Question
Chromium metal can be plated out from an acidic solution containing $CrO_3$ according to the following equation:
$\text{CrO}_3(\text{aq})+6\text{H}^+(\text{aq})+6\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{Cr(s)}+3\text{H}_2\text{O}$
Calculate:
  1. How many grams of chromium will be plated out by $24,000$ coulombs.
  2. How long will it take to plate out $1.5g$ of chromium by using $12.5A$ current?
(At. mass of Cr = 52).

Answer

  1. 6 × 96500 coulomb deposit Cr = 1 mole = 52g
$\therefore$ 24000 coulomb deposit Cr $=\frac{52\times24000}{6\times96500}\text{g}=2.1554\text{g}$
  1. 52g of Cr is deposited by electricity = 6 × 96500C
$\therefore$ 1.5g require electricity = 1.5C = 16071.9C
$\therefore$ Time for which the current is required to be passed $=\frac{16071.9\text{C}}{12.5\text{A}}=1336\text{s}$

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