Explanation:
A disaccharide on hydrolysis gives two molecules of the same or one molecule each of two different monosaccharides since a disaccharide is defined as the sugar formed by the glycosidic linkage of 2 monosaccharides.
Consider, for example, maltose. It hydrolyses to give 2 molecules of glucose.
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$(I)$ Bromine water
$(II)$ Tollen's reagents
$(III)$ Shiff's reagent
$\mathrm{Pt}(s) \mid \mathrm{H}_2(g, 1 \text { bar })\left|\mathrm{H}^{+}(a q, 1 \mathrm{M}) \| \mathrm{M}^{4+}(a q), \mathrm{M}^{2+}(a q)\right| \mathrm{Pt}(s)$
$E_{\text {cell }}=0.092 \mathrm{~V} \text { when } \frac{\left[\mathrm{M}^{2+}(a q)\right]}{\left[\mathrm{M}^{4+}(a q)\right]}=10^x$
Given : $ E_{\mathrm{M}^4 / \mathrm{M}^{2+}}^0=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}$
The value of $x$ is