A current of 9.65 A is drawn from a Daniel cell for exactly 1 hr. The…

The correct answer is option 2. 11.77 g, 11.43 g.

To find the loss in mass at the anode and the gain in mass at the cathode of a Daniel cell, we'll use Faraday's laws of electrolysis.

Daniel Cell Reaction

In a Daniel cell, the reactions at the electrodes are:

Anode Reaction: Zn(s) → Zn²⁺(aq) + 2e⁻

Cathode Reaction: Cu²⁺(aq) + 2e⁻ → Cu(s)

The cell reaction is:

\(\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}\)

Given Data

Current (\(I\)) = 9.65 A

Time (\(t\)) = 1 hour = 3600 seconds

Faraday's constant (\(F\)) = 96500 C/mol (approximately)

Total Charge (Q): 

\( Q = I \times t \)

\( Q = 9.65 \text{ A} \times 3600 \text{ s} \)  

\(Q = 34,740 \text{ C}\)

The number of moles of electrons transferred is given by:

\(\text{Moles of electrons} = \frac{Q}{F} \)

\(\text{Moles of electrons} = \frac{34,740 \text{ C}}{96,500 \text{ C/mol}} \)

\(\text{Moles of electrons} \approx 0.36 \text{ mol}\)

The molar mass of zinc is 65.38 g/mol.

The loss in mass at the anode can be calculated using:

\(\text{Mass loss} = \text{Moles of electrons} \times \text{Molar mass of Zn}\)

Since 1 mole of Zn releases 2 moles of electrons:

\(\text{Moles of Zn} = \frac{0.36 \text{ mol}}{2} \)

\(\text{Moles of Zn} = 0.18 \text{ mol}\)

\(\text{Mass loss} = 0.18 \text{ mol} \times 65.38 \text{ g/mol}\)

\(\text{Mass loss} \approx 11.77 \text{ g}\)

The molar mass of copper is 63.55 g/mol. The gain in mass at the cathode can be calculated similarly:

\(\text{Mass gain} = \text{Moles of electrons} \times \text{Molar mass of Cu}\)

Since 1 mole of Cu²⁺ accepts 2 moles of electrons:

\(\text{Moles of Cu} = \frac{0.36 \text{ mol}}{2}\)

\(\text{Moles of Cu} = 0.18 \text{ mol}\)

\(\text{Mass gain} = 0.18 \text{ mol} \times 63.55 \text{ g/mol}\)

\(\text{Mass gain} \approx 11.43 \text{ g} \)

Summary

Loss in mass at anode (Zinc): 11.77 g

Gain in mass at cathode (Copper): 11.43 g

Thus, the correct answer is: 11.77 g, 11.43 g.