(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) \[\underset{Lead\mathrm{ }nitrate}{\mathop{Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\left( aq \right)+\underset{Potassium\mathrm{ }iodide}{\mathop{2Kl\left( aq \right)}}\,\mathrm{ }\to \underset{\begin{smallmatrix} Lead\mathrm{ }iodide \\ (yellow\mathrm{ }ppt) \end{smallmatrix}}{\mathop{Pb{{I}_{2}}(s)}}\,+2KN{{O}_{3}}(aq)\]
You need to login to perform this action.
You will be redirected in
3 sec