Answer to Question #281674 in Inorganic Chemistry for kioo

Question

An aqueous solution of a platinum salt is electrolyzed at a current of 2.498 A for 3.00 h. As a result, 9.09 g of metallic Pt are formed at the cathode. Following this half reaction: Ptn+ + ne- → Pt, calculate the charge (represented by “n+” on the half reaction) on the Pt ions in this solution.

Answer 1

Quantity of electricity $Q=current (I)×time(t)$

$Q=2.498×3×3600\\=26978.4C$

1 Faraday$=96485C$

Number of Faradays$=\frac{26978.4}{96485}=0.279612$

Molar mass of Platinum$=195.084g$

$n$ Faradays discharges $195.084g \>of \>Pt^{n+} \>$

But $0.279612 \>Faradays\> discharges\> 9.09g$

$\therefore\>\frac{n}{0.279612}=\frac{195.084}{9.09}$

$n=6.00086$

Charge $(n+)=+6$