For the reaction 2Fe3- aq - 2I- aq - 2Fe2- aq - I2sThe magnitude of the standard molar free energy change-rG0m - -kJ Round off the Nearest Integer-E0Fe2-Fes-0-440 VE0Fe3-Fes-0-036 VE0I2-2I-0-539 VF - 96500 C

Answer: 45 kJ 2Fe^3+(aq) + 2I^–(aq) → 2Fe^+2(aq) + I_2(s) Δ G^0 = nFE^0 Fe^3+(aq) + e^ → Fe^2+(aq)....(1); Δ G^0_1 = 1F E^0 Fe^2+(aq) + 2e^ → Fe(s)...... ...

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Byju's Answer

Standard XII

Chemistry

Chemical Reactivity

For the react...

Question

For the reaction $2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s)$
The magnitude of the standard molar free energy change,
$Δ_{r} G_{m}^{0} = -$ ______kJ (Round off the Nearest Integer).
$E_{F e^{2 +} / F e (s)}^{0} = - 0.440 V E_{F e^{3 +} / F e (s)}^{0} = - 0.036 V E_{I_{2} / 2 I^{-}}^{0} = 0.539 V F = 96500 C$

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Solution

Answer: 45 kJ

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

$Δ G^{0} = - n F E^{0}$

$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q) . . . . (1); Δ G_{1}^{0} = - 1 F E^{0}$
$F e^{2 +} (a q) + 2 e^{-} \to F e (s) . . . . . . (2); Δ G_{2}^{0} = + 2 F \times 0.44$

Adding equation (1) and (2), we get,
$F e^{3 +} (a q) + 3 e^{-} \to F e (s) . . . . . . . (3); Δ G_{3}^{0} = - 3 F E^{0}$
Thus,
$Δ G_{3}^{0} = Δ G_{1}^{0} + Δ G_{2}^{0}$
$- 3 F \times - 0.036 = - 1 F E^{0} + 2 F \times 0.44$

$- E_{F e^{+ 3} / F e^{+ 2}}^{0} = 3 \times (0.036) - 2 \times (0.44)$
$E_{F e^{+ 3} / F e^{+ 2}}^{0} = - 3 \times (0.036) + 2 \times (0.44)$

$= - 0.108 + 0.88$

$= 0.772 V$

$E^{0}$ of the given reaction is
$E_{c e l l}^{o} = E_{F e^{+ 3} / F e^{+ 2}}^{o} + E_{I^{-} / I_{2}}^{o}$

$= 0.772 - 0.539 = 0.233 V$

$Δ G ° = - n F E_{c e l l}^{0}$

$= - 2 \times 96500 \times 0.233$

$= - 44969 J = - 44.9 k J ≃ - 45 k J$
Thus, answer is 45

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Similar questions

2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)

2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)

In the following reaction, which element is oxidized?

Q. The standard potential for the reaction,

{A g}^{+} (a q .) + {F e}^{2 +} (a q .) ⟶ {F e}^{3 +} (a q .) + A g (s)

0.028 V

. What is the standard free energy change for this reaction?

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

298 K

Calculate the standard Gibbs energy of the cell reaction. (Given :

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

Q. For the cell reaction

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

The standard Gibbs energy

△ G^{o}

of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]

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For the reaction 2Fe3+(aq)+2I–(aq)→2Fe2+(aq)+I2(s) The magnitude of the standard molar free energy change, ΔrG0m=–______kJ (Round off the Nearest Integer). E0Fe2+/Fe(s)=−0.440 VE0Fe3+/Fe(s)=−0.036 VE0I2/2I−=0.539 VF=96500 C