Q.61. Which of the following sets includes all the species that will change the orange colour of KCr

Q: What is the correct answer for this exam Chemistry Q61?

The correct answer is A. Potassium dichromate (K$_2$Cr$_2$O$_7$) in acidic medium is a strong oxidizing agent. It is orange in color and gets reduced to green Cr$^{3+}$ ions. Therefore, any species that can be oxidized by K$_2$Cr$_2$O$_7$ in acidic medium will cause the color change from orange to green. The standard reduction potential for dichromate is: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarro

Q: Q.61. Which of the following sets includes all the species that will change the orange colour of KCrO in acidic medium ?

The correct answer is A. Potassium dichromate (K$_2$Cr$_2$O$_7$) in acidic medium is a strong oxidizing agent. It is orange in color and gets reduced to green Cr$^{3+}$ ions. Therefore, any species that can be oxidized by K$_2$Cr$_2$O$_7$ in acidic medium will cause the color change from orange to green. The standard reduction potential for dichromate is: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarro

Q: What are the options for this this exam Chemistry question?

A. Fe$^{2+}$, Sn$^{2+}$, I$^{-}$, S$^{2-}$ · B. S$^{2-}$, Fe$^{3+}$, I$^{-}$, C$_2$O$_4^{2-}$ · C. Fe$^{2+}$, NO$_2$, SO$_2$, Sn$^{2+}$ · D. Fe$^{3+}$, SO$_4^{2-}$, S$^{2-}$, Sn$^{4+}$

Question

Q.61. Which of the following sets includes all the species that will change the orange colour of KCrO in acidic medium ?

FirstInTest · Accepted Answer

Potassium dichromate (K$_2$Cr$_2$O$_7$) in acidic medium is a strong oxidizing agent. It is orange in color and gets reduced to green Cr$^{3+}$ ions. Therefore, any species that can be oxidized by K$_2$Cr$_2$O$_7$ in acidic medium will cause the color change from orange to green.

The standard reduction potential for dichromate is: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}$ (E$^{\circ}$ = +1.33 V).
We need to identify species that have a standard oxidation potential greater than -1.33 V (or standard reduction potential less than +1.33 V) and can be oxidized.

Let's examine the species in option (1): Fe$^{2+}$, Sn$^{2+}$, I$^{-}$, S$^{2-}$.
*   **Fe$^{2+}$:** Can be oxidized to Fe$^{3+}$ ($\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}$, E$^{\circ}_{ox}$ = -0.77 V). Since -0.77 V > -1.33 V, Fe$^{2+}$ will be oxidized by Cr$_2$O$_7^{2-}$.
*   **Sn$^{2+}$:** Can be oxidized to Sn$^{4+}$ ($\mathrm{Sn^{2+}} \rightarrow \mathrm{Sn^{4+}} + 2\mathrm{e^-}$, E$^{\circ}_{ox}$ = -0.15 V). Since -0.15 V > -1.33 V, Sn$^{2+}$ will be oxidized by Cr$_2$O$_7^{2-}$.
*   **I$^{-}$:** Can be oxidized to I$_2$ ($\mathrm{2I^-} \rightarrow \mathrm{I_2} + 2\mathrm{e^-}$, E$^{\circ}_{ox}$ = -0.54 V). Since -0.54 V > -1.33 V, I$^{-}$ will be oxidized by Cr$_2$O$_7^{2-}$.
*   **S$^{2-}$:** Can be oxidized to S ($\mathrm{S^{2-}} \rightarrow \mathrm{S} + 2\mathrm{e^-}$, E$^{\circ}_{ox}$ = +0.48 V). Since +0.48 V > -1.33 V, S$^{2-}$ will be oxidized by Cr$_2$O$_7^{2-}$.
All species in option (1) can be oxidized by K$_2$Cr$_2$O$_7$ in acidic medium, thus changing its color.

Let's quickly check other options for species that cannot be oxidized:
*   Option (2): Fe$^{3+}$ (already oxidized), C$_2$O$_4^{2-}$ (oxalate, can be oxidized, but Fe$^{3+}$ cannot).
*   Option (3): NO$_2$ (can be oxidized), SO$_2$ (can be oxidized), but Fe$^{2+}$ and Sn$^{2+}$ are missing.
*   Option (4): Fe$^{3+}$ (already oxidized), S$^{2-}$ (can be oxidized), Sn$^{4+}$ (already oxidized).

Therefore, option (1) is the correct set.

Answer

Fe$^{2+}$, Sn$^{2+}$, I$^{-}$, S$^{2-}$

Answer

S$^{2-}$, Fe$^{3+}$, I$^{-}$, C$_2$O$_4^{2-}$

Answer

Fe$^{2+}$, NO$_2$, SO$_2$, Sn$^{2+}$

Answer

Fe$^{3+}$, SO$_4^{2-}$, S$^{2-}$, Sn$^{4+}$

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