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Application of Electrolysis

Construct gal...

Question

Construct galvanic cells from the following pairs of half cells and calculate their emf at $25^{o} C .$

$C d^{2 +} (a q, 1 M) ∣ C d (s); E_{C d^{2 +} / C d}^{0} = - 0.40 V$

$H g_{2} C l_{2} (s) ∣ H g (l) ∣ C l^{-} (a q, 0.1 M); E_{C l^{-} / H g_{2} C l_{2} / H g}^{0} = 0.28 V$

E_{c e l l} = - 0.01 V

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E_{c e l l} = - 0.739 V

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E_{c e l l} = 0.01 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E_{c e l l} = 0.739 V

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Solution

From the standard reduction potential values we can see $H g_{2} C l_{2} (s)$ will act as cathode in the galvanic cell. Applying nernst eq. for individual half cells.
For the half - cell reaction,
$C d^{2 +} (a q) + 2 e^{-} ⇌ C d (s)$
By Nernst equation,
$E_{C d^{2 +} / C d} = E_{C d^{2 +} / C d}^{0} - \frac{0.0591}{2} l o g \frac{1}{[C d^{2 +}]}$
$E_{C d^{2 +} / C d} = - 0.40 - \frac{0.0591}{2} l o g \frac{1}{1} = - 0.40 V$
And, for the half - cell reaction,
$H g_{2} C l_{2} (s) + 2 e^{-} ⇌ 2 H g (l) + 2 C l^{-} (a q)$
$E_{C l^{-} / H g_{2} C l_{2} / H g} = E_{C l^{-} / H g_{2} C l_{2} / H g}^{0} - \frac{0.0591}{2} l o g {[C I^{-}]}^{2}$
$E_{C l^{-} / H g_{2} C l_{2} / H g} = 0.28 - \frac{0.0591}{2} \times 2 l o g 0.1$
$E_{C l^{-} / H g_{2} C l_{2} / H g} = 0.28 - \frac{0.0591}{2} \times 2 l o g 10^{-}$
$E_{C l^{-} / H g_{2} C l_{2} / H g} = 0.28 + \frac{0.0591}{2} \times 2$
$E_{C l^{-} / H g_{2} C l_{2} / H g} = 0.28 + 0.0591 = 0.339 V$
The cell may be represented as
$C d (s) ∣ C d^{2 +} (a q) ∣∣ C l^{-} (a q) ∣ H g_{2} C l_{2} (s) ∣ H g (l)$
$E_{c e l l} = reduction potential of RHS electrode - reduction potential of LHS electrode$
$E_{c e l l} = 0.339 - (- 0.40) = 0.739 V$

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