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Standard XII

Chemistry

Redox Reactions

10.0 g pure ...

Question

$10.0$ g pure $V_{2} O_{5}$ is brought in solution state by dissolving it in acid medium. Now, the solution is reduced to $V^{2 +}$ by quantitative mass $Z n$ metal. If $I_{2}$ is now added to this solution, how much moles of $I_{2}$ will be reduced if $V^{2 +}$ is oxidized to $V^{4 +}$ ions?
Given that atomic mass of $V = 51$ .

220

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147

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211

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None of these

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Solution

The correct option is A $220$
The half reactions are as follows:
$6 e + (V^{5 +}) ⟶ 2 V^{2 +}$
$V^{2 +} ⟶ V^{4 +} + 2 e$
$2 e + I_{2}^{0} ⟶ 2 I^{-}$
Let Meq. of $V^{2 +} (V . f = 2)$ be $a$ .
Meq. of $I_{2} =$ Meq. of $V^{2 +} (V . f = 2) = a$ $=$ Meq. of $V^{2 +} (V . f = 3) = \frac{3 a}{2}$ $=$ Meq. of $V^{5 +} = \frac{3 a}{2}$
or, $\frac{w}{E_{V_{2} O_{5}}} \times 1000 = \frac{3 a}{2}$
$a = \frac{2 \times 10 \times 1000 \times 6}{3 \times 182}$ $= 219.78$ $[E_{V_{2} O_{5}} = \frac{182}{6}]$
Hence, moles of $I_{2}$ that will be reduced if $V^{2 +}$ is oxidized to $V^{4 +}$ ions is $219.78 \approx 220$ mol.

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Similar questions

Q. If

10

g of

V_{2} O_{5}

is dissolved in acid and is reduced to

V^{2 +}

by zinc metal, how many moles of

I_{2}

could be reduced by the resulting solution if it is further oxidized to

V O^{2 +}

ions? (Atomic mass of

V

51

g/mol)

Q. If 10 g of

V_{2} O_{5}

is dissolved in acid and is reduced to

V^{2 +}

by zinc metal, how many moles

I_{2}

could be reduced by the resulting solution if it is further oxidised to

V O^{2 +}

ions?

[Assume no change in state of

Z n^{2 +}

ions]

(V = 51, O = 16, I = 127)

Q. If

10

g of

V_{2} O_{5}

is dissolved in acid and is reduced to

V^{2 +}

by zinc metal, how many moles of

I_{2}

could be reduced by the resulting solution if it is further oxidised to

V O^{2 +}

ions?

10.0

g sample of

C u_{2} O

is dissolved in

d i l . H_{2} S O_{4}

where it undergoes disproportionation quantitatively. The solution is filtered off and

8.3

g pure

K I

crystals are added to clear filtrate in order to precipitate

C u I

with the liberation of

I_{2}

. The solution is again filtered and boiled till all the

I_{2}

is removed. Now excess of an oxidising agent is added to the filtrate which liberates

I_{2}

again. The liberated

I_{2}

now requires

10

mL of

1.0 M

N a_{2} S_{2} O_{3}

solution. The percentage by mass of

C u_{2} O

in the sample is :

Q. A

3.0

g sample of

C u_{2} O

is dissolved in

d i l . H_{2} S O_{4}

where it undergoes disproportionation quantitatively. The solution is filtered off and

8.3

g pure

K I

crystals are added to clear filtrate in order to precipitate

C u I

with evolution of

I_{2}

. The solution is again filtered and boiled till all the

I_{2}

is expelled. Now excess of an oxidizing agent is added to filtrate, which liberates

I_{2}

again. The liberated

I_{2}

this time requires

10

mL of

1.0 N

N a_{2} S_{2} O_{3}

solution. Calculate

%

by mass of

C u_{2} O

in sample.

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10.0 g pure V2O5 is brought in solution state by dissolving it in acid medium. Now, the solution is reduced to V2+ by quantitative mass Zn metal. If I2 is now added to this solution, how much moles of I2 will be reduced if V2+ is oxidized to V4+ ions? Given that atomic mass of V=51.