In the reaction between and , is liberated. This is then titrated with . The -factor of in this sequence is:

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Question

In the reaction between $Cu^{2+}$ and $I^-$, $I_2$ is liberated. This $I_2$ is then titrated with $Na_2S_2O_3$. The $n$-factor of $CuSO_4$ in this sequence is:

Prashant Kotian · Accepted Answer

Correct answer: A. Reaction: .
Copper goes from +2 to +1. The change in oxidation state is 1. Thus, the -factor of is 1. [11, 13]

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