Value of coefficients a, b and c in the following balanced reaction is :-aCr2O7–2 + bI – + cH+ → xCr+3 + yI2 + zH2O1, 6, 71, 6, 141, 3, 14None of these

To determine the coefficients $a$, $b$, and $c$ for the balanced chemical reaction

$-a \text{Cr}_2\text{O}_7^{2-} + b \text{I}^- + c \text{H}^+ \rightarrow x \text{Cr}^{3+} + y \text{I}_2 + z \text{H}_2\text{O}$

we will follow these steps:

1. Break Down the Problem

We are given a redox reaction involving dichromate ions ($\text{Cr}_2\text{O}_7^{2-}$), iodide ions ($\text{I}^-$), and protons ($\text{H}^+$). We need to assign numerical values to $a$, $b$, and $c$ such that the reaction is balanced in terms of atoms and charges.

2. Relevant Concepts

For balancing redox reactions:

• The oxidation states of elements must be conserved.
• The total charge must equal on both sides of the equation.
• We'll divide the process into half-reactions for oxidation and reduction.

3. Analysis and Detail

Dichromate Reduction Half-Reaction: $\text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O}$

Iodide Oxidation Half-Reaction: $2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^-$

Combining Half-Reactions:

• From the reduction half-reaction, we have 6 electrons.
• From the oxidation half-reaction, we have 2 electrons.

To equalize the electrons transferred, we multiply the iodide oxidation half-reaction by 3: $3 \left( 2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^- \right) \Rightarrow 6 \text{I}^- \rightarrow 3 \text{I}_2 + 6 \text{e}^-$

Now combining both half-reactions, we have: $\text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{I}^- \rightarrow 2 \text{Cr}^{3+} + 3 \text{I}_2 + 7 \text{H}_2\text{O}$

4. Verify and Summarize

From the balanced equation, we can see:

• $a = 1$ (1 $\text{Cr}_2\text{O}_7^{2-}$)
• $b = 6$ (6 $\text{I}^-$)
• $c = 14$ (14 $\text{H}^+$)

Final Answer

The values of coefficients are:

• $a = 1$
• $b = 6$
• $c = 14$