315426
An acid solution of 0.2 mol of \({\mathrm{\mathrm{KReO}_{4}}}\) was reduced with Zn and then titrated with 1.6 Eq of acidic \({\mathrm{\mathrm{KMnO}_{4}}}\) solution. For the reoxidation of all the rhenium Re to the perrhenate ion \({\mathrm{\left(\mathrm{ReO}_{4}{ }^{\ominus}\right)}}\). Assuming that rhenium was the only element reduced, what is the oxidate state to which rhenium was reduced by Zn ?
1 10
2 \({ - 1}\)
3 \(\frac{1}{5}\)
4 1.6
Explanation:
Eq of \({\mathrm{\mathrm{KReO}_{4} \equiv \mathrm{Eq}}}\) of \({\mathrm{\mathrm{MnO}_{4}{ }^{\ominus}}}\) \({0.2 \times {\rm{n}} = 1.6{\mkern 1mu} }\) \({{\rm{n}} = \frac{{1.6}}{{0.2}} = 8}\) Hence, there is an \({\mathrm{8 e^{-}}}\)reduction. \({8{{\rm{e}}^ - } + \mathop {{\rm{ReO}}_4^ \ominus }\limits_{{\rm{x}} = 7}^{ + 7} \to \mathop {{\rm{R}}{{\rm{e}}^{ - 1}}}\limits_{{\rm{x}} = - 1} }\) Oxidation state of \({\mathrm{\mathrm{Re}=-1}}\)