When Cu^{2+} ion is treated with KI,a white p | vedclass

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(B) When $Cu^{2+}$ ions react with $KI$,$CuI_2$ is initially formed,which is unstable and decomposes to form a white precipitate of $Cu_2I_2$ and iodi

Vedclass · Accepted Answer

$X = Cu_2I_2, Y = Na_2S_4O_6$

Vedclass · Answer

(B) When $Cu^{2+}$ ions react with $KI$,$CuI_2$ is initially formed,which is unstable and decomposes to form a white precipitate of $Cu_2I_2$ and iodine $(I_2)$: $2Cu^{2+} + 4I^-$ $\rightarrow 2CuI_2$ $\rightarrow Cu_2I_2 \downarrow + I_2$The liberated $I_2$ dissolves in excess $KI$ to form $KI_3$ (triiodide ion),which gives a brown color to the solution: $I_2 + I^- \rightarrow I_3^-$This solution is then titrated against sodium thiosulphate $(Na_2S_2O_3)$,where $I_2$ is reduced to $I^-$ and thiosulphate is oxidized to tetrathionate $(Na_2S_4O_6)$: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$Thus,$X$ is $Cu_2I_2$ and $Y$ is $Na_2S_4O_6$.

When $Cu^{2+}$ ion is treated with $KI$,a white precipitate,$X$ appears in solution. The solution is titrated with sodium thiosulphate,the compound $Y$ is formed. $X$ and $Y$ respectively are

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