When $K M n O_{4}$ is reduced with oxalic acid in acidic solution, the change in oxidation number of $M n$ is

5.00

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

5.0

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

Open in App

Solution

Reaction involved:
$K M n O_{4} + H_{2} C_{2} O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + C O_{2} + H_{2} O$
Assign the oxidation number to Mn
Let the oxidation number of Mn be 'x'
In $K M n O_{4}; 1 + x + 4 \times (- 2) = 0 \Rightarrow x + 1 - 8 = 0 \Rightarrow x = + 7$
In $M n S O_{4} : x + (- 2) = 0 x = + 2$
In the reaction thus:
$K + 7 M n O_{4} + H_{2} C_{2} O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + + 2 M n S O_{4} + C O_{2} + H_{2} O$
When $K M n O_{4}$ is reduced with oxallic acid in acidic solution, the oxidation umbee of Mn changes from + 7 to +2.
Change in oxidation state = 7-2 = 5

Suggest Corrections

Similar questions

K M n O_{4}

K M n O_{4}

M n

K M n O_{4}

M n

K M n O_{4}

K M n O_{4}

M n

Join BYJU'S Learning Program