aqueous neutral
aqueous acidic
both aqueous acidic and neutral
both aqueous acidic and faintly alkaline
The reaction of permanganate \( (\text{MnO}_4^-) \) with thiosulphate \( (\text{S}_2\text{O}_3^{2-}) \) depends on the pH of the medium. Permanganate can be reduced to different manganese species depending on the conditions.
In a neutral or weakly alkaline medium, permanganate is reduced to manganese(IV) oxide \( (\text{MnO}_2) \):
\( \text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3e^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)} \)
In this reaction, the oxidation state of manganese changes from +7 in \( \text{MnO}_4^- \) to +4 in \( \text{MnO}_2 \), a change of 3 units.
In an acidic medium, permanganate is reduced to manganese(II) ions \( (\text{Mn}^{2+}) \):
\( \text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5e^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)} \)
Here, the oxidation state of manganese changes from +7 to +2, a change of 5 units.
Since the question specifies a change in oxidation state of 3, the reaction must be occurring in a neutral or weakly alkaline medium.

The correct sequence of reagents for the above conversion of X to Y is :
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]