When connected to a silver voltameter, the potential difference...

The terminal potential difference of the battery is given by $V = E - Ir$.

Substituting the given values, $10 = 12 - I(2)$.

$2I = 2 \Rightarrow I = 1 \text{ A}$.

The time for which the current flows is $t = 30 \text{ min} = 1800 \text{ s}$.

The total charge passed through the voltameter is $Q = It = 1 \times 1800 = 1800 \text{ C}$.

According to Faraday's first law of electrolysis, the mass of silver deposited is $m = ZQ$, where $Z$ is the electrochemical equivalent.

$Z = \dfrac{\text{Atomic weight}}{\text{Valency} \times F} = \dfrac{107.9}{1 \times 96500} \text{ g C}^{-1}$.

$m = \dfrac{107.9}{96500} \times 1800 = \dfrac{107.9 \times 18}{965} \approx 2.01 \text{ g}$.

The amount of silver deposited is approximately $2 \text{ g}$.