$C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to C r^{+ 3} + S O_{4}^{2 -}$

Open in App

Solution

$C r_{2} O_{7}^{2} - + S O_{3}^{2} - ⟶ C r^{+} 3 + S O_{4}^{2} -$
In the above reaction, in $C r_{2} O_{7}^{2} -, C r$ is in $+ 6$ oxidation state which get convert to $+ 3$ oxidation state.
As there are two atoms of $C r$ , so total $e -$ transfer will be $2 * 3 = 6$
In $S O_{3}^{2} -, S$ is in $+ 4$ oxidation state
In $S O_{4}^{2} -, S$ is in $+ 6$ oxidation state
Therefore; total number of $e -$ transfer is $2$ .

Suggest Corrections

Similar questions

C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to S O_{4}^{2 -} + C r^{3 +}

C r_{2} O_{7}^{2 -} (a q) + S O_{3}^{2 -} (a q) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)

H^{+} + C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to C r^{3 +} + S O_{4}^{2 -}

H^{+}

H_{2} O

C r_{2} O_{7}^{2 -} + H^{+} + I^{-} \to C r^{3 +} + I_{2} + H_{2} O

C r_{2} O_{7}^{2 -} + F e_{2}^{+} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O

M n O^{4 -} + S O_{3}^{2 -} + H^{+} \to M n^{2 +} + S O_{4}^{2 -} + H_{2} O

M n O^{4 -} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O

P

P O_{4}^{3 -}

S

S O_{4}^{2 -}

C r

C r_{2} O_{7}^{2 -}

Join BYJU'S Learning Program