Observe the following unbalanced equation \(aS_{8}+b~OH^{-}(aq)\rightarrow cS^{2-}(aq)+dS_{2}O_{3}^{2-}(aq)+eH_{2}O(l)\). In the balanced equation, the ratio of c and d is:
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In disproportionation reactions, ensure the atom being disproportionated is balanced in both the reduced and oxidized products before finalizing the electron count.
Step 1: Spot the type of reaction. The same element sulfur goes both down to $S^{2-}$ and up to $S_2O_3^{2-}$. When one element is both oxidised and reduced, it is a disproportionation reaction. Step 2: Write the reduction half. Sulfur drops to the $-2$ state: $S_8+16e^-\rightarrow 8S^{2-}$. Eight sulfur atoms each gain 2 electrons, so 16 electrons. Step 3: Write the oxidation half. In $S_2O_3^{2-}$ the average state of sulfur is $+2$. Building it in base: $S_8+24OH^-\rightarrow 4S_2O_3^{2-}+12H_2O+24e^-$. Step 4: Balance the electrons. Multiply the reduction half by 3 (gives 48 electrons) and the oxidation half by 2 (gives 48 electrons) so they cancel. Step 5: Add the halves. This gives $5S_8+48OH^-\rightarrow 24S^{2-}+8S_2O_3^{2-}+24H_2O$. So $c=24$ and $d=8$. Step 6: Form the ratio. \[ \frac{c}{d}=\frac{24}{8}=\frac{3}{1} \] \[ \boxed{c:d=3:1} \]