# How to Balance Redox Equations in Alkaline (and Acid) Solutions?

The Fuel Cell of Hydrogen and Oxygen can take place in Acid or Alkaline Solution.

In either situation we first balance the redox equations in acid solution. The product 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 on RHS so add 2e to the RHS

H₂O + H₂ ----> H₂O + 2H⁺ + **2e⁻**

4. Cancel H₂O

The oxidation half equation is 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. OH⁻ and H⁺ combine to form H₂O

**H₂ + 2 OH⁻ -----> 2H₂O + 2e⁻**

The above oxidation half equation is 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**t

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⁻

6. 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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