Standard electrode potentials are Fe2+/Fe (E° = - 0.44) Fe3+/Fe2+ (E° = 0.77) Fe2+, Fe3+ and Fe blocks are kept together, then

Question

Standard electrode potentials are Fe2+/Fe (E° = - 0.44) Fe3+/Fe2+ (E° = 0.77) Fe2+, Fe3+ and Fe blocks are kept together, then (A) Fe3+ increases (B) Fe2+ increases (C) Fe2+/Fe3+ remains unchanged (D) Fe2+ decreases

Tardigrade Admin · Accepted Answer

Positive reduction potential for Fe3+/ Fe2+ couple suggest that Fe3+ would reduce to Fe2+ whereas negative reduction potential of Fe2+/Fe couple suggest that Fe would oxidise to Fe2+.Therefore Fe3+ and Fe decrease whereas Fe2+ increases.

Q. Standard electrode potentials are $F e^{2 +} / F e (E^{\circ} = - 0.44)$
$F e^{3 +} / F e^{2 +} (E^{\circ} = 0.77)$
$F e^{2 +}, F e^{3 +}$ and $F e$ blocks are kept together, then

$F e^{3 +}$ increases

$F e^{2 +}$ increases

$F e^{2 +} / F e^{3 +}$ remains unchanged

$F e^{2 +}$ decreases

Positive reduction potential for $F e^{3 +} / F e^{2 +}$ couple suggest that $F e^{3 +}$ would reduce to $F e^{2 +}$ whereas negative reduction potential of $F e^{2 +} / F e$ couple suggest that Fe would oxidise to $F e^{2 +}$ .Therefore $F e^{3 +}$ and $F e$ decrease whereas $F e^{2 +}$ increases.

Q. Standard electrode potentials are Fe2+/Fe(E∘=−0.44) Fe3+/Fe2+(E∘=0.77) Fe2+,Fe3+ and Fe blocks are kept together, then

Fe3+ increases

Fe2+ increases

Fe2+/Fe3+ remains unchanged

Fe2+ decreases