Question #3d7dc
1 Answer
Explanation:
Start by assigning oxidation numbers to all the atoms that take part in the reaction
#stackrel(color(blue)(+1))("H") stackrel(color(blue)(-1))("Br")_text((aq]) + stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_text(4(aq]) -> stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O")_text(2(g]) + stackrel(color(blue)(0))("Br"_text(2(l])#
Notice that sulfur's oxidation number changes from
Bromine's oxidation number changes from
The two half-equations will look like this
- oxidation half-equation
#2stackrel(color(blue)(-1))("Br"^(-)) -> stackrel(color(blue)(0))"Br"_2 + 2e^(-)#
Two bromine atoms will lose a total of two electrons.
- reduction half-equation
#stackrel(color(blue)(+6))("S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S")O_2#
Notice that the oxygen atoms are not balanced. Since you're in acidic solution, you can use water molecules to balance oxygen atoms and protons,
#stackrel(color(blue)(+6))("S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S")O_2 + 2"H"_2"O"#
Now balance the hydrogen atoms
#4"H"^(+) + stackrel(color(blue)(+6))("S")O_4^(2-) + 2e^(-) -> stackrel(color(blue)(+4))("S")O_2 + 2"H"_2"O"#
In any redox reaction, the number of electrons lost during oxidation must be equal to the number of electrons gained during reduction.
In your case, the numbe of electrons gained is equal to the number of electrons lost, so you don't have to change the half-equations.
So, you have
#{(color(white)(xxxxxxxxxx)2"Br"^(-) -> "Br"_2 + 2e^(-)), (4"H"^(+) + "SO"_4^(2-) + 2e^(-) -> "SO"_2 + 2"H"_2"O") :}#
Add these two half-equations to get
#2"Br"^(-) + 4"H"^(+) + "SO"_4^(2-) + color(red)(cancel(color(black)(2e^(-)))) -> "Br"_2 + color(red)(cancel(color(black)(2e^(-)))) + "SO"_2 + 2"H"_2"O"#
Finally, the balanced chemical equation will be - keep an eye out for those
#2"HBr"_text((aq]) + 2"H"_2"SO"_text(4(aq]) -> "Br"_text(2(l]) + "SO"_text(2(g]) + 2"H"_2"O"_text((l])#