Consider the following list of reagents : Acidified K2Cr2O7, alkaline KMnO4, CuSO4, H2O2, Cl2, O3, F

Q: Why doesn't alkaline $KMnO_4$ produce $I_2$ like it does in an acidic medium?

In an alkaline medium, the oxidation process is more vigorous for iodide. Instead of stopping at $I_2$ (oxidation state 0), the strong alkaline conditions drive the oxidation all the way to iodate, $IO_3^-$ (oxidation state +5). This is a classic trap in competitive exams!

Q: I thought $Cu^{2+}$ and $I^-$ just form $CuI_2$. Why does it produce $I_2$?

Cupric iodide ($CuI_2$) is highly unstable. As soon as it forms, it spontaneously decomposes into the highly insoluble white precipitate of cuprous iodide ($Cu_2I_2$) and releases free iodine ($I_2$). The precipitation drives the forward reaction.

Q: Can ozone ($O_3$) really oxidize iodide?

Absolutely! Ozone is a very powerful oxidizing agent. In fact, passing ozone through a solution of potassium iodide is a standard chemical test. It oxidizes $I^-$ to $I_2$, which can then be detected using starch paper (turning it blue-black).

Q: What is the role of $Na_2S_2O_3$ in this list?

Sodium thiosulfate is a distractor. It is a well-known reducing agent used in iodometric titrations to determine the amount of iodine. It reacts with $I_2$ to convert it back to $I^-$, so it cannot possibly oxidize $I^-$ to $I_2$.

Q: Does $FeCl_3$ have enough oxidizing power to react with iodide?

Yes, the $Fe^{3+}/Fe^{2+}$ redox couple has a higher standard reduction potential (+0.77 V) compared to the $I_2/I^-$ couple (+0.54 V). Therefore, $Fe^{3+}$ can spontaneously oxidize $I^-$ to $I_2$ while being reduced to $Fe^{2+}$.

Question

Consider the following list of reagents :
Acidified $	ext{K}_2	ext{Cr}_2	ext{O}_7$, alkaline $	ext{KMnO}_4$, $	ext{CuSO}_4$, $	ext{H}_2	ext{O}_2$, $	ext{Cl}_2$, $	ext{O}_3$, $	ext{FeCl}_3$, $	ext{HNO}_3$ and $	ext{Na}_2	ext{S}_2	ext{O}_3$.
The total number of reagents that can oxidise aqueous iodide to iodine is

Sigma Percentile · Accepted Answer

The beauty of inorganic chemistry lies in the predictable yet fascinating dance of electrons during redox reactions. In this problem, we are tasked with identifying which reagents from a given list possess the oxidative muscle to strip electrons from aqueous iodide ($I^-$) and convert it into elemental iodine ($I_2$).

Let's embark on a journey through these nine reagents, analyzing their chemical behavior and uncovering the classic traps set by the examiners.

### The Setup: A Test Tube of Iodide
Imagine a test tube containing a colorless solution of potassium iodide. The iodide ion ($I^-$) is a relatively mild reducing agent. To oxidize it to iodine ($I_2$), we need a reagent with a standard reduction potential higher than that of the $I_2/I^-$ couple, which is $+0.54	ext{ V}$. The reaction we are looking for is:

$$2I^- ightarrow I_2 + 2e^-$$

Let's test our reagents one by one.

### The Strong Oxidizers: Dichromate and The Permanganate Trap
First, we introduce **acidified potassium dichromate** ($K_2Cr_2O_7$). This is a textbook strong oxidizing agent. The dichromate ion ($Cr_2O_7^{2-}$) eagerly accepts electrons in an acidic medium, reducing itself to the green $Cr^{3+}$ ion while liberating iodine:

$$Cr_2O_7^{2-} + 14H^+ + 6I^- ightarrow 2Cr^{3+} + 3I_2 + 7H_2O$$

So, dichromate is a definitive **yes**.

Next, we encounter **alkaline potassium permanganate** ($KMnO_4$). Here lies a brilliant trap! While acidic $KMnO_4$ would indeed produce $I_2$, the alkaline medium changes the game entirely. In basic conditions, the oxidation is so vigorous that the iodide is over-oxidized past the zero oxidation state of $I_2$, all the way up to the $+5$ oxidation state of the iodate ion ($IO_3^-$):

$$2MnO_4^- + H_2O + I^- ightarrow 2MnO_2 + IO_3^- + 2OH^-$$

Because it produces iodate instead of iodine, alkaline $KMnO_4$ is a **no**.

### The Copper Anomaly
Moving on to **copper sulfate** ($CuSO_4$). At first glance, one might think $Cu^{2+}$ and $I^-$ simply form cupric iodide ($CuI_2$). However, $CuI_2$ is highly unstable. The $Cu^{2+}$ ion oxidizes $I^-$ to $I_2$, and in the process, gets reduced to $Cu^+$, which immediately precipitates as the highly insoluble white cuprous iodide ($Cu_2I_2$). This precipitation drives the reaction forward:

$$2Cu^{2+} + 4I^- ightarrow Cu_2I_2 \downarrow + I_2$$

Thus, $CuSO_4$ is a **yes**.

