Balance these equations for disproportionation reactions. (a) $\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{MnO}_{4}^{-}$ (basic solution) (b) $\mathrm{P}_{4}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+\mathrm{PH}_{3}(\mathrm{g})$ (basic solution) (c) $\mathrm{S}_{8}(\mathrm{s}) \longrightarrow \mathrm{S}^{2-}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ (basic solution) (d) $\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{AsO}_{4}^{3-}+\mathrm{SO}_{4}^{2-}$

In reaction (a), balance the Mn by adding a 2 in front of MnO4- on the right side. This gives: \(MnO_{4}^{2-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Next, balance the oxygens by adding 2H2O to the left side: \(2 MnO_{4}^{2-} + 2H2O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Then, balance the hydrogens by adding 4H+ to the right side: \(2 MnO_{4}^{2-} + 2H_{2}O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H^{+}\). Since the solution is in basic medium, neutralize H+ ions by adding 4OH- ions on both sides: \(2 MnO_{4}^{2-} + 2H_{2}O + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H_{2}O\). Simplify by cancelling 2H2O from both sides to get the balanced reaction: \(2 MnO_{4}^{2-} + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 2H_{2}O\).

Should apply similar steps as in equation (a), starting with balancing the P by adding a 4 in front of H2PO2- and 1 in front of PH3. Continue with balancing oxygen and hydrogen, and neutralizing H+ with OH-. Remember to simplify if necessary.

The process is similar as in previous steps - balance S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- accordingly. Always simplify when possible.

Similar process as before - balance As and S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- respectively. Remember to simplify where it's necessary.