Two voltameters, one containing a solution of silver salt and the...

According to Faraday's first law of electrolysis, the mass of a substance deposited at an electrode is given by:

$m = \dfrac{E I t}{F}$

For silver, the valency is $1$, so its equivalent weight $E_{\text{Ag}}$ is equal to its atomic weight:

$E_{\text{Ag}} = 107.9 \text{ g eq}^{-1}$

The total charge passed through the series circuit is:

$Q = I \times t = 2 \text{ A} \times (1.50 \times 3600) \text{ s} = 10800 \text{ C}$

Substituting the values into the formula:

$m_{\text{Ag}} = \dfrac{107.9 \times 10800}{96500}$

$m_{\text{Ag}} = \dfrac{1165320}{96500} \approx 12.075 \text{ g}$

Rounding off to one decimal place, the mass of silver deposited is $12.1 \text{ g}$. The data regarding the trivalent metal is extraneous.