Write the balanced ionic equation for the oxidation of iodide ions to iodine by acidified potassium dichromate

Exercise · Q24

Q. Write the balanced ionic equation for the oxidation of iodide ions to iodine by acidified potassium dichromate.

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The reduction half-reaction for dichromate in acidic medium (established earlier in this chapter) is

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

requiring 6 electrons.

Iodide ion is oxidized to iodine according to the half-reaction

2\text{I}^- \longrightarrow \text{I}_2 + 2e^-

a two-electron change. To supply the 6 electrons the dichromate half-reaction requires, this iodide half-reaction must be multiplied by 3:

6\text{I}^- \longrightarrow 3\text{I}_2 + 6e^-

Adding the two half-reactions (the 6 electrons cancel exactly):

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \longrightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}

Checking the balance. Cr: 2 each side. I: 6 each side ( $3\text{I}_2$ has 6 iodine atoms). O: 7 each side (7 in $\text{Cr}_2\text{O}_7^{2-}$ , 7 in $7\text{H}_2\text{O}$ ). H: 14 each side. Charge: left, $(-2) + 14(+1) + 6(-1) = -2+14-6 = +6$ ; right, $2(+3) + 0 + 0 = +6$ -- balanced. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 56/5/11 markMCQ
$Q. While doing qualitative analysis in chemistry lab, Abhishek added yellow coloured potassium chromate solution into a test tube. He was surprised to see the colour of the solution changing immediately to orange. He realised that the test tube was not clean and contained a few drops of some liquid. Which of the following substances will be the most likely liquid to be present in the test tube before adding potassium chromate solution ? (A) Sodium hydrogen carbonate solution (B) Methyl orange solution (C) Sodium hydroxide solution (D) HCl solution$
›Reveal solutionSolution
The colour change from yellow (chromate, $\ce{CrO4^{2-}}$ ) to orange (dichromate, $\ce{Cr2O7^{2-}}$ ) is caused by adding an acid. The only acidic substance among the options is HCl, so the correct answer is (D).

The key here is the chromate–dichromate equilibrium. In aqueous solution, the yellow chromate ion ( $\ce{CrO4^{2-}}$ ) and the orange dichromate ion ( $\ce{Cr2O7^{2-}}$ ) exist in a pH-dependent equilibrium:
$2\ce{CrO4^{2-}} + 2\ce{H+} \rightleftharpoons \ce{Cr2O7^{2-}} + \ce{H2O}$
When you add an acid (increase $\ce{H+}$ concentration), the equilibrium shifts to the right, producing more orange dichromate. When you add a base (decrease $\ce{H+}$ ), it shifts left, giving yellow chromate. So the liquid left in the test tube must have been acidic to cause the immediate colour change.

Let’s check each option:

Sodium hydrogen carbonate solution — This is mildly basic (it hydrolyses to give $\ce{OH-}$ ). It would shift the equilibrium left, keeping the solution yellow, not orange. So not this.

Methyl orange solution — This is an indicator, not an acid or base. It would not change the pH significantly. The colour change from yellow to orange is not due to methyl orange itself (which is orange in acid, yellow in base — but here the chromate colour dominates). So not this.

Sodium hydroxide solution — This is a strong base. It would consume $\ce{H+}$ , shifting the equilibrium left, making the solution stay yellow. So not this. …
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Q.Write the balanced ionic equation for the oxidation of iodide ions to iodine by acidified potassium dichromate.

Concept understanding — Preparation and Properties of Potassium Dichromate

Q.Write the balanced ionic equation for the oxidation of iodide ions to iodine by acidified potassium dichromate.

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