Question

The gas which has no effect on acidified ${K_2}C{r_2}{O_7}$ or $KMn{O_4}$ is:

Answer 1

When carbon dioxide is passed through calcium hydroxide or limewater, it forms an insoluble and a white precipitate due to the formation of calcium carbonate, but when excess of carbon dioxide is passed through lime water or calcium hydroxide, the milkiness disappears because calcium carbonate, the insoluble precipitate and the cause of milkiness is converted to calcium bicarbonate which is soluble hence the milkiness of the limewater disappears.

Complete answer:
Potassium dichromate with molecular formula ${K_2}C{r_2}{O_7}$ is a very bright red-orange, crystalline ionic solid and an oxidizing agent used in laboratory and industrial applications. While, potassium permanganate with molecular formula $KMn{O_4}$ is a purplish-black crystalline salt.
Acidified potassium dichromate solution changes to green colour from red-orange colour when alcohol is oxidized. Similarly, acidified ${K_2}C{r_2}{O_7}$ oxidized $S{O_2}$to sulphuric acid and the colour of the solution changes from orange to green. Potassium permanganate gives a pink or purple colour solution when dissolved in water.
When $S{O_2}$ gas is passed through acidified $KMn{O_4}$, the colour changes from purple to colorless.
Carbon dioxide gas when passed through lime water i.e. calcium hydroxide, turns it milky due to the formation of calcium carbonate, a white and insoluble precipitate.
$Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O$
But, when carbon dioxide is passed through acidified potassium dichromate or potassium permanganate there is no change. Therefore, carbon dioxide has no effect on acidified potassium dichromate or potassium permanganate.
So,$C{O_2}$ gas has no effect on acidified ${K_2}C{r_2}{O_7}$ or $KMn{O_4}$.

Note:
Potassium permanganate is also used in the laboratory in qualitative analysis. Potassium permanganate titration are carried out only in the presence of dilute sulfuric acid because oxygen produced from the reaction of $KMn{O_4}$ with dilute ${H_2}S{O_4}$ is used for oxidizing the reducing agent. $KMn{O_4}$ does not produce any oxygen of its own to act as an oxidizing agent. Hydrochloric acid cannot be used in place of dilute ${H_2}S{O_4}$.