How To Balance Redox Equations In Alkaline And Acidic Solutions | A-Level Chemistry
- Jun 6, 2022
- 3 min read
Updated: 6 days ago
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This video explains the same redox balancing method using a different A-Level Chemistry example.
Many students find balancing redox equations in acidic and alkaline solutions difficult because most examples begin by balancing the equation in acidic conditions.
This step-by-step guide shows how to balance redox equations in both acidic and alkaline solutions using the hydrogen-oxygen fuel cell as an example.
The method is suitable for OCR, AQA and Edexcel A-Level Chemistry.
The hydrogen-oxygen fuel cell can operate in either acidic or alkaline conditions.
In either situation we first balance the redox equations in acid solution.
The product of the reaction is water.
Oxidation Half Equation (Acid Solution)
H₂ -----> H₂O
1. Add H₂O to balance O atoms
H₂O + H₂ -----> H₂O
2. Balance H atoms with H+
H₂O + H₂ -----> H₂O + 2H⁺
3. Balance charge with e⁻
There is 0 charge on LHS and +2 charge on RHS so add 2e to the RHS:
H₂O + H₂ ----> H₂O + 2H⁺ + 2e⁻
4. Cancel H₂O on both sides
The oxidation half equation is now balanced in acid solution: H₂ ----> 2H⁺ + 2e⁻
Now Balance in Alkaline Solution
1. Add the same amount of OH⁻ as H⁺ to both sides of the equation.
H₂ + 2 OH⁻ -----> 2 H⁺ + 2 OH⁻ + 2e⁻
2. 2 OH⁻ and 2 H⁺ on the RHS combine to form 2 H₂O
H₂ + 2 OH⁻ -----> 2H₂O + 2e⁻
The above oxidation half equation is now balanced in alkaline solution!
Reduction Half Equation (Acid Solution)
O₂ ---------> H₂O
1. Balance the oxygen atoms with H₂O
O₂ ---------> 2 H₂O
2. Balance the H atoms with H⁺
O₂ + 4 H+ -----> 2 H₂O
3. Balance the charge using e⁻
O₂ + 4 H⁺ + 4e- --> 2 H₂O
The above reduction half equation is balanced in acid solution!
Now balance in alkaline solution
Start with the reduction equation in acid solution
O₂ + 4 H⁺ + 4e- --> 2 H₂O
1. Add the same amount of OH⁻ as H+ to both sides of the half equation
O₂+4H⁺ +4OH⁻+4e- > 2H₂O + 4OH⁻
2. OH⁻ and H⁺ form H₂O
O₂+4H₂O +4e- > 2H₂O + 4OH⁻
3. Cancel 2H₂O each side
O₂+2H₂O +4e- > 4OH⁻
The reduction half equation is balanced in alkaline solution!
Full Redox Equation
Acid Solution:
Combine the 2 half equations
H₂ ----> 2H⁺ + 2e⁻ (X2)
O₂ + 4H⁺ + 4e- --> 2 H₂O
1. Multiply so electrons balance..
2H₂ ----> 4H⁺ + 4e⁻
O₂ + 4 H⁺ + 4e- --> 2 H₂O
2. Cancel 4e⁻ and 4H+....
2 H₂ + O₂ -----> 2 H₂O (Full redox equation)
Alkaline Solution:
Combine the 2 half equations:
O₂ + 2H₂O + 4e⁻ ----> 4OH⁻
H₂ + 2OH⁻ -----> 2H₂O + 2e⁻ (×2)
1. Multiply so electrons balance..
O₂+2H₂O +4e- ----> 4OH⁻
2H₂ + 4OH⁻ -----> 4H₂O + 4e⁻
2. Cancel 4e⁻ and 4OH⁻ and 2H₂O....
2 H₂ + O₂ -----> 2 H₂O (Full redox equation)
Note: The full redox equation is the same in both acid and alkaline solution.
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