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Standard X

Chemistry

Electrolysis and Electrolytes

Estimate the ...

Question

Estimate the standard reduction potential for the copper/copper sulphide electrode.
For $C u S, K_{s p} = 8.5 \times 10^{- 36}, E_{C u | C u^{2 +}}^{⊖} = - 0.34 V$ .

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Solution

$K_{s p} (C u S) = 8.5 \times 10^{- 36}$
We know,
Anode : $C u + S^{-} ⇌ C u S + {2 e}^{-}$
Cathode : ${C u}^{2 +} + {2 e}^{-} ⇌ C u - ------------------ -$
over all : ${C u}^{2 +} + S^{-} ⇌ C u S$
$E_{c e l l} = 0 = E^{0} - \frac{0.0591}{n} l o g \frac{[C u S]}{({C u}^{2 +}) (S^{-})}$
$E^{0} = \frac{0.0591}{2} l o g \frac{1}{K_{s p}}$
$[- E_{C u / C u S} + E_{{C u}^{2 +}} / C u] = - 0.0591 2 l o g K_{s p}$
$0.34 - E_{C u / C u S} = \frac{- 0.0591}{2} l o g (8.5 \times 10^{- 36})$
$0.34 - E_{C u / C u S} = 1.0363$
$E_{C u / C u S} = - 0.697$

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Estimate the standard reduction potential for the copper/copper sulphide electrode. For CuS,Ksp=8.5×10−36,E⊖Cu|Cu2+=−0.34 V.

Estimate the standard reduction potential for the copper/copper sulphide electrode.
For $C u S, K_{s p} = 8.5 \times 10^{- 36}, E_{C u | C u^{2 +}}^{⊖} = - 0.34 V$ .