QUESTION 9 (Start on a new page.) The simplified diagram below represents the cell used for electroplating ornaments with silver, Ag, P and Q are the two terminals of the battery - NSC Physical Sciences - Question 9 - 2024 - Paper 2

Terminal Q of the battery is negative.

The half-cell reaction at the cathode can be written as: $Ag^{+} + e^{-} \rightarrow Ag$

To calculate the current needed, follow these steps:

1. Calculate moles of silver (Ag) deposited:

   • Molar mass of Ag = 108 g/mol.
   • Moles of Ag = ( \frac{3.25 g}{108 g/mol} = 0.0300 mol )

2. Calculate the total charge required:

   • Each mole of Ag requires 1 mole of electrons, so for 0.0300 mol, we need:
   • 0.0300 mol ( \times 96500 C/mol = 2895 C )

3. Determine time in seconds:

   • 30 minutes = 30 ( \times 60 = 1800 seconds )

4. Calculate current (I):

   • Using the formula ( Q = I \times t ), we rearrange it to find:
   • ( I = \frac{Q}{t} = \frac{2895 C}{1800 s} \approx 1.61 A )

Thus, the current needed is approximately 1.61 A.