(a) The cell in which the following reaction occurs:
$2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s)$ has $E_{cell}^{⊝} = 0.236 V$ at 298 K.
Calculate the standard Gibbs energy of the cell reaction. (Given: 2F= 96500 C $m o l^{- 1}$ )
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F= 96500 C $m o l^{- 1}$ )

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Solution

(a) The two half cell reactions for the cell are:
$2 F e^{3 +} (a q) + 2 e^{-} \to 2 F e^{2 +} (Reduction)$
$2 I^{-} \to I_{2} + 2 e^{-} (Oxidation)$
No. of electrons transferred = 2
$Δ G^{⊝} = - n F E_{cell}^{⊝}$
$= - 2 \times 96500 \times 0.236$
$= - 45.55 k J$

(b) $Q = I \times t = 0.5 \times 2 \times 60 \times 60 = 3600 C$
No. of electrons flowing through wire on passing charge of $1 F = 6.023 \times 10^{23}$
No. of electrons flowing through wire on passing charge of 3600 C $= \frac{6.022 \times 10^{23} \times 3600}{96500}$
$= 2.246 \times 10^{22}$

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Similar questions

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

E_{cell}^{⊝} = 0.236 V

m o l^{- 1}

m o l^{- 1}

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

E_{c e l l}^{\circ} = 0.236 V

298 K

1 F = 96, 500 C m o l^{- 1})

2 {Fe}^{3 +} (aq) + 2 I^{-} (aq) \to 2 {Fe}^{2 +} (aq) + I_{2} (aq) E °_{cell} = 0.24 V at 298 K

△ G^{o}

2 F e^{3 +} (aq) + 2 I^{-} (aq) \to 2 F e^{2 +} (aq) + I_{2} (s) has E_{(cell)}^{⊙} = 0.236 V

2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} \to 2 F e_{(a q)}^{2 +} + I_{2 (s)}

E_{c e l l}^{o} = 0.236 V

298 K

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