H^{+} + Cr_2O_7^{2-} + SO_3^{2-} to Cr^{3+} + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The unbalanced equation is: $H^{+} + Cr_2O_7^{2-} + SO_3^{2-} 	o Cr^{3+} + SO_4^{2-} + H_2O$. Step $1$: Identify oxidation and reduction half-rea

Vedclass · Accepted Answer

$8, 3$

Vedclass · Answer

(A) The unbalanced equation is: $H^{+} + Cr_2O_7^{2-} + SO_3^{2-} 	o Cr^{3+} + SO_4^{2-} + H_2O$. Step $1$: Identify oxidation and reduction half-reactions. Reduction: $Cr_2O_7^{2-} 	o Cr^{3+}$. Oxidation: $SO_3^{2-} 	o SO_4^{2-}$. Step $2$: Balance atoms and charges. Reduction half: $Cr_2O_7^{2-} + 14H^{+} + 6e^{-} 	o 2Cr^{3+} + 7H_2O$. Oxidation half: $SO_3^{2-} + H_2O 	o SO_4^{2-} + 2H^{+} + 2e^{-}$. Step $3$: Multiply oxidation half by $3$ to balance electrons: $3SO_3^{2-} + 3H_2O 	o 3SO_4^{2-} + 6H^{+} + 6e^{-}$. Step $4$: Add the two half-reactions: $Cr_2O_7^{2-} + 14H^{+} + 3SO_3^{2-} + 3H_2O 	o 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 6H^{+}$. Step $5$: Simplify: $Cr_2O_7^{2-} + 8H^{+} + 3SO_3^{2-} 	o 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$. The coefficient of $H^{+}$ is $8$ and the coefficient of $SO_4^{2-}$ is $3$.

$H^{+} + Cr_2O_7^{2-} + SO_3^{2-} \to Cr^{3+} + SO_4^{2-} + H_2O$. In the balanced reaction,the coefficients of $H^{+}$ and $SO_4^{2-}$ will be respectively:

Similar Questions

The oxidation state of chromium in the final product formed in the reaction between $KI$ and acidified $K_{2}Cr_{2}O_{7}$ solution is:

When $H_2S$ is passed through acidified $KMnO_4,$ we get

In the redox reaction
$x \ MnO + y \ PbO_2 + z \ HNO_3 \to a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$
Find the values of $x, y, z$.

Number of moles of $H^{+}$ ions required by $1 \ mole$ of $MnO_4^{-}$ to oxidise oxalate ion to $CO_2$ is . . . . . .

When sulphur dioxide is passed in an acidified $K_{2}Cr_{2}O_{7}$ solution,the oxidation state of sulphur is changed from

Vedclass Test Series

Exam Paper Generator

Online Exam Module