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CHEM - Faraday and Mole Calculation Example 1

Chemistry, Faraday, Mole Calculations - IGCSE | AP Chemistry

· chemistry,question analysis,mole calculations,Faraday

When dealing with questions related to mole calculations, following a few simple steps can make it easier to solve them. 🤔

🤓Here is an example:

**Step 1:**

Start by writing out the equation

**Cu ^{2+} + 2e > Cu**

**Step 2:**

Find out the coulombs of electrons flowing

**Charge (coulomb, C) = Current (ampere, A) × Time (second, s)**.

*Remember to convert the time into seconds !

**Therefore, Charge = 0.2 × 7200 = 1440 coulombs **

**Step 3:**

Convert Coulombs into moles of electrons

**Moles = Coulombs / faraday**

**Moles = 1440/96500 = 0.015**

**Step 4:**

Find the moles of the product using the **scale factor**

*From the equation, for every **2 moles of electrons**, there will be** 1 copper.**

**Scale factor = moles of product/moles of electrons = 1/2**

**Thus, there are 0.015 × 1/2 = 0.0075 moles of Cu**

**LAST STEP**👏

Convert moles into mass

**Mass = Moles × **

**The mass of Cu = 0.0075 × 63.5 = 0.48 g**

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