Given- Co3- - e- - Co2- - Eo - - 1-81 V Pb4- - 2e- - Pb2- - Eo - - 1-67 V Ce4- - e- - Ce3- - Eo - - 1-61 V Bi3- - 3e- - Bi - Eo - - 0-20 VOxidizing power of the species will increase in the order -

The correct option is D Bi^3+ As the reduction potential (E^o_red) increase, oxidizing power also increases. #160;So, the correct order is : #160; ...

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Byju's Answer

Standard XII

Chemistry

Representation of Cell

Given: Co3+ ...

Question

Given:
$C o^{3 +} + e^{-} \to C o^{2 +}; E^{o} = + 1.81 V$
$P b^{4 +} + 2 e^{-} \to P b^{2 +}; E^{o} = + 1.67 V$
$C e^{4 +} + e^{-} \to C e^{3 +}; E^{o} = + 1.61 V$
$B i^{3 +} + 3 e^{-} \to B i; E^{o} = + 0.20 V$
Oxidizing power of the species will increase in the order :

C e^{4 +} < P b^{4 +} < B i^{3 +} < C o^{3 +}

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C o^{3 +} < P b^{4 +} < C e^{4 +} < B i^{3 +}

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C o^{3 +} < C e^{4 +} < B i^{3 +} < P b^{4 +}

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B i^{3 +} < C e^{4 +} < P b^{4 +} < C o^{3 +}

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Solution

The correct option is D $B i^{3 +} < C e^{4 +} < P b^{4 +} < C o^{3 +}$
As the reduction potential $(E_{r e d}^{o})$ increase, oxidizing power also increases.

So, the correct order is : $B i^{3 +} < C e^{4 +} < P b^{4 +} < C o^{3 +}$ .

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Similar questions

Q. Given standard reduction potentials:

C o^{3 +} + e^{-} \to C o^{2 +}; E^{o} = + 1.81 V

P b^{4 +} + 2 e^{-} \to P b^{2 +}; E^{o} = + 1.67 V

C e^{4 +} + e^{-} \to C e^{3 +}; E^{o} = + 1.61 V

B i^{3 +} + 3 e^{-} \to B i; E^{o} = + 0.20 V

oxidizing power of the species will increase in the order:

Q. If the following half-reactions are combined in a galvanic cell, what will be the standard potential?
(Equation 1)

{C o}^{3 +} (a q) + e^{-} \to {C o}^{2 +} (a q)

;

E^{o} = 1.82 V

(Equation 2)

{N a}^{+} (a q) + e^{-} \to N a (s)

;

E^{o} = - 2.71 V

For the given cell reaction:

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

. The equilibrium constant is:

(Given:

E_{{C e}^{4 +} / C e^{3 +}}^{\circ} = 1.44 V; E_{F e^{2 +} / F e^{3 +}}^{\circ} = 0.68 V

)

Q. The correct code for stability of oxidation states of Sn and Pb is :

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

Q. The standard redox potentials

E^{0}

of the following system are

I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V

I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V

I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V

I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V

The oxidizing power of the various species are related as

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Given:Co3++e−→Co2+;Eo=+1.81VPb4++2e−→Pb2+;Eo=+1.67VCe4++e−→Ce3+;Eo=+1.61VBi3++3e−→Bi;Eo=+0.20VOxidizing power of the species will increase in the order :

The correct option is D Bi3+<Ce4+<Pb4+<Co3+As the reduction potential (Eored) increase, oxidizing power also increases. So, the correct order is : Bi3+<Ce4+<Pb4+<Co3+.

Given:
$C o^{3 +} + e^{-} \to C o^{2 +}; E^{o} = + 1.81 V$
$P b^{4 +} + 2 e^{-} \to P b^{2 +}; E^{o} = + 1.67 V$
$C e^{4 +} + e^{-} \to C e^{3 +}; E^{o} = + 1.61 V$
$B i^{3 +} + 3 e^{-} \to B i; E^{o} = + 0.20 V$
Oxidizing power of the species will increase in the order :

The correct option is D $B i^{3 +} < C e^{4 +} < P b^{4 +} < C o^{3 +}$
As the reduction potential $(E_{r e d}^{o})$ increase, oxidizing power also increases.

So, the correct order is : $B i^{3 +} < C e^{4 +} < P b^{4 +} < C o^{3 +}$ .