Answer:
85.2 grams.
Explanation:
Look up the relative atomic mass data on a modern periodic table.
The Faraday's Constant [tex]F[/tex] gives the size of the charge (in Coulombs) on one mole of electrons. Also, keep in mind that [tex]\rm 1 \; Ampere = 1\; Coulomb/ second[/tex].
Charge supplied to the iodine:
[tex]\begin{aligned}Q &= I \cdot t = \rm 1.50 \; C \cdot s^{-1} \times (12.0 \times 3600) \; s\\ &= \rm 64800\; C\end{aligned}[/tex]
How many moles of electrons does that much charge correspond to?
[tex]\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} = \rm \frac{64800\; C}{96500\; C\cdot mol^{-1}}= 0.67150\; mol\end{aligned}[/tex].
Consider the reduction half-equation for iodine:
[tex]\rm I_2 + 2 \; e^{-} \to 2 \; I^{-}[/tex].
Divide both sides by 2 to make sure that the coefficient in front of [tex]\rm e^{-}[/tex] is equal to 1.
[tex]\displaystyle \rm \frac{1}{2}\; I_2 + e^{-} \to I^{-}[/tex].
The ratio between the coefficient in front of [tex]\rm e^{-}[/tex] and that of [tex]\rm I^{-}[/tex] is equal to 1. In other words, while gaseous iodine is in excess, each mole of electrons will produce one mole of iodine ions.
[tex]\rm 0.67150\; mol[/tex] of electrons will produce [tex]\rm 0.67150\; mol[/tex] of iodine ions.
The mass of one mole of iodine ions is approximately the same as that of one mole of iodine atoms.
[tex]M(\mathrm{I^{-}}) \approx M(\mathrm{I})= \rm 126.904\; g\cdot mol^{-1}[/tex].
[tex]m(\mathrm{I}^{-}) = n \cdot M \approx \rm 85.2\; g[/tex].
