(D) When $SO_{2}$ is passed through an acidified $K_{2}Cr_{2}O_{7}$ solution,the orange dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green ch

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green $Cr_{2}(SO_{4})_{3}$ is formed

(D) When $SO_{2}$ is passed through an acidified $K_{2}Cr_{2}O_{7}$ solution,the orange dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green chromium$(III)$ ion $(Cr^{3+})$.The balanced chemical equation is:$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$As a result,the solution turns green due to the formation of $Cr_{2}(SO_{4})_{3}$.

Class 11 Chemistry Redox Reactions Redox reaction and Method for balancing Redox reaction

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When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$,the process that takes place is

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A
the solution turns blue
B
the solution is decolourised
C
$SO_{2}$ is reduced
D
green $Cr_{2}(SO_{4})_{3}$ is formed

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Redox Reactions

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Redox reaction and Method for balancing Redox reaction

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When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$,the process that takes place is

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What is the ratio of the stoichiometric coefficient of $N_2H_4$ to $Cl^-$ in the following balanced redox reaction?$ClO_3^- + N_2H_4 \to NO_3^- + Cl^-$

In balancing the half-reaction $S_2O_3^{2-} \to S_{(s)}$,the number of electrons that must be added is:

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Assign $A, B, C, D$ from the given type of reaction.$Na_2CO_3 + PbSO_4 \longrightarrow PbCO_3 \downarrow + Na_2SO_4$

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What is the ratio of the stoichiometric coefficient of $N_2H_4$ to $Cl^-$ in the following balanced redox reaction?
$ClO_3^- + N_2H_4 \to NO_3^- + Cl^-$

Assign $A, B, C, D$ from the given type of reaction.
$Na_2CO_3 + PbSO_4 \longrightarrow PbCO_3 \downarrow + Na_2SO_4$

Find the values of $x$ and $y$ for the following balanced equation:
$5H_2O_2 + xClO_2 + 2OH^- \to xCl^- + yO_2 + 6H_2O$