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_{\left(Cu\right)^{2 +} / Cu}^{o} = 0 .34 \textrm{ } V, E_{\left(Zn\right)^{2 +} / Zn}^{o} = - 0 .76 \textrm{ } V \frac{2 .303 \textrm{ } RT}{F} = 0 .06, log 2 = 0.3]

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]

Linked Question 1

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)

1.35

13.5

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

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Linked Question 2

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.

1.1

1.04

1.16

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

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Linked Question 3

After making this calculation, the student discarded the solutions from both the compartments and washed the cell thoroughly. She refilled the compartments with 100 cm³ of the original CuSO₄ and ZnSO₄ solutions. A current of 0.965 A was drawn from the cell for a period of 20 min.What will be the cell potential after 20 min.?

1.082 V

1.118 V

1.091 V

1.109 V

ne = $\frac{0.965 \times 20 \times 60}{96500}$ = 0.012

E = 1.1 – $\frac{0 .06}{2} \log \frac{0 .01 + \frac{n_{e}}{2}}{0 .01 - \frac{n_{e}}{2}}$

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