Answer
0.054 mol
Work Step by Step
Given half reaction is $Ag^{+}$ (aq) + e- → Ag(s)
So, we can conclude that 1 mol of electrons can be used to electroplate 1 mole of Ag.
1 mol of Ag= 107.8g
So to electroplate 5.8g of Ag, number of moles electrons required is:
= 5.8 g x( 1 mole of electrons/ 107.8 g )
= 0.054 mol