ok so glucose gets oxidized while reducing copper. the solution is hot and alcalic.
Write the formula for the reaction. copper ions look like Cu(OH)2 and Cu+ ions as Cu2O.
My attempt at the solution:
CH2OH(CHOH)4CHO + Cu(OH)2 ---I--> Cu2O + CH2OH(CHOH)4COOH
Reasoning; aldehyd into carboxylic acid. My next step is to remove particles that or not involved in the reaction.
CHO + Cu2+ --I--> Cu+ + COO-
oxidation method gives me Carbon goes from +1 to +3.
copper goes from 2+ to 1+.
so 2 copper ions total. and glucose stays the same
CHO + 2Cu2+ ----I---> 2Cu+ + COO-
Next i neutrulize the charge.
3H+ on the right side which gives me 1 water molecule on the left side. Then i add Hydroxide ions since its a basic solution. So 3 extra on both sides.
CHO + 2Cu2+ +H2O + 3OH- ----I---> 2Cu+ + COO- + 3H+ + 3OH-
CHO + 2Cu2+ + 3OH- -------> 2Cu+ + COO- + 2H2O
From here i think it looks balanced. i add the compounds that were not involved in the reaction and get
CH2OH(CHOH)4CHO + 2CU(OH)2 + ---> Cu2O + CH2OH(CHOH)4COOH + 2H2O
Which I find extremely well balanced.
My text book states
CH2OH(CHOH)4CHO + 2CU2+ + 5OH- ---> Cu2O + CH2OH(CHOH)4COO- + 3H2O
who is right?