Q10 / 90 · JEE Main 2019 (09 Jan Shift 2) · Chemistry

If the standard electrode potential for a cell is $2 V$ at $300 K,$ the equilibrium constant $(K)$ for the reaction.

$Z n (s) + C u^{2 +} (a q) ⇌ Z n^{2 +} (a q) + C u (s)$

at $300 K$ is approximately:

$(R = 8 J K^{- 1} m o l^{- 1}, F = 96000 C m o l^{- 1})$

A
$e^{- 160}$
B
$e^{- 80}$
Correct answer
C
$e^{160}$
D
$e^{320}$

Solution

$∆ G = ∆ G^{o} + R T l n Q$
At equilibrium $∆ G = 0, Q = k$
$0 = ∆ G^{o} + R T l n K$
$∆ G^{o} = - R T l n K (∴ ∆ G^{o} = - n F E^{o})$
So $- n F E^{o} = - R T l n k$
$' n'$ factor for the above cell $= 2$
So $- 2 \times 96000 \times 2 = - 8 \times 300 l n k$
$l n k = 160$
$k = e^{160}$

Hence, the correct answer is option: 'C'.

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Solution

Relevant Questions

If the standard electrode potential for a cell is $2 V$ at $300 K,$ the equilibrium constant $(K)$ for the reaction.

$Z n (s) + C u^{2 +} (a q) ⇌ Z n^{2 +} (a q) + C u (s)$

at $300 K$ is approximately:

$(R = 8 J K^{- 1} m o l^{- 1}, F = 96000 C m o l^{- 1})$