Select the best oxidising and reducing agent.

F e, {F e}^{2 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e O_{4}^{2 -}, {F e}^{3 +}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e O_{4}^{2 -}, {F e}^{2 +}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F e O_{4}^{2 -}, F e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $F e O_{4}^{2 -}, {F e}^{2 +}$
According to the given table, it can be seen that $F e O_{4}^{2 -}$ has the highest reduction potential.

Therefore, it has the highest tendency to reduce and it is the best oxidising agent.
${F e}^{2 +}$ have lowest reduction potential.

Therefore, it will be the best reducing agent.

Suggest Corrections

Similar questions

F e O_{4}^{2 -}, {F e}^{3 +}, {F e}^{2 +}

F e

F e O_{4}^{2 -} E^{o} = + 2.20 V - ------ \to F e^{3 +} E^{o} = 0.77 V - ----- \to F e^{2 +} F e^{2 +} E^{o} = 0.445 V - ------ \to F e^{0}

E_{F e O_{4}^{2 -} / F e^{2 +}}^{o}

F e O_{4}^{2 -}, F e {C l}_{3}, F e {C l}_{2}

F e

F e | F e^{2 +} | | F e^{3 +} | F e

F e | F e^{2 +} | | F e^{3 +}, F e^{2 +} | P t

F e | F e^{3 +} | | F e^{3 +}, F e^{2 +} | P t

E_{F e^{2 +} / F e}^{\circ} = - 0.44 V and E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V

E_{F e^{3 +} / F e}^{\circ}

Join BYJU'S Learning Program