Question

34. 500 ml of 0.3 M AgNO, are added to 600 ml of 0.5 M KI
solution. The ions which will move towards the cathode
and anode respectively are
(1) Agl/Ag and NO, (2) Agl/l- and K
(3) K and Agl/l
(4) Agl/K* and I
​

Answer 1

Answer:

AgI/Ag and No,

Explanation:

Ag/l-and k

Answer 2

Answer:

The correct option is (3), $\[{K^ + }{\text{ and }}AgI/{I^ - }\]$.

Explanation:

The overall chemical equation of the given reaction is given as

$\[AgN{O_3} + KI \to AgI + KN{O_3}\]$

Volume 0.3M silver nitrate = 500ml

Milliequivalance of silver nitrate = $0.3\times500=150$

Volume of 0.5m potassium iodide = 600ml

Milliequivalance of potassium iodide = $0.5\times600= 300$

Since the milliequivalent silver nitrate is less, therefore, silver nitrate is a limiting reagent. 150 milliequivalent of potassium iodide will be remaining at the end of the reaction.

Therefore, silver iodide sol will form. It will adsorb iodide ion to form negatively charged sol which further attracts potassium ion and form a mobile ion.

Therefore, a mobile layer of potassium ions will be attracted toward the cathode and a fixed layer  $AgI/I^-$ will be attracted towards the anode.