Answer:
0.0164 g
Explanation:
Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.
Ag⁺(aq) + 1 e⁻ → Ag(s)
We can establish the following relations.
The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:
[tex]19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g[/tex]