You visited us 1 times! Enjoying our articles? Unlock Full Access!

Nature of Reactant and Rate of Reaction

It requires 4...

Question

It requires $40.05 m l$ of $1 M C e^{4 +}$ to titrate 20 ml of 1 M $S n^{2 +}$ to $S n^{4 +} .$ What is the oxidation state of the cerium in the product ?

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

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

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

Open in App

Solution

The correct option is C 3+
$Number of equivalents of C e^{4 +} = Number of equivalents of S n^{2 +}$
$40.05 \times 1 \times n-factor = 20 \times 1 \times 2$
$\Rightarrow n-factor of C e^{4 +} = 1$
Therefore oxidation state change will be $C e^{4 +} \to C e^{3 +}$
Hence, the correct answer is option (c).

Suggest Corrections

Similar questions

40 mL

1 M C e^{4 +}

20 mL

1 M

S n^{2 +} to S n^{4 +}

40.05

1 M C e^{4 +}

20

1 M S n^{2 +}

S n^{4 +}

40 m l

C e^{4 +}

S n^{2 +}

S n^{4 +}

100 {c m}^{3}

0.1

M

{S n}^{2 +}

{S n}^{4 +}

50

{c m}^{3}

0.40

M

{C e}^{4 +}

100 {c m}^{3}

0.1

M

{S n}^{2 +}

{S n}^{4 +}

50

{c m}^{3}

0.40

M

{C e}^{4 +}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program