SATHEE: Electrochemistry - Result Question 41

Electrochemistry - Result Question 41

####43. Given:

[2009]

(i) $C u^{2 +} + 2 e^{-} \to C u, E^{0} = 0.337 V$

(ii) $C u^{2 +} + e^{-} \to C u^{+}, E^{\circ} = 0.153 V$

Electrode potential, $E^{\circ}$ for the reaction, $C u^{+} + e^{-} \to C u$ , will be :

(a) $0.90 V$

(b) $0.30 V$

(d) $0.52 V$

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Solution:

(d) $C u^{2 +} + 2 e^{-} \to C u; Δ G^{0} = - n E^{o} F$

$= - 2 \times F \times 0.337 = - 0.674 F$

$C u^{+} \to C u^{2 +} + e^{-}; Δ G^{0} = - n E^{0} F$

$= - 1 \times F \times - 0.153 = 0.153 F$

On adding eqn (i) & (ii)

$C u^{+} + e^{-} \to C u$ ;

$Δ G^{0} = - 0.521 F = - n E^{\circ} F;$

Here $n = 1 ∴ E^{\circ} = + 0.52 V$

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Electrochemistry - Result Question 41

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