ELECTROCHEMISTRY (Calculations involving Faraday's laws of electrolysis continued)

Faraday's laws of electrolysis provide a framework for quantitative calculations involving the amount of electricity and the mass of substances deposited or liberated during electrolysis.

Here are the two main laws and some examples of calculations you can perform:

**Faraday's First Law:**

This law states that the **mass (m)** of a substance deposited or liberated at an electrode is **directly proportional** to the **amount of charge (Q)** passed through the electrolyte. Mathematically, it can be expressed as:

```
m ∝ Q
```

where the symbol "∝" represents proportionality.

**Example 1:**

* You electrolyze copper sulfate (CuSO4) solution for 30 minutes with a current of 2 amps.
* You need to calculate the mass of copper deposited at the cathode.

**Steps:**

1. **Convert time to seconds:** 30 minutes * 60 seconds/minute = 1800 seconds.
2. **Calculate the total charge:** Q = I * t = 2 A * 1800 s = 3600 C.
3. **Look up the electrochemical equivalent (ECE) of copper:** This value is typically found in chemistry handbooks or online resources. Let's assume the ECE of copper is 0.0329 mg/C.
4. **Calculate the mass of copper deposited:** m = Q * ECE = 3600 C * 0.0329 mg/C = 118.44 mg.

**Faraday's Second Law:**

This law states that the **masses of different ions liberated at the electrodes** when the **same amount of electricity (Q)** is passed through **different electrolytes** are **directly proportional to their chemical equivalents (CE)**. Mathematically, it can be expressed as:

```
m1 / m2 = CE1 / CE2
```

where:

* m1 and m2 are the masses of the deposited/liberated substances
* CE1 and CE2 are their respective chemical equivalents

**Example 2:**

* You electrolyze silver nitrate (AgNO3) and copper sulfate (CuSO4) solutions simultaneously with the same amount of charge (Q).
* You need to determine the ratio of the masses of silver and copper deposited.

**Steps:**

1. **Look up the chemical equivalents of silver and copper:** Let's assume the CE of silver (Ag) is 107.87 g/eq and the CE of copper (Cu) is 31.78 g/eq.
2. **Apply the second law equation:** m_Ag / m_Cu = CE_Ag / CE_Cu = 107.87 g/eq / 31.78 g/eq ≈ 3.4.

Therefore, the mass of silver deposited will be approximately 3.4 times the mass of copper deposited under the same conditions.

These are just two examples of calculations involving Faraday's laws. Remember, you'll need additional information like current, time, electrochemical equivalents, and chemical equivalents to perform these calculations effectively.