Inroganic chemsitry lab question--please only answer if you understand it and yo
ID: 475228 • Letter: I
Question
Inroganic chemsitry lab question--please only answer if you understand it and you can answer it CORRECTLY. I ONLY need help with number 26, buy included it all if it helps
**Number 21 explains what I saw on each ends of the alligator clips (positive end versus negative end)
(D) Exploring Electrolysis a) obtain an electrolysis setup from your instructor or assistant. The setup consists of a 2-hole stopper fitted with nichrome wires sealed in plastic tubes, a small beaker that the stopper should fit into, two alligator clip wire leads and a 9V battery as shown below: Figure 4: An Electrolytic Cell b Add 10 mL of 0.2 M KI to the beaker. The KI solution contains a small amt. of the acid-base indicator phenolphthalein. swirl the solution in the beaker and place a small piece of white paper underneath the beaker c) Insert the stopper assembly onto the beaker and connect the wires to the battery terminals with the alligator clips. Answer the questions that follow: 21. Carefully describe what you observe from the moment the clips are connected Be sure to note which wire is attached to the battery's (t) terminal and which is attached to the terminal. The positive side turned yellow The negative side was yellow and turned pink, and was bubblyExplanation / Answer
Ions present in solution are K+, I- and H2O
Possible reactions at anode:- Oxidation reaction of I- dominates over oxidation of H2O as oxidation potential of I- is greater than the oxidation potential of H2O.
2 I- (aq) --> I2 (g) + 2 e- E0 = -0.53 V
2 H2O (l) ---> 4 H+ (aq) + O2 + 4e- E0 = -1.23 V
Possible cathode reactions: Reduction potential of the reaction of H2O is greater than that of K+. Hence reduction reaction of K+ takes place at the cathode.
K+ (aq) + e- --> K (s) E0 = – 2.93V
2 H2O (l) + 2e- --> H2 (g) + 2 OH- (aq) E0 = – 0.83V
Actual reaction at anode: 2 I- (aq) --> I2 (g) + 2 e- E0 = -0.53 V
Actual reaction at cathode: 2 H2O (l) + 2e- --> H2 (g) + 2 OH- (aq) E0 = – 0.83V
Overall reaction: 2 I- (aq) + 2 H2O (l)--> I2 (g) + H2 (g) + 2 OH- (aq)