Redox Reactions: Balancing in Alkaline Medium
Master the dual strategies to balance complex redox processes in an Alkaline (Basic) Medium using the Oxidation Number Method! Abhishek Sengar Sir details structural transitions, electronic equalizations, and how to safely utilize $\text{OH}^-$ and $\text{H}_2\text{O}$ to manage elements and charges.
Video Lecture Broadcast
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In-Depth Lecture Notes & Summary
Understanding Alkaline Medium Demands
In alkaline (basic) media, balancing equations poses a unique challenge because the abundance of free hydroxide ions ($\text{OH}^-$) restricts the use of free protonic hydrogen ($\text{H}^+$). Therefore, visual adjustments must preserve basicity while strictly obeying the Conservation of Mass and Charge.
The Two Approaches for Alkaline Balancing
Abhishek Sengar Sir explains that students can confidently use either of the following two standard pathways:
Method A: Acidic first, then convert
- Balance skeletal atoms, then O with $\text{H}_2\text{O}$ and H with $\text{H}^+$.
- Count the exact number of $\text{H}^+$ present.
- Add an equivalent count of $\text{OH}^-$ to both LHS and RHS.
- Combine LHS $\text{H}^+ + \text{OH}^-$ to form $\text{H}_2\text{O}$.
- Simplify water molecules on opposite sides.
Method B: Direct Alkaline Balancing
- Assign oxidation numbers and equalize overall electron transfers.
- Balance elements other than oxygen and hydrogen.
- For Oxygen deficit: Add $1\text{H}_2\text{O}$ per O deficit to the deficient side.
- For Hydrogen deficit: Add $1\text{H}_2\text{O}$ to the H-deficient side and an equal number of $\text{OH}^-$ to the opposite side simultaneously.
Worked Disproportionation Mechanism
Let's look at chlorine disproportionating in a hot alkaline solution—an extremely frequent board and competitive exam question:
Disproportionation of Chlorine ($\text{Cl}_2 \to \text{Cl}^- + \text{ClO}_3^-$)
Step 1: Write Separate Changes. Here, Chlorine is simultaneously oxidized ($0 \to +5$ in $\text{ClO}_3^-$) and reduced ($0 \to -1$ in $\text{Cl}^-$).
Step 2: Determine Electron shifts.
• Reduction: $\text{Cl}_2^0 \to 2\text{Cl}^-$ (Gain of $2e^-$ per $\text{Cl}_2$ molecule)
• Oxidation: $\text{Cl}_2^0 \to 2\text{ClO}_3^-$ (Loss of $10e^-$ per $\text{Cl}_2$ molecule)
Step 3: Equalize Shifts. Multiply the reduction reactant pair by 5 (total $10e^-$ gain) to balance the oxidation reaction's $10e^-$ loss. This gives:
Step 4: Balance Oxygen and Hydrogen in Alkaline Medium.
• There are 3 Oxygen atoms on RHS and 0 on LHS. To balance, add $6\text{OH}^-$ to LHS and $3\text{H}_2\text{O}$ to RHS. Let's verify:
Verification check: LHS Atoms: $\text{Cl}=6, \text{O}=6, \text{H}=6$. RHS Atoms: $\text{Cl}=6, \text{O}=6, \text{H}=6$. LHS Charge: $-6$. RHS Charge: $5(-1) + (-1) = -6$. Perfectly Balanced!
Redox Solver & Step-Simulator
Select an alkaline reaction template below and progress step-by-step to visualize the balancing mechanics live!
Computing floating states...
Charge: -2
Charge: -1
Alkaline Medium Rules Grid
Review the protocol sequences taught by Abhishek Sir for managing Oxygen/Hydrogen in basic conditions.
Add $\text{OH}^-$ to both sides
Balance H deficit
$$\sum \text{Charge}_{\text{LHS}} = \sum \text{Charge}_{\text{RHS}}$$
Lecture Supplementary Quiz
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Doubt with Basic Mediums?
If you have doubts regarding disproportionations, $\text{OH}^-$ distributions, or basic coefficients, email Abhishek Sir directly!
Email abhishek.sengar@chemca.in →
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