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 = ?
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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.
To solve this question, we need to apply some basic principles of electrochemistry. Let's break it down step by step:
a. Predicting the potential of the cell under standard conditions:
The potential of a cell can be determined by the difference in reduction potentials of the two half-cells. The reduction potential of a half-cell is a measure of its tendency to gain electrons.
In this case, we have the reduction potential of the Al3+(aq)/Al(s) half-cell and the Zn2+(aq)/Zn(s) half-cell. The reduction potential for the Al3+(aq)/Al(s) half-cell is +1.66 V, and for the Zn2+(aq)/Zn(s) half-cell is -0.76 V (values can be found in standard tables).
To calculate the potential of the cell, we subtract the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs):
Ecell = E(cathode) - E(anode)
Ecell = 0.00 V - (-0.76 V) = 0.76 V
So, the potential of the cell under standard conditions is +0.76 V.
b. Predicting whether the reaction will occur spontaneously:
The spontaneity of a reaction is determined by the sign of the cell potential. If the cell potential (Ecell) is positive, the reaction will occur spontaneously. If Ecell is negative, an external source of electricity will be required for the reaction to occur.
In this case, the potential of the cell is +0.76 V (positive). Therefore, the reaction will occur spontaneously.
c. Predicting the direction of electron flow in the reaction:
Since the reduction potential of the Al3+ half-cell (+1.66 V) is greater than the reduction potential of the Zn2+ half-cell (-0.76 V), electrons will flow from the zinc electrode to the aluminum electrode. Electrons flow from the anode to the cathode in an electrochemical cell.
d. Predicting which electrode will lose mass and which will gain mass:
During the reaction, oxidation occurs at the anode (where electrons are lost) and reduction occurs at the cathode (where electrons are gained). The anode is where the metal loses mass, and the cathode is where the metal gains mass.
In this case, the aluminum electrode is the anode, where aluminum metal oxidizes to form Al3+ ions. As a result, the aluminum electrode will lose mass. The zinc electrode is the cathode, where Zn2+ ions gain electrons to form zinc metal. Therefore, the zinc electrode will gain mass.
Overall, the aluminum electrode will lose mass, and the zinc electrode will gain mass.