question_answer

(a) In electrolysis of water, why is the volume of gas collected over one electrode double that of gas collected over the other electrode?

(b) (i) What is observed when 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.

Answer:

(a) In electrolysis of water \[({{H}_{2}}O)\], the hydrogen goes to one test tube and oxygen goes 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 to lead nitrate solution, then a precipitate of lead iodide is produced with potassium nitrate solution.

(ii) This is double displacement reaction.

(iii) \[\underset{Lead\mathrm{ }nitrate}{\mathop{Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\left( aq \right)+\underset{Potassium\mathrm{ }iodide}{\mathop{2Kl\left( aq \right)}}\,\mathrm{ }\to \underset{\begin{smallmatrix}  Lead\mathrm{ }iodide \\  (yellow\mathrm{ }ppt) \end{smallmatrix}}{\mathop{Pb{{I}_{2}}(s)}}\,+2KN{{O}_{3}}(aq)\]