Question 9 of 29intermediate🔧 ApplyNumerical2 marks

Consider the reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O₇²⁻?

Correct Answer

The quantity of electricity needed is 578922 C (or approximately 579000 C).

Exercise: Intext Questions | Q: 2.12 | (Page Page 24)
For More Understanding

Explanation

From the given reduction reaction, 6 moles of electrons are required to reduce 1 mole of Cr₂O₇²⁻. Using Faraday's constant (96487 C mol⁻¹), the total charge is calculated as 6×964876 \times 96487 = 578922 C.

Solution Steps

  1. Step 1: From the reaction Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, we observe that 6 moles of electrons are required to reduce 1 mol of Cr₂O₇²⁻.

  2. Step 2: From the context, charge on one mole of electrons (1 Faraday) = 96487 C mol⁻¹ (approximately 96500 C mol⁻¹).

  3. Step 3: Quantity of electricity = Number of moles of electrons × Faraday = 6×964876 \times 96487 C = 578922 C (or approximately 6×965006 \times 96500 = 579000 C).

PreviousNext