This is a reduction-oxidation reaction.
You can find the general technique for balancing redox equations here.
We can use the method of oxidation numbers to balance this equation.
We start with the unbalanced equation:
#"C"_12"H"_22"O"_11 + "H"_2"O" → "C"_2"H"_5"OH" + "CO"_2#
Step 1. Identify the atoms that change oxidation number
Determine the oxidation numbers of every atom in the equation.
#stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+1")("H")_22stackrelcolor(blue)("-2")("O")_11 + color(white)(l)stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O")_2#
#color(white)(stackrelcolor(blue)(0)("C")_12stackrelcolor(blue)("+22")("H")_22stackrelcolor(blue)("-22")("O")_11 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-4")("C")_2stackrelcolor(blue)("+5")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-4")("O")_2)#
We see that the oxidation number of #"C"# in sucrose is reduced to -2 in #"C"_2"H"_5"OH"# and increased to +4 in #"CO"_2#.
This is a disproportionation reaction.
The changes in oxidation number are:
#"C: 0 → -2";color(white)(l) "Change ="color(white)(m) "-2 (oxidation)"#
#"C: 0 → +4"; "Change ="color(white)(l) "+4 (reduction)"#
Step 2. Equalize the changes in oxidation number
We need 2 atoms of #"C"# that become ethanol for every 1 atom of #"C"# that becomes #"CO"_2#.
We must also have a total of 12 #"C"# atoms from the sucrose.
That means we need 1 molecule of sucrose, 4 of ethanol, and 4 of #"CO"_2#.
Step 3. Insert coefficients to get these numbers
#color(red)(1)"C"_12"H"_22"O"_11 + "H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#
Step 4. Balance #"O"#
We have fixed 12 #"O"# atoms on the right and 11 #"O"# atoms on the right, so we need 1 more #"O"# atoms on the left. Put a 1 before #"H"_2"O"#.
#color(red)(1)"C"_12"H"_22"O"_11 + color(blue)(1)"H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO"_2#
Every formula now has a coefficient. The equation should be balanced.
Step 5. Check that all atoms are balanced.
#bb("On the left"color(white)(l) "On the right")#
#color(white)(mm)"12 C"color(white)(mmmml) "12 C"#
#color(white)(mm)"24 H"color(white)(mmmml) "24 H"#
#color(white)(mm)"12 O"color(white)(mmmml) "12 O"#
The balanced equation is
#color(blue)("C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO"_2)#
