The number of equivalents of Na_2S_2O_3 requi | vedclass

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(A) The reaction of $Cu^{2+}$ with iodide ions is given by:$2Cu^{2+} + 4I^- 	o 2CuI + I_2$The liberated $I_2$ is then titrated against sodium thiosul

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(A) The reaction of $Cu^{2+}$ with iodide ions is given by:$2Cu^{2+} + 4I^- 	o 2CuI + I_2$The liberated $I_2$ is then titrated against sodium thiosulfate $(Na_2S_2O_3)$:$I_2 + 2Na_2S_2O_3 	o 2NaI + Na_2S_4O_6$From the stoichiometry,$2$ moles of $Cu^{2+}$ produce $1$ mole of $I_2$,which reacts with $2$ moles of $Na_2S_2O_3$.Thus,$1$ mole of $Cu^{2+}$ reacts with $1$ mole of $Na_2S_2O_3$.Since the $n$-factor for $Cu^{2+}$ in this reaction is $1$ (reduction from $+2$ to $+1$) and the $n$-factor for $Na_2S_2O_3$ is $1$ (oxidation from $+2$ to $+2.5$ average),the equivalent ratio is $1:1$.However,the standard volumetric estimation of $Cu^{2+}$ involves the reaction $2Cu^{2+} + 4I^- 	o 2CuI + I_2$ followed by $I_2 + 2S_2O_3^{2-} 	o 2I^- + S_4O_6^{2-}$.For $1$ equivalent of $Cu^{2+}$,$1$ equivalent of $Na_2S_2O_3$ is required.

The number of equivalents of $Na_2S_2O_3$ required for the volumetric estimation of one equivalent of $Cu^{2+}$ is

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