SATHEE: Electrochemistry - Result Question 28

Electrochemistry - Result Question 28

####30. For the cell reaction

[2019]

$2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (a q)$

$0.24 V$ at $298 K$ . The standard Gibbs energy ( $Δ$ , $G^{\circ}$ ) of the cell reaction is:

[Given that Faraday constant $F = 96500 C m o l^{- 1}$ ]

(a) $- 46.32 k J m o l^{- 1}$

(b) $- 23.16 k J m o l^{- 1}$

(d) $23.16 k J m o l^{- 1}$

Show Answer

Solution:

(a) $Δ G = - n F E^{\circ}$

$= - 2 \times 96500 \times 0.24 = - 46320 J / m o l$

$= - 46.32 k J / m o l$

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Electrochemistry - Result Question 28

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