### The Halogen and Oxygen Family
Now we look at **hydrogen peroxide** ($H_2O_2$), **chlorine** ($Cl_2$), and **ozone** ($O_3$). All three are classic, powerful oxidizing agents. They have high reduction potentials and will readily accept electrons from iodide.

$$H_2O_2 + 2I^- + 2H^+ ightarrow I_2 + 2H_2O$$
$$Cl_2 + 2I^- ightarrow 2Cl^- + I_2$$
$$O_3 + 2I^- + H_2O ightarrow O_2 + I_2 + 2OH^-$$

All three of these non-metallic oxidizers successfully produce iodine. That's three more **yes** votes.

### The Metal Cations and Nitric Acid
What about **ferric chloride** ($FeCl_3$) and **nitric acid** ($HNO_3$)? The ferric ion ($Fe^{3+}$) has a reduction potential of $+0.77	ext{ V}$, which is sufficient to oxidize iodide ($+0.54	ext{ V}$). It reduces to the ferrous ion ($Fe^{2+}$):

$$2Fe^{3+} + 2I^- ightarrow 2Fe^{2+} + I_2$$

Nitric acid, especially when concentrated, is a notorious oxidizing acid. It easily oxidizes iodide to iodine while releasing nitrogen oxides (like $NO_2$ or $NO$):

$$2HNO_3 + 2I^- ightarrow I_2 + 2NO_2 + 2H_2O$$

Both $FeCl_3$ and $HNO_3$ are a **yes**.

### The Thiosulfate Reversal
Finally, we arrive at **sodium thiosulfate** ($Na_2S_2O_3$). If you've ever performed an iodometric titration in the lab, you know exactly what this does. Thiosulfate is a **reducing agent**. We use it to titrate *against* iodine, reducing the $I_2$ back to $I^-$ while the thiosulfate oxidizes to tetrathionate ($S_4O_6^{2-}$):

$$I_2 + 2S_2O_3^{2-} ightarrow 2I^- + S_4O_6^{2-}$$

Since it reduces iodine, it cannot possibly oxidize iodide. Thiosulfate is a definitive **no**.

### Final Tally
Let's count our successful reagents:
1. Acidified $K_2Cr_2O_7$
2. $CuSO_4$
3. $H_2O_2$
4. $Cl_2$
5. $O_3$
6. $FeCl_3$
7. $HNO_3$

We have exactly **7** reagents capable of oxidizing aqueous iodide to iodine. This problem beautifully tests your memory of standard reactions, your awareness of medium-dependent products (the $KMnO_4$ trap), and your practical lab knowledge (the thiosulfate titration).

Enter Numerical Value:

Visualized Solution

$Oxidation of I^{-} to I_{2}$

$Acidified K_{2} Cr_{2} O_{7}$

$Alkaline KMnO_{4}$

$CuSO_{4}$

$H_{2} O_{2}, Cl_{2}, O_{3}$

$FeCl_{3} and HNO_{3}$

$Na_{2} S_{2} O_{3}$

$Final Count$

The Sigma Insight: d-block Elements

The Setup

The Strong Oxidizers

The Copper Anomaly Moving on to copper sulfate ( $C u S O_{4}$ )

The Halogen and Oxygen Family Now we look at hydrogen peroxide ( $H_{2} O_{2}$ ), chlorine ( $C l_{2}$ ), and ozone ( $O_{3}$ )

The Metal Cations and Nitric Acid What about ferric chloride ( $F e C l_{3}$ ) and nitric acid ( $H N O_{3}$ )? The ferric ion ( $F e^{3 +}$ ) has a reduction potential of $+ 0.77 V$ , which is sufficient to oxidize iodide ( $+ 0.54 V$ )

The Thiosulfate Reversal Finally, we arrive at sodium thiosulfate ( $N a_{2} S_{2} O_{3}$ )

Final Tally Let's count our successful reagents: 1

Similar Questions

One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

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In mild alkaline medium, thiosulphate ion is oxidised by $MnO_{4}^{-}$ to "A". The oxidation state of sulphur in "A" is ......... .

The correct order of following $3 d$ -metal oxides, according to their oxidation numbers is (A) $CrO_{3}$ (B) $Fe_{2} O_{3}$ (C) $MnO_{2}$ (D) $V_{2} O_{5}$ (E) $Cu_{2} O$

In the given chemical reaction, colours of the $Fe^{2 +}$ and $Fe^{3 +}$ ions, are respectively $5 Fe^{2 +} + MnO_{4}^{-} + 8 H^{+} ⟶ Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}$

A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation of I− to I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Acidified K2​Cr2​O7​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alkaline KMnO4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CuSO4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​O2​,Cl2​,O3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FeCl3​ and HNO3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Na2​S2​O3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Count