In the reaction X^- + XO_3^- + H^+ longrighta | vedclass

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

Question

(B) The given reaction is a comproportionation reaction where $X$ is oxidized from $-1$ to $0$ and reduced from $+5$ to $0$.Step $1$: Write the half-r

Vedclass · Accepted Answer

$5 : 1$

Vedclass · Answer

(B) The given reaction is a comproportionation reaction where $X$ is oxidized from $-1$ to $0$ and reduced from $+5$ to $0$.Step $1$: Write the half-reactions:Oxidation: $X^- \longrightarrow X_2 + e^-$Reduction: $XO_3^- + 6H^+ + 5e^- \longrightarrow X_2 + 3H_2O$Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $5$:$5X^- \longrightarrow 2.5X_2 + 5e^-$Step $3$: Add the two half-reactions:$5X^- + XO_3^- + 6H^+ \longrightarrow 3X_2 + 3H_2O$From the balanced equation,the molar ratio of $X^-$ to $XO_3^-$ is $5 : 1$.

In the reaction $X^- + XO_3^- + H^+ \longrightarrow X_2 + H_2O$,the molar ratio in which $X^-$ and $XO_3^-$ react is:

Similar Questions

The equivalent weight of $Na_2S_2O_3$ (Gram molecular weight $= M$) in the given reaction is $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.

In the reaction $2[Fe(CN)_6]^{3-} + H_2O_2 + 2OH^{-} \rightarrow 2[Fe(CN)_6]^{4-} + 2H_2O + O_2$,what is the equivalent weight of the reductant? $(Fe = 56, C = 12, N = 14, O = 16, H = 1)$

Aqueous solutions of hydrogen sulphide and sulphur dioxide when mixed together,yield

$KMnO_4$ oxidises oxalic acid in acidic medium. The number of $CO_2$ molecules produced per mole of $KMnO_4$ is

$H_2O_2$ reduces $MnO_4^-$ ion to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module