) ---> I2(aq) + 2e

These Electrical Potentials DEPEND STRONGLY ON THE pH's LEVELs, so you will be able to distinguish AN ACIDIC TREATISE (at pH = 0) AND AN ALKALINE's ONE
(at pH = 14).


ACIDIC MEDIA
By means of the Standard Electrical Potentials stated that
IN pH = 0's MEDIA, POTASSIUM IODIDE's
HALF-REACTION LIES BETWEEN HYDROGEN PEROXIDE's ONEs :

i) H2O2(aq) + 2 H+(aq) + 2e ---> 2 H2O(aq)
E,i(pH = 0) = +1.77 V

ii) 2 I-(aq) + 2 H2O(aq) ---> 2 IO-(aq) + 4 H+(aq) + 4e
E,ii(pH = 0) = +1.00 V

iii) H2O2(aq) ---> 2 H+(aq) + O2(g) + 2e
E,iii(pH = 0) = +0.68 V

iv) 2 I-(aq) ---> I2(aq) + 2e
E,iv(pH = 0) = +0.55 V

AT pH = 0, IODINE DERIVES FROM THE LEAST NOBLE EVENT : IODINE FORMS.
POTASSIUM IODIDE ACTS AS CATALYST ON HYDROGEN PEROXIDE's DECOMPOSITION WHILE IODIDE UNDERGOES OXIDIZATION TO IODINE.


ALKALINE MEDIA
By means of the Standard Electrical Potentials stated that
IN pH = 14's MEDIA, POTASSIUM IODIDE TAKE PART INDIFFERENTLY TO ITS OWN HALF-REACTIONs:

i) H2O2(aq) + 2e ---> 2 OH-(aq)
E,i(pH = 14) = +0.88 V

iv) 2 I-(aq) ---> I2(aq) + 2e
E,iv(pH = 14) = +0.55 V

ii) 2 I-(aq) + 2 OH-(aq) ---> 2 IO-(aq) + 2 H2O(aq) + 4e
E,ii(pH = 14) = +0.49 V

iii) H2O2(aq) + 2 OH-(aq) ---> 2 H2O(aq) + O2(g) + 2e
E,iii(pH = 14) = -0.08 V

AT pH = 14, IODINE RESULTS NO MORE AS THE LEAST NOBLE's ONE : NOW, OXYGEN DOES IT.
POTASSIUM IODIDE PREFERS CATALYTIC ROLE ON HYDROGEN PEROXIDE's DECOMPOSITION INSTEAD TO FORM IODINE.


CONCLUSIONs
Potassium Iodide undergoes a Chemical Equilibrium toward Hypoiodite Ion : in this fashion, Iodide result able to decompose Hydrogen Peroxide by means of leaving Oxygen bubbles.
Potassium Iodide may undergo Chemical Oxidization to form Iodine : the greater the Acidity's medium, the greater the action of Hydrogen Peroxide in order to form Iodine.