Question

the dichromate ion is reduced by the chloride ion in acidic solution. balance the chloride half-reaction from the skeleton equation.
cl⁻ → cl₂
how many electrons must be transferred in the balanced half-reaction?

Explanation:

Step1: Balance the number of Cl atoms

To balance the Cl atoms, we need 2 moles of \( \text{Cl}^- \) on the left side because there are 2 Cl atoms in \( \text{Cl}_2 \) on the right side. So the equation becomes:
\( 2\text{Cl}^-
ightarrow \text{Cl}_2 \)

Step2: Determine the change in oxidation state

The oxidation state of Cl in \( \text{Cl}^- \) is -1, and in \( \text{Cl}_2 \) it is 0. Each Cl atom loses 1 electron (since it goes from -1 to 0). Since there are 2 Cl atoms (from 2 \( \text{Cl}^- \) ions), the total number of electrons lost is \( 2\times1 = 2 \).

Step3: Balance the charge (optional for electron transfer count, but confirms)

The left side has a charge of \( 2\times(-1) = -2 \), and the right side has a charge of 0. To balance the charge, we need to lose 2 electrons (since losing electrons increases the charge). So the balanced half - reaction is:
\( 2\text{Cl}^-
ightarrow \text{Cl}_2 + 2e^- \)

Answer:

2