- Go for the reaction- Cu2-Fe- Fe2-Cu is-Given- EoCu2-Cu-0-34V- EoFe2-Fe-0-44V-

$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{A g^{2 +} / g}^{\circ} = 0.80 V$ . Will Fe displace Cu from $C u S O_{4}$ solution? Will Ag displace Cu from $C u S O_{4}$ solution?

1. Yes, Yes

2. No, No

3. Yes, No

4. No, Yes

Q. Calculate the equilibrium constant for the reaction:

F e + C u S O_{4} ⟶ F e S O_{4} + C u

25^{o}

[Given :

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

;

E_{(C u / C u^{2 +})}^{0} = - 0.337 V

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ΔGo for the reaction, Cu2++Fe→Fe2++Cu is:(Given: EoCu2+|Cu=+0.34V, EoFe2+|Fe=−0.44V).

The correct option is D 150.5 kJWe know that,E0cell=E0Cu2+/Cu−E0Fe2+/Fe =0.34−(−0.44)V=0.78VNow,ΔG=−nFE0cell =−2×96500×0.78=150540 J or 150.5 kJ

$Δ G^{o}$ for the reaction, $C u^{2 +} + F e \to F e^{2 +} + C u$ is:
(Given: $E_{C u^{2 +} | C u}^{o} = + 0.34$ V, $E_{F e^{2 +} | F e}^{o} = - 0.44$ V).

The correct option is D $150.5 k J$
We know that,

$E_{c e l l}^{0} = E_{C u^{2 +} / C u}^{0} - E_{F e^{2 +} / F e}^{0}$

$= 0.34 - (- 0.44) V = 0.78 V$

Now,

$Δ G = - n F E_{c e l l}^{0}$

$= - 2 \times 96500 \times 0.78 = 150540 J$ or $150.5 k J$