The standard free energy change in kJ for the reaction- 3Fe2-aq-2Crs-2Cr3-aq-3Fes given EoFe2-Fe-0-44V and ECr3-Cro-0-74V is F-96500C-

The correct option is C 173,700 E^0_cell=E^o_Fe^2+/Fe E_Cr^3+/Cr^o= 0.44 ( 0.74)=0.3 V In the balance reaction, 6 electrons are involved. Δ G= n ...

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Gibb's Energy and Nernst Equation

The standard ...

Question

The standard free energy change (in kJ) for the reaction,

$3 F e^{2 +} (a q) + 2 C r (s) = 2 C r^{3 +} (a q) + 3 F e (s)$

given $E_{F e^{2 +} / F e}^{o} = - 0.44$ V and $E_{C r^{3 +} / C r}^{o} = - 0.74$ V is $(F = 96500 C)$ .

57, 900

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- 173, 700

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173, 700

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298 K

2 C r (s) + 3 {F e}^{2 +} (0.1 M) \to 2 {C r}^{3 +} (0.01 M) + 3 F e (s)

E ({C r}^{3 +} | C r) = - 0.74 V E ({F e}^{2 +} | F e) = - 0.44 V

2 C r_{(s)} + 3 C d_{(a q)}^{2 +} \to 2 C r_{(a q)}^{3 +} + 3 C d_{(s)}

E_{C r^{3 +} / C r}^{o} = - 0.74

E_{C d^{2 +} / C d}^{o} = - 0.40

E_{F e (s) / F e^{2 +} (a q)}^{0} = 0.44 V

E_{F e^{2 +} (a q) / F e^{3 +} (a q)}^{0} = - 0.77 V

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

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

Δ_{r} G_{m}^{0} = -

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

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

0.028 V

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