The oxidation state of chromium in the final product formed by the reaction between KI and acidified

Q: Why do we need an acidic medium for this reaction?

In neutral or basic media, dichromate ($Cr_2O_7^{2-}$) converts into chromate ($CrO_4^{2-}$), which is a much weaker oxidizing agent. The $H^+$ ions are essential to bond with the oxygen atoms from the dichromate ion to form water, driving the reduction forward.

Q: How do I know that $Cr^{3+}$ is the final product?

This is a standard factual property of the dichromate ion in acidic conditions. It consistently reduces to the stable $Cr^{3+}$ state when reacting with common reducing agents like $I^-$, $Fe^{2+}$, or $H_2S$.

Q: What happens to the potassium ions ($K^+$)?

The potassium ions are spectator ions. They do not participate in the electron transfer process and remain unchanged in their $+1$ oxidation state throughout the reaction.

Q: Why is the oxidation state of oxygen taken as $-2$?

In most of its compounds (except peroxides and superoxides), oxygen is highly electronegative and pulls two electrons towards itself, resulting in a standard oxidation state of $-2$.

Q: Can this reaction happen in a basic medium?

No, the chemistry changes significantly. In a basic medium, adding $OH^-$ shifts the equilibrium to form the yellow chromate ion ($CrO_4^{2-}$), which does not readily oxidize iodide in the same manner.

Question

The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is

Options:

(A) + 3

(B) + 2

(C) + 6

(D) + 4

Sigma Percentile · Accepted Answer

The chemistry of transition metals is often accompanied by brilliant displays of color, and the reaction between potassium dichromate and potassium iodide is a perfect example. This problem tests your fundamental understanding of redox reactions, specifically the behavior of strong oxidizing agents in acidic media.

### Analyzing the Setup

When we mix potassium dichromate ($K_2Cr_2O_7$) with potassium iodide ($KI$) in the presence of an acid (like dilute sulfuric acid), a classic redox reaction is initiated.

**Potassium dichromate** is a well-known, powerful oxidizing agent. In an acidic environment, it is eager to accept electrons. On the other hand, the **iodide ion** ($I^-$) from potassium iodide acts as a reducing agent, ready to give up its extra electron.

### The Master Equation

To understand what happens to the chromium atoms, we need to look at the reduction half-reaction. The dichromate ion ($Cr_2O_7^{2-}$) reacts with hydrogen ions ($H^+$) from the acid and electrons ($e^-$) provided by the iodide ions:

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$$

Simultaneously, the iodide ions are oxidized to elemental iodine:

$$6I^- \longrightarrow 3I_2 + 6e^-$$

### Tracking the Oxidation States

Let's trace the journey of the chromium atom. Initially, it is part of the dichromate ion. We can calculate its oxidation state by setting up a simple algebraic equation. Knowing that oxygen typically has an oxidation state of $-2$, and the overall charge of the ion is $-2$:

$$2x + 7(-2) = -2$$
$$2x - 14 = -2$$
$$2x = +12 \implies x = +6$$

So, chromium starts in the **$+6$ oxidation state**.

### Final Calculation

As the reaction proceeds, the dichromate ion is reduced, meaning it gains electrons. The product of this reduction is the chromium(III) ion, $Cr^{3+}$.

For any monatomic ion, its oxidation state is simply equal to its net charge. Therefore, the oxidation state of chromium in the final product is **$+3$**.

This chemical transformation is not just a theoretical exercise; it is visually striking. The initial dichromate solution is a vibrant **orange**, but as the $Cr^{3+}$ ions are formed, the solution turns a distinct **green**. This color change is a hallmark of dichromate reduction and is frequently used in laboratory tests to detect the presence of reducing agents!

Animated Solution for Chemistry - Redox Reactions: The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is

Select Answer:

Visualized Solution

\text{The Reactants}

\text{Redox Principle}

\text{Reduction Half-Reaction}

\text{Initial Oxidation State}

\text{Final Oxidation State}

\text{Visual Confirmation}

The Sigma Insight: Oxidation and Reduction

Analyzing the Setup

The Master Equation

Tracking the Oxidation States

Final Calculation

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