Question

The most popular electrochemical cell, Daniel cell was originally developed by the English chemist John F. Daniel. The general assembly of the cell is as given below

An undergraduate student up a Daniel cell using 100 cm³ of 0.100 M CuSO₄ and 0.100 M ZnSO₄ solution respectively. The two compartments are connected by suitable salt bridge.

[Given : $E_{{Cu}^{2 +} / Cu}^{o} = 0 .34 V$ , $E_{{Zn}^{2 +} / Zn}^{o} = - 0 .76 V$ $\frac{2 .303 RT}{F} = 0 .06$ , log 2 = 0.3]

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Linked Question 1

Calculate the emf (in Volt) of the above cell. Assume the temperature of the laboratory to be 25°C and the solutions are dilute enough so that molar concentrations can be used instead of activities of the solutions.

1.04

1.1

1.16

E = 1.1 – $\frac{0.06}{2} \log \frac{0.1}{0.1}$

Linked Question 2

A lab mate of the student asked her for some solid CuCl₂. While she was lifting the bottle from a shelf, the lid of the bottle slipped and some amount of CuCl₂ fell in the CuSO₄ compartment at constant volume. She measured the emf of the cell again and found that it had increased by 9 mV. She used this data to calculate the amount of CuCl₂ that had split in the compartment. Calculate the mass of CuCl₂ in grams that had split into the Daniel cell?
(Molecular mass of CuCl₂ = 135 g)

13.5

1.35

2.7

1.109 = 1.100 – $\frac{0 .06}{2} \log \frac{0 .1}{[{Cu}^{2 +}]}$

[Ci²⁺] = 0.2 M
Δn = 0.2 × 0.1 – 0.1 × 0.1
= 0.01
w = 0.01 × 135 = 1.35 gm