I am honestly lost on this question and you can't proceed without solving the first on, so if anyone would be so inclines to help me out
Aluminum metal reacts with zinc(II) ion in an aqueous solution by the following half-cell reactions:
Al(s) → Al3+(aq) + 3e−
Zn2+(aq) + 2e− → Zn(s)
a. Predict the potential of the cell under standard conditions.
b. Predict whether the reaction will occur spontaneously, or whether a source of electricity will be required for the reaction. Justify your answer.
c. In terms of the metals involved, predict the direction electrons will flow in the reaction.
d. Predict which electrode will lose mass and which will gain mass.
Use the standard reduction tables to find the Eo value of each half reaction. Remember that the Al half rxn is written as an oxidation; therefore, you must change the sign of the Eo reduction you find in the table.
Al ==> Al^3+ + 3e Eo = ?
Zn^2+ + 2e ==> Zn Eo = ?
make the electrons equal (multiply dqn 1 by 2 and eqn 2 by 3) and add to find the final equation.
2Al + 3Zn^2+ ==> you finish.
Now ADD the two Eo values to find Ecell. DO NOT MULTIPLY THE Eo VALUES FOR THE HALF CELL BY THAT MULTIPLIER USED TO BALANCED THE EQUATION. The Eocell is #1 answer.
2. If the Eo cell is + the rxn is spontaneous. If negative it is not spontaneous.
3. The build up of electrons is on the Al electrode (that's where the electrons are coming from) so the direction of flow is from Al to Zn electrodes.
4. So you should see that Al is dissolving and Zn is increasing in mass since Zn^2+ is becoming Zn.
posted by DrBob222