Question

can someone help me write the half reactions for part A and B.

2. Electrolytic Cells A. In a small beaker, mix 1 mL of 0.1 M sodium iodide solution, 9 mL of water, 1 mL of fresh starch solution, and two drops of phenolphthalein indicator. Lead wires from a dry cell or storage battery into this solution and observe the region around each wire (9). Write half- reaction equations for the reactions (10). What would happen if the reaction products were to come into contact (11)? Why (12)? B. Polish a 1 x 3-cm copper strip until it is very bright and attach to it a wire leading to the negative pole of a 6-V de source. Fasten another wire to a small carbon electrode and lead this wire to the positive pole of the de source (see Fig. 29.2). Place the electrodes in a 50-ml beaker and add 35 mL of the "nickel solution for electroplating." To prepare, use 90 g of NiS04.6H20, 30 g of NiCl2, 6 H20, 15 g of boric acid, and 1 L of water. Allow the action to proceed until a good coating is obtained on the copper strip. Record your observations (13). Disconnect the wires and wash the plated electrode thoroughly, finally, polish it with a clean cloth. Describe the deposit (14). Write equations for the half-reactions, assuming that hydrogen is evolved at the positive electrode (15). Suggest some conditions to be maintained during electroplating to avoid a coating that is porous and nonadherent (16).

Answer 1

A Sodium iodide (NaI aqueous solution will contain Nah and I ims along with ho molecules. Ho molecules can undergo both onidation and reduchion To predict the product of electrolysis, we need to find out which onidation and reduchom reachions occurs more readily. - Onidation peachon Reduction meachion = Bethe I and Hyo can undergo Both nat and H20 can undergo sed onidation reduction 21- (aq) → Iz (aq) + 2e Na * +2e- Na (s) E = -2:214 - E' = 0.54v 242064+260 +H2(g) + 2014 (2) 2460 (1) 2(g) + 4H+ (aq) + 4e (con-] 2200M) . It is easier to extract an electron from It is easier for a less negative a lessen posihive electrode potenhal sa electrode foolembial to recept not onidation of I will be favored over electrons. So reduchon of a HO ho will be favored over Nata I " At anode At cathode : 21 lag) : 240 (1) + 26 Cag) + ae- H₂ (9) 204- (aq) 21 + 2 H2O (e) Is Caq) + H₂(g) + 204- (aq) No neaction would take place if the reachon products were to come in contact because the product formed at anode is rodime gas while the product formed at cathode is hydrogen gas and they don't react with each other. !