Give that, EFe2+/Fe° =-0.44V, EFe2+/Fe3+° =-0.77V. if Fe2+,Fe3+ and Fe solid are kept together, then

Question

Give that, EFe2+/Fe° =-0.44V, EFe2+/Fe3+° =-0.77V. if Fe2+,Fe3+ and Fe solid are kept together, then (A)  Fe3+ decreases (B)  Fe3+ increases (C)  Fe2+ decreases (D)  Fe2+,Fe3+ remain unchanged

Tardigrade Admin · Accepted Answer

Cell reaction, 2Fe3++Fe→ 3Fe2+ ECell° =0.77-(-0.44)=1.21 textV ⇒ Fe3+ and Fe will reduce.

Q. Give that, $E_{F e^{2 +} / F e}^{\circ} = - 0.44 V,$ $E_{F e^{2 +} / F e^{3 +}}^{\circ} = - 0.77 V .$ if $F e^{2 +}, F e^{3 +}$ and Fe solid are kept together, then

$F e^{3 +}$ decreases

$F e^{3 +}$ increases

$F e^{2 +}$ decreases

$F e^{2 +}, F e^{3 +}$ remain unchanged

Cell reaction, $2 F e^{3 +} + F e \to 3 F e^{2 +}$ $E_{C e ll}^{\circ} = 0.77 - (- 0.44) = 1.21 V$ $\Rightarrow$ $F e^{3 +}$ and Fe will reduce.

Q. Give that, EFe2+/Fe∘​=−0.44V, EFe2+/Fe3+∘​=−0.77V. if Fe2+,Fe3+ and Fe solid are kept together, then

Fe3+ decreases

Fe3+ increases

Fe2+ decreases

Fe2+,Fe3+ remain unchanged