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Faraday's First Law

a State Farad...

Question

(a) State Faraday's first law of electrolysis. How much charge in terms of faraday is required for the reduction of $1$ mole of ${C u}^{2 +}$ to $C u$ ?
(b) Calculate emf of the following cell at $298 K$ :
$M g (s) | {M g}^{2 +} (0.1 M) | | {C u}^{2 +} (0.01 M) | C u (s)$
[Given $E_{c e l l}^{0} = + 2.71 V, 1 F = 96500 C {m o l}^{- 1}$ ]

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Solution

(a) Faraday's first law of electrolysis: The amount of chemical reaction and hence the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte (solution or melt).
$C u^{2 +} + 2 e^{-} ⟶ C u$
So, two Faraday of charge is required

(b) $E = E^{o} - \frac{0.0591}{n} l o g K_{c}$

$= 2.71 - \frac{0.0591}{2} l o g \frac{0.1}{0.01}$

$= 2.71 - 0.02955$
$= 2.68$ volts

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Δ_{r} G^{0}

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E_{c e l l}^{0} = 2.71 V, 1 F = 96500 C {m o l}^{- 1}

△_{r} G^{0}

M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)

E_{c e l l}^{0} = + 2.71 V, 1 F = 96500 C m o l^{- 1}

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Z n | Z n^{2 +} (0.01 M) | | C u^{2 +}] (0.1 M) | C u

298 K

298 K

{M g}_{(s)} + {C u}^{2 +} (0.0001 M) \to {M g}^{2 +} (0.001 M) + {C u}_{(s)}

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Δ_{r} G^{0}

M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)

E_{c e l l}^{0} = + 2.71 V, 1 F = 96500 C m o l^{- 1}

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