Consider the ions $F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}$ . Strongest reducing ion is ?

F e^{2 +}

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M n^{2 +}

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C r^{2 +}

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M n^{3 +}

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Solution

The correct option is D $F e^{2 +}$
$F e^{+ 2}$ will remove one electron and will get electronic configuration $d^{5}$ that is more stable.
hence $F e^{+ 2}$ reacts as reducing agent.

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Similar questions

Q. Consider the ions

F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}

. Strongest oxidizing ion is ?

Q. For

M^{2 +} / M

and

M^{3 +} / M^{2 +}

systems the

E^{⊖}

values for some metals are as follows:

$C r^{2 +} / C r$	$- 0.9 V$ $	$C r^{3} / C r^{2 +}$	$- 0.4 V$
$M n^{2 +} / M n$	$- 1.2 V$	$M n^{3 +} / M n^{2 +}$	$+ 1.5 V$
$F e^{2 +} / F e$	$- 0.4 V$	$F e^{3 +} / F e^{2 +}$	$+ 0.8 V$

Use this data to comment upon:
(i) the stability of

F e^{3 +}

in acid solution as compared to that of

C r^{3 +}

M n^{3 +}

and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

Q. Give reasons.

E^{o}

value for

M n^{3 +} / M n^{2 +}

couple is much more positive than that for

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

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Consider the ions Fe2+,Fe3+,Cr2+,Cr3+,Mn2+,Mn3+. Strongest reducing ion is ?

The correct option is D Fe2+Fe+2 will remove one electron and will get electronic configuration d5 that is more stable.hence Fe+2 reacts as reducing agent.

Consider the ions $F e^{2 +}, F e^{3 +}, C r^{2 +}, C r^{3 +}, M n^{2 +}, M n^{3 +}$ . Strongest reducing ion is ?

The correct option is D $F e^{2 +}$
$F e^{+ 2}$ will remove one electron and will get electronic configuration $d^{5}$ that is more stable.
hence $F e^{+ 2}$ reacts as reducing agent.