Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe^3+ oxidises the iodide ion B. Fe^3+ oxidises the persulphate ion C. Fe^2+ reduces the iodide ion D. Fe^2+ reduces the persulphate ion Choose the most appropriate answer from the options given below:

D and F Block Elements PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q4 with Solution

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Correct answer: C

Overall reaction being catalysed

The reaction between iodide and persulphate ions is:

\mathrm{S_2O_8^{2-} + 2I^- \rightarrow I_2 + 2SO_4^{2-}}

Here:

$\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$
$\mathrm{S_2O_8^{2-}}$ is reduced to $\mathrm{SO_4^{2-}}$

Role of $\mathrm{Fe^{3+}/Fe^{2+}}$ as catalyst

Iron(III) provides an alternate pathway through the redox couple $\mathrm{Fe^{3+}/Fe^{2+}}$ .

Step 1: $\mathrm{Fe^{3+}}$ reacts with iodide

Since $\mathrm{Fe^{3+}}$ is an oxidising agent, it oxidises iodide to iodine while itself gets reduced to $\mathrm{Fe^{2+}}$ :

2\mathrm{Fe^{3+}} + 2\mathrm{I^-} \rightarrow 2\mathrm{Fe^{2+}} + \mathrm{I_2}

So statement A is correct:

A. $\mathrm{Fe^{3+}}$ oxidises the iodide ion ✅

Now $\mathrm{Fe^{2+}}$ formed must be converted back to $\mathrm{Fe^{3+}}$

Persulphate is a strong oxidising agent, so it oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ . Therefore $\mathrm{Fe^{2+}}$ reduces persulphate:

\mathrm{S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}}

So statement D is correct:

D. $\mathrm{Fe^{2+}}$ reduces the persulphate ion ✅

Check the other statements

B. $\mathrm{Fe^{3+}}$ oxidises the persulphate ion ❌
Persulphate is already a strong oxidising agent; it is not oxidised here.
C. $\mathrm{Fe^{2+}}$ reduces the iodide ion ❌
Iodide is itself a reducing agent; it is oxidised, not reduced.

Correct option

Correct statements are A and D only.

So the correct option is:

\boxed{\text{C}}

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View written solutionFree

Correct answer: C

Overall reaction being catalysed

The reaction between iodide and persulphate ions is:

\mathrm{S_2O_8^{2-} + 2I^- \rightarrow I_2 + 2SO_4^{2-}}

Here:

$\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$
$\mathrm{S_2O_8^{2-}}$ is reduced to $\mathrm{SO_4^{2-}}$

Role of $\mathrm{Fe^{3+}/Fe^{2+}}$ as catalyst

Iron(III) provides an alternate pathway through the redox couple $\mathrm{Fe^{3+}/Fe^{2+}}$ .

Step 1: $\mathrm{Fe^{3+}}$ reacts with iodide

Since $\mathrm{Fe^{3+}}$ is an oxidising agent, it oxidises iodide to iodine while itself gets reduced to $\mathrm{Fe^{2+}}$ :

2\mathrm{Fe^{3+}} + 2\mathrm{I^-} \rightarrow 2\mathrm{Fe^{2+}} + \mathrm{I_2}

So statement A is correct:

A. $\mathrm{Fe^{3+}}$ oxidises the iodide ion ✅

Now $\mathrm{Fe^{2+}}$ formed must be converted back to $\mathrm{Fe^{3+}}$

Persulphate is a strong oxidising agent, so it oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ . Therefore $\mathrm{Fe^{2+}}$ reduces persulphate:

\mathrm{S_2O_8^{2-} + 2Fe^{2+} \rightarrow 2SO_4^{2-} + 2Fe^{3+}}

So statement D is correct:

D. $\mathrm{Fe^{2+}}$ reduces the persulphate ion ✅

Check the other statements

B. $\mathrm{Fe^{3+}}$ oxidises the persulphate ion ❌
Persulphate is already a strong oxidising agent; it is not oxidised here.
C. $\mathrm{Fe^{2+}}$ reduces the iodide ion ❌
Iodide is itself a reducing agent; it is oxidised, not reduced.

Correct option

Correct statements are A and D only.

So the correct option is:

\boxed{\text{C}}

D and F Block Elements question

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