A current of 9.65 A is drawn from a Daniel cell for exactly 1 hr. The loss in mass at anode and gain in mass at cathode, respectively are

11.77 g, 11.43 g

To calculate the loss in mass at the anode and gain in mass at the cathode, we need to consider the reactions occurring in the Daniel cell and use Faraday's laws of electrolysis.
The Daniel cell reaction can be represented as follows:
At the anode: Zn(s) → Zn²⁺  +  2e^-
At the cathode: Cu²⁺ + 2e^-  → Cu(s)
Given:
Current = 9.65 A
Time = 1 hr = 3600 s

First, we calculate the quantity of electricity used using the formula:
Quantity of electricity (Coulombs) = Current (Amperes) × Time (seconds)

Quantity of electricity = 9.65 A × 3600 s = 34740 C

Using Faraday's constant:
1 Faraday = 96500 Coulombs

The moles of electrons transferred (n) can be calculated using the equation:
n = Quantity of electricity (Coulombs) / Faraday's constant

n = 34740 C / 96500 C/mol = 0.36 mol

Since the stoichiometry of the reactions shows that 2 moles of electrons correspond to 1 mole of Zn, the moles of Zn involved in the reaction can be calculated as:
Moles of Zn = n / 2 = 0.36 mol / 2 = 0.18 mol

To find the mass loss at the anode, we can multiply the moles of Zn by the molar mass of Zn:
Mass loss at anode = Moles of Zn × Molar mass of Zn

The molar mass of Zn is 65.38 g/mol:
Mass loss at anode = 0.18 mol × 65.38 g/mol = 11.77 g

Since the reaction at the cathode involves the deposition of Cu, the gain in mass at the cathode is equal to the mass loss at the anode:

Gain in mass at cathode = 11.77 g

Therefore, the correct answer is (2) 11.77 g, 11.43 g.