One Faraday of electricity liberates gram atom of copper from copper sulphate, isv

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Solution

Q.
One Faraday of electricity liberates $x \times 10^{- 1}$ gram atom of copper from copper sulphate, $x$ is ____. [2024]

Ans.
(5)

Gram atoms of copper means mols of Cu.

   ${Cu}^{2 +} (aq) + 2 e^{-} \to Cu(s)$

1 F is charge carried by 1 mol $e^{-}$ .

  Thus number of electrons $(n_{e^{-}})$ passed through the solution = 1 mol

  By stoichiometry:

$\frac{n_{Cu}}{1} = \frac{n_{e^{-}}}{2}$

$\frac{n_{Cu}}{1} = \frac{1}{2}$

$n_{Cu} = 0.5 mol = 5 \times 10^{- 1} mol$

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