engineering chemistry MCQ #2779

The EMF of the cell: Zn / Zn²⁺ (0.1 M) // Cu²⁺ (0.01 M) / Cu, given E⁰_Zn²⁺/Zn = -0.76 V and E⁰_Cu²⁺/Cu = 0.34 V, is

engineering chemistry MCQ #2779

Question 1
Q1. The EMF of the cell: Zn / Zn²⁺ (0.1 M) // Cu²⁺ (0.01 M) / Cu, given E⁰_Zn²⁺/Zn = -0.76 V and E⁰_Cu²⁺/Cu = 0.34 V, is
- A) 1.13 V
- B) 1.07 V
- C) 1.10 V
- D) 1.03 V
Answer: 1.07 V
Explanation: Using E_cell = E⁰_cell - 0.0591 / n log([Zn²⁺] / [Cu²⁺]), where E⁰_cell = E⁰_Cu²⁺/Cu - E⁰_Zn²⁺/Zn = 1.10 V.