state whether oxidation or reduction will occur, determine the number of electrons gained or lost, and write a balanced half-reaction.
a. Sn4+(aq) →Sn2+(aq)
oxidation, 2 electrons lost
b. Na+(aq) →Na(s)
reduction, no electrons lost
c. S(s) →S2-(aq)
reduction, 2 electrons gained
I might be a bit mixed up on some parts (science is not my niche) and if someone could help with the balanceing that would be very appreciated.
state whether oxidation or reduction will occur, determine the number of electrons gained or lost, and write a balanced half-reaction.
a. Sn4+(aq) →Sn2+(aq)
oxidation, 2 electrons lost
b. Na+(aq) →Na(s)
reduction, no electrons lost
c. S(s) →S2-(aq)
reduction, 2 electrons gained
Yes you are a little mixed up. Remember the definitions. Oxidation is the loss of electrons. Reduction is the gain of electrons.Here is a mnemonic to help you remember.
LEO the lion goes GER.
Loss Electrons Oxidation; Gain Electrons Reduction.
In balancing equations remember that you must make sure to balance THREE things.
1. The NUMBER and kind of atoms must be the same on both sides.
2. The CHARGE must balance left and right side.
3. The electron change must balance with the loss or gain of electrons.
Here is your first one with your answer.
a. Sn4+(aq) →Sn2+(aq)
oxidation, 2 electrons lost
Not correct. If you balanced the equation first you may have answered corectly; i.e.,
Sn^+4 + 2e ==> Sn^2+. Note Sn atoms balance, charge balances [+4 +(-2)] = +2 on the left and +2 on the right. Third if Sn is +4 it must gain 2e to become +2. The equation TELLS you the 2e are GAINED (you ADD them to Sn^+4) so Sn^+4 must be reduced. Your next equation is
b. Na+(aq) →Na(s)
reduction, no electrons lost
Balance it like this. Na^+ + e ==> Na(s).
Na balances. The charge on the left is 0 and on the right is zero. Electron change balanced. The equation TELLS you that an electron is added (gained) so that must be reduction
I will leave you with the others. I'll be happy to check your answers.