To derive the half-equation for the reduction of dichromate ions (\( \text{Cr}_2\text{O}_7^{2-} \)) to chromium ions (\( \text{Cr}^{3+} \)), we start by considering the elements involved, their oxidation states, and the stoichiometry of the reaction.
Step 1: Identify Oxidation States
In the dichromate ion, \( \text{Cr}_2\text{O}_7^{2-} \):
- Each oxygen is typically assigned an oxidation state of -2.
- Let \( x \) be the oxidation state of chromium. The overall charge states must balance to -2.
The equation for the oxidation states in \( \text{Cr}_2\text{O}_7^{2-} \) can be set up as follows: \[ 2x + 7(-2) = -2 \] Simplifying it: \[ 2x - 14 = -2 \ 2x = 12 \ x = 6 \] So, in \( \text{Cr}_2\text{O}_7^{2-} \), chromium has an oxidation state of +6.
In \( \text{Cr}^{3+} \), chromium has an oxidation state of +3.
Step 2: Determine the Change in Oxidation States
Now, we assess the change in oxidation state:
- Chromium changes from +6 to +3, indicating a reduction process (gain of electrons).
Step 3: Write the Basic Half-Equation
To balance the reduction of chromium, we note that for each chromium atom reduced (from +6 to +3), three electrons are gained: \[ \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} \]
Step 4: Balance the Oxygen Atoms
Each dichromate ion contains 7 oxygen atoms. To balance these oxygen atoms during the reduction reaction, we usually consider adding water (\( \text{H}_2\text{O} \)) and hydroxide ions (\( \text{OH}^- \)), especially in an alkaline medium, or hydrogen ions (\( \text{H}^+ \)) if in an acidic medium.
Assuming the reaction occurs in acidic conditions (common for chromium reduction), we will add \( \text{H}^+ \): \[ \text{Cr}_2\text{O}_7^{2-} + 6e^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
Step 5: Verify Mass and Charge Balance
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Atoms Count:
- Left: 2 Cr, 14 H, 7 O
- Right: 2 Cr, 14 H (from 7 \( \text{H}_2\text{O} \)), 7 O
- Both sides have the same number of chromium, hydrogen, and oxygen atoms.
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Charge Balance:
- Left: \( 2- + 6(-1) + 14(+1) = 2 - 6 + 14 = +10 \)
- Right: \( 2(+3) = +6 \)
Clearly, in terms of charge, the left side is (+10) and the right side is (+6). Since it does not balance, we make modifications: We can introduce the negative charges which means we balance for \(7OH^- \) as we add both sides.
Overall, the reaction in basic medium would yield: \[ \text{Cr}_2\text{O}_7^{2-} + 3\text{H}_2\text{O} + 6e^- \rightarrow 2\text{Cr}^{3+} + 14\text{OH}^- \]
Conclusion
Thus, the complete balanced half-equation for the reduction of dichromate \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \) in an acidic medium is: \[ \text{Cr}_2\text{O}_7^{2-} + 6e^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \] Or in a basic medium: \[ \text{Cr}_2\text{O}_7^{2-} + 3\text{H}_2\text{O} + 6e^- \rightarrow 2\text{Cr}^{3+} + 14\text{OH}^- \]
This preparation accounts for the conservation of mass (atoms) and charge, and it is a crucial step in redox chemistry involving chromium compounds.