Given: (i) Cu 2++2 e - arrow Cu , E o =0.337 V (i) Cu 2++ e - arrow Cu +, E o -0.153 V Electrode potential E ° for the reaction, Cu ++ e - arrow Cu, will be :

Question

Given: (i) Cu 2++2 e - arrow Cu , E o =0.337 V (i) Cu 2++ e - arrow Cu +, E o -0.153 V Electrode potential E ° for the reaction, Cu ++ e - arrow Cu, will be : (A) 0.52 V (B) 0.90 V (C) 0.30 V (D) 0.38 V

Tardigrade Admin · Accepted Answer

The option marked is incorrect.
The correct option is

B

i.e

0.52 V

Solution:

Δ G^{\circ} = - n FE^{\circ}

So for the reaction,

C u^{2 +} + 2 e^{-} \to C u

Δ G^{\circ} = 2 \times F \times 0.337 \dots\dots\dots . (1)

for the reaction,

C u^{+} \to C u^{2 +} + e^{-}

Δ G^{\circ} = 1 \times F \times 0.153 \dots\dots\dots . (2)

So adding both the equations we get,

C u^{+} + e^{-} \to C u; Δ G^{\circ} = - 0.521 F

Δ G^{\circ} = - n FE

Δ E^{\circ} = 0.52 V

Q. Given: (i) Cu2++2e−→Cu,Eo=0.337V (i) Cu2++e−→Cu+,Eo−0.153V Electrode potential E∘ for the reaction, Cu++e−→Cu, will be :

0.52 V

0.90 V

0.30 V

0.38 V

Q. Given:
(i) $C u^{2 +} + 2 e^{-} \to C u, E^{o} = 0.337 V$
(i) $C u^{2 +} + e^{-} \to C u^{+}, E^{o} - 0.153 V$
Electrode potential $E^{\circ}$ for the reaction,
$C u^{+} + e^{-} \to C u$ , will be :