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Chelation

An electroche...

Question

An electrochemical cell consists of two half-cell reactions:
$A g C l (s) + e^{-} \to A g (s) + C l^{-} (a q)$
$C u (s) \to C u^{2 +} (a q) + 2 e^{-}$
The mass of copper (in grams) dissolved on passing $0.5 A$ current for $1 h$ is:
(Given: atomic mass of $C u$ is $63.6$ , $F = 96500 C m o l^{- 1}$ )

0.88

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1.18

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0.29

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0.59

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Solution

The correct option is D $0.59$
Mass of copper is deposited is given as :
$m = z i t$
$[z = E C E = \frac{E}{F}]$
Where, $E = Equivalent weight = \frac{M}{n}$
$\Rightarrow m = \frac{M}{n F} i t$
$\Rightarrow m = \frac{M}{n F} i t [Number of electrons liberated]$
From question, we have
$M = 63.6 g m o l^{- 1}$
$F = 96500 C m o l^{- 1}$
$n = 2$
$i = 0.5 A$
$t = 1 h = 3600 s$
$m = \frac{63.6}{2 \times 96500} \times 0.5 \times 3600 g = 0.59 g$

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Similar questions

C u (s) | C u^{2 +} (a q) | | A g^{+} (a q) | A g (s)

C u (s) | C u^{2 +} (a q) | | A g^{+} (a q) | A g (s)

E^{\circ}

Z n (s) | Z n^{2 +} 1 M (a q .) ∥ C u^{2 +} 1 M (a q .) | C u (s)

1.10 V

25^{\circ} C

Z n (s) + C u^{2 +} (a q .) ⇌ Z n^{2 +} (a q .) + C u (s)

C l^{-}

C u^{2 +}

25^{0} C

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C l^{-}

C u^{2 +}

C l^{-}

\Leftrightarrow

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C u^{2 +} + 2 e^{-} \to C u

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E^{⊖}

C u^{\oplus} + e^{-} \to C u

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