Question

Which of the following colour changes dshown during redox titration is not correct?

• A. $C{r}_2{O}_7^{2-}$ oxidises the indicator diphenylamine to produce blue colour showing end point.
• B. Iodine formed by oxidation of $I^{-}$ ions gives blue colour with starch showing end point.
• C. $KMn{O}_4$ in the form of $Mn{O}_4^{-}$ ions gives pink colour showing end point.
• D. Thiosulphate ions $\left(S_2{O}_3^{2-}\right)$ give blue colour showing end point.

Answer 1

The correct option is D Thiosulphate ions $\left(S_2{O}_3^{2-}\right)$ give blue colour showing end point.

A. Diphenylamine is an indicator used in titrations that involve redox reactions, especially those in which chromate ($Cr{O}_4^{2-}$) or dichromate ($C{r}_2{O}_7^{2-}$) is used as the oxidizing agent. It shows a color change from green to violet when end point of the titration is reached.

B. Iodine element in the presence of iodide ion will give the characteristic blue black color with starch which is the end-point of iodometric titrations.

C. In the presence of a reducing agent, $KMn{O}_4$ acts as self-indicator and changes color from pink to colorless. Because in $Mn{O}_4^{-}$ Mn is in +7 oxidation state and is in highest oxidation state. Thus will tend to get reduced and easily takes electrons and being a charge transfer complex, shows intense color. After titration it is $Mn{O}_2$ in +4 state and has no electronic charge transfer, therefore colorless.

D. Thiosuplate $S_2{O}_3^{2-}$ is not a indicator thus has no color change as end point in titrations.