Question 5 of 29intermediate🔧 ApplyNumerical3 marks

Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.160 M) + 2Ag(s). Given that E°cell = 1.05 V

Correct Answer

The emf of the cell is 0.91 V.

Exercise: Intext Questions | Q: 2.5 | (Page Page 11)
For More Understanding

Explanation

This question tests the application of the Nernst equation to calculate the cell potential under non-standard conditions. The textbook provides the formula E(cell) = E°(cell) - (RT/2F) ln [Ni²⁺]/[Ag⁺]², which at 298 K simplifies to E(cell) = E°(cell) - (0.059/2) log [Ni²⁺]/[Ag⁺]². The concentration of Ag⁺ is squared because 2 moles of Ag⁺ are involved in the reaction.

Solution Steps

  1. Step 1: Write the Nernst equation for the given cell reaction. E(cell) = E°(cell) - (0.059/2) log [Ni²⁺]/[Ag⁺]²

  2. Step 2: Substitute the given values. E°(cell) = 1.05 V [Ni²⁺] = 0.160 M [Ag⁺] = 0.002 M

  3. Step 3: Calculate the concentration ratio. [Ni²⁺]/[Ag⁺]² = 0.160/(0.002)² = 0.160/0.000004 = 40000

  4. Step 4: Calculate log of the ratio. log(40000) = log(4×104 \times 10⁴) = 4.602

  5. Step 5: Apply the Nernst equation. E(cell) = 1.05 - (0.059/2) × 4.602 E(cell) = 1.05 - 0.0295×4.6020.0295 \times 4.602 E(cell) = 1.05 - 0.136 E(cell) = 0.914 V ≈\approx 0.91 V

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