(a) In electrolysis of water, why is the volume of gas collected over one electrode double that gas collected over the other electrode ?
(b) (i) What is observed whern a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube ?
(ii) What type of reaction is this ?
(iii) Write a balanced chemical equation to represent the above reaction.

Explanation

(a) In electrolysis of water `(H_(2)O)`, the hydrogen goes to one test tube and oxygen goes to to another. The two electrodes collect H and O separately.
Since water `(H_(2)O)` consists of 2 parts of hydrogen and 1 part of oxygen, so, the volume of hydrogen gas `(H_(2))` collected over cathode (negative electrode) is double the volume of oxygen gas `(O_(2))` collected over anode (positive electrode).
(b) (i) When potassium iodide solution is added to lead nitrate solution, then a yellow precipitate of lead iodide is produced along with potassium nitrate solution.
(ii) This is a double displacement reaction.
(iii) `underset("Lead nitrate")(Pb(NO_(3))_(2)(aq))+underset("Potassium iodide")(2KI(aq)) rarr underset("Lead iodide (yellow ppt)")(PbI_(2)(s)) + 2KNO_(3)(aq)`