Răspuns:
a. [tex]\bf\underline{Mn^{+4}}O_2^{-2}+H^{+1}\underline{\underline{Cl^{-1}}}\rightarrow \underline{Mn^{+2}}Cl_2^{-1}+H_2^{+1}O^{-2}+\underline{\underline{Cl_2^0}}[/tex]
[tex]\bf Mn^{+4}\xrightarrow{+2e^-}Mn^{+2}|\cdot2\!\!\!/ \:\:1\:\:\: reducere[/tex]
[tex]\bf 2Cl^{-1}\xrightarrow{-2e^-}Cl_2^{0}|\cdot2\!\!\!/ \:\:1\:\:\: oxidare[/tex]
Din egalarea redox nu avem coeficienți, deci vom egala algebric:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂↑
b. [tex]\bf \underline{Al^0}+\underline{\underline{H_2^{+1}}}S^{+6}O_4^{-2} \rightarrow\underline{Al_2^{+3}}(S^{+6}O_4^{-2})_3+\underline{\underline{H_2^0}}[/tex]
[tex]\bf Al^{0}\xrightarrow{-3e^-}Al^{+3}|\cdot2\:\:\: oxidare[/tex]
[tex]\bf 2H^{+1}\xrightarrow{+2e^-}H_2^0|\cdot3\:\:\:reducere[/tex]
Din egalarea redox avem coeficientul 2 la aluminiu, 3 la acid sulfuric și 3 la hidrogen:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂↑
