In a titration H 2 O 2 is oxidized to O 2 by MnO 4- .24 mL of 0.1 M H 2 O 2 of require 16 mL of 0.1 M MnO 4- solution. Hence, MnO 4-changes to

Question

In a titration H 2 O 2 is oxidized to O 2 by MnO 4- .24 mL of 0.1 M H 2 O 2 of require 16 mL of 0.1 M MnO 4- solution. Hence, MnO 4-changes to (A) MnO 42- (B) MnO 2 (C) MnO 42- (D) Mn 2 O 7

Tardigrade Admin · Accepted Answer

H 2 O 2 arrow 2 H ++ O 2+2 e - 0.1 M =0.2 N MnO 4- ⇒ Mn x+ 0.1 M =0.1(7= x ) N N1 V2=N2 V2 0.2 × 24=0.1(7- x ) 16 (7- x )=3 x=4 Thus, change in oxidation number of MnO 4-is 3 . Thus, MnO 4-changes to MnO 2

Q. In a titration $H_{2} O_{2}$ is oxidized to $O_{2}$ by $M n O_{4}^{-} .24 m L$ of $0.1 M H_{2} O_{2}$ of require $16 m L$ of $0.1 M M n O_{4}^{-}$ solution. Hence, $M n O_{4}^{-}$ changes to

$M n O_{4}^{2 -}$

$M n O_{2}$

$M n O_{4}^{2 -}$

$M n_{2} O_{7}$

Q. In a titration H2​O2​ is oxidized to O2​ by MnO4−​.24mL of 0.1MH2​O2​ of require 16mL of 0.1MMnO4−​ solution. Hence, MnO4−​changes to

MnO42−​

MnO2​

MnO42−​

Mn2​O7​

H2​O2​→2H++O2​+2e−0.1M=0.2N MnO4−​⇒Mnx+0.1M=0.1(7=x)N N1​V2​=N2​V2​ 0.2×24=0.1(7−x)16 (7−x)=3 x=4 Thus, change in oxidation number of MnO4−​is 3 . Thus, MnO4−​changes to MnO2​

Q. In a titration $H_{2} O_{2}$ is oxidized to $O_{2}$ by $M n O_{4}^{-} .24 m L$ of $0.1 M H_{2} O_{2}$ of require $16 m L$ of $0.1 M M n O_{4}^{-}$ solution. Hence, $M n O_{4}^{-}$ changes to

$M n O_{4}^{2 -}$

$M n O_{2}$

$M n O_{4}^{2 -}$

$M n_{2} O_{7}$