The Eo reduction for Cu-CuS electrode is -x- Find the value of x-Given- Ksp for CuS-8-0 - 10-36- EoCu-Cu2-0-34 V- log 2-0-3- 2-303RT-F-0-06

Anode: nbsp; Cu ⇌ Cu^2++2e^ Cathode:CuS + 2e^ ⇌ Cu+S^2 Cell reaction:CuS nbsp; ⇌ nbsp; Cu^2+ +S^2 Cell is in equilibrium, therefore nbsp;E_cell=0 ∴ ...

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Byju's Answer

Standard XI

Chemistry

Electrode Potential

The Eo reduct...

Question

The $E^{o}$ reduction for $C u / C u S$ electrode is $- x$ . Find the value of $x$ .
(Given: $K_{s p}$ for $C u S = 8.0 \times 10^{- 36}; E_{(C u / C u^{2 +})}^{o} = - 0.34 V; l o g 2 = 0.3; \frac{2.303 R T}{F} = 0.06$ )

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Solution

$A n o d e : C u ⇌ C u^{2 +} + 2 e^{-}$
$C a t h o d e : C u S + 2 e^{-} ⇌ C u + S^{2 -}$
$C e l l r e a c t i o n : C u S ⇌ C u^{2 +} + S^{2 -}$

Cell is in equilibrium, therefore $E_{c e l l} = 0$
$∴ E_{c e l l} = E_{c e l l}^{o} - \frac{0.06}{2} l o g K_{s p}$

$E_{C u / C u S}^{o} - 0.34 = - 1.053$
$E_{C u / C u S}^{o} = - 0.713 V$

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The Eo reduction for Cu/CuS electrode is −x. Find the value of x. (Given: Ksp for CuS=8.0×10−36;Eo(Cu/Cu2+)=−0.34 V;log 2=0.3;2.303RTF=0.06)

Anode:Cu⇌Cu2++2e− Cathode:CuS+2e−⇌Cu+S2− Cell reaction:CuS⇌Cu2++S2− Cell is in equilibrium, therefore Ecell=0 ∴Ecell=Eocell−0.062log Ksp EoCu/CuS−0.34=−1.053 EoCu/CuS=−0.713 V

The $E^{o}$ reduction for $C u / C u S$ electrode is $- x$ . Find the value of $x$ .
(Given: $K_{s p}$ for $C u S = 8.0 \times 10^{- 36}; E_{(C u / C u^{2 +})}^{o} = - 0.34 V; l o g 2 = 0.3; \frac{2.303 R T}{F} = 0.06$ )