Question

(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 1

(a) In electrolysis of water $\left(H_{2}O\right)$, the hydrogen goes to one electrode and oxygen goes to another. The two electrodes collect H and O separately. Since water $\left(H_{2}O\right)$ consists of 2 parts of hydrogen and 1 part of oxygen, the volume of hydrogen gas $\left(H_2\right)$ collected over cathode (negative electrode) is double the volume of oxygen gas $\left(O_2\right)$ collected over anode (positive electrode).

(b) (i)When potassium iodide solution $\left(KI\right)$ is added to lead nitrate solution $\left(Pb\right(N{O}_3{)}_2)$, then a yellow precipitate of lead iodide $\left(Pb{I}_2\right)$ is produced along with potassium nitrate solution $\left(KN{O}_3\right)$.

(ii) This is double displacement reaction.

(iii)
$\begin{array}{cccc}& & & \\ Pb\left(N{O}_3{)}_2\right(aq)& +& 2KI\left(aq\right)⟶& \\ \text{Lead}& & \text{Potassium}& \\ \text{nitrate}& & \text{iodide}& \\ & & Pb{I}_2\left(s\right)+& 2KN{O}_3\left(aq\right)\\ & & \text{Lead iodide}& \\ & & \text{(yellow ppt)}& \end{array}$