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Gibb's Energy and Nernst Equation

Question

The value of $Δ G (k J m o l^{- 1})$ for the given cell is (take $1 F = 96500 C m o l^{- 1})$ :
-5.7
5.7
11.4
-11.4

A

-11.4

B

5.7

C

11.4

D

-5.7

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Solution

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$M | M^{2 +}$ (saturated solution of a sparingly soluble salt $M X_{2}) | | M^{2 +} 0.001 m o l d m^{- 3} | M$
Reactions of cell
At anode : $M (s) \to M^{2 +} (a q .) + 2 e^{-}$
At cathode : $M^{2 +} (a q .) + 2 e^{-} \to M (s)$
So no of reactions taking part in reaction is 2
Also, as we know
$Δ G = - n F E$
$Δ G = - n F E_{c e l l} = - 2 \times 96500 \times 0.059 = - 113873 k J / m o l$
$= - 11.387 k J / m o l = - 11.4 k J / m o l e$

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