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Question

On the basis of the following $$ E^o $$ values, the strongest oxidizing agent is
$$ [Fe(CN)_6]^{4-} \rightarrow [Fe(CN)_6]^{3-} +e^{-} ; E^o = -0.35 V $$
$$ Fe^{2+} \rightarrow Fe^{3+} +e^- ; E^o = -0.77V $$

A
$$ Fe^{3+} $$
B
$$ [Fe(CN)_6]^{3-} $$
C
$$ [Fe(CN)_6]^{4-} $$
D
$$ Fe^{2+} $$
Solution
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Correct option is A. $$ Fe^{3+} $$

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Similar Questions
Q1
On the basis of the following Eo values, the strongest oxidising agent is:
[Fe(CN)6]4[Fe(CN)6]3++e,Eo=0.35V
Fe2+Fe3++e1;
Fe2+Fe3++e1;Eo=0.77V
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Q2
On the basis of the following E values, the strongest oxidizing agent is:

[Fe(CN)6]4[Fe(CN)6]3+e1,E0=0.35V

Fe2+Fe3++e1,E0=0.77V

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Q3
FerrocyanideFerricyanide ion+e Eo=0.35V
Ferrous ion Ferric ion +e Eo=0.77V
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Q4
If the following half cells have the $${E}^{o}$$ values as
$${ Fe }^{ 3+ }+e\rightarrow { Fe }^{ 2+ };{ E }^{ o }=+0.77V$$ and $${ Fe }^{ 2+ }+2e\rightarrow Fe;{ E }^{ o }=-0.44V$$; the $${E}^{o}$$ of the half cell $${ Fe }^{ 3+ }+e\rightarrow Fe$$ will be
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Q5
Given: $${ Fe }^{ 3+ }+e\rightleftharpoons { Fe }^{ 2+ };{ E }^{ o }=0.77V$$
$${ Fe }^{ 2+ }+2e\rightleftharpoons Fe;{ E }^{ o }=-0.44V$$

What will be the $${E}^{o}$$ value for the following half cell?

$${ Fe }^{ 3+ }+3e\rightleftharpoons { Fe }$$
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