2. 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. (4 points)
b. Predict whether the reaction will occur spontaneously, or whether a source of electricity will be required for the reaction. Justify your answer. (2 points)
c. Predict the direction electrons will flow in the reaction. (2 points)
d. Predict which electrode will lose mass and which will gain mass. (2 points)
a. To predict the potential of the cell under standard conditions, we can use the Nernst equation:
Ecell = E°cell - (0.0592 V/n) * log(Q)
Where Ecell is the cell potential, E°cell is the standard cell potential, n is the number of moles of electrons transferred in the balanced equation, and Q is the reaction quotient.
In this case, the balanced equation for the cell reaction is:
2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s)
From the given half-cell reactions, we know that the standard reduction potential of the Al3+(aq) + 3e− → Al(s) reaction is -1.66 V, and the standard reduction potential of the Zn2+(aq) + 2e− → Zn(s) reaction is -0.76 V.
To calculate the standard cell potential, we need to subtract the reduction potential of the oxidation half-reaction (Al3+(aq) + 3e− → Al(s)) from the reduction potential of the reduction half-reaction (Zn2+(aq) + 2e− → Zn(s)):
E°cell = E°reduction - E°oxidation
E°cell = -0.76 V - (-1.66 V)
E°cell = 0.90 V
Therefore, the predicted potential of the cell under standard conditions is 0.90 V.
b. To determine if the reaction will occur spontaneously or if a source of electricity will be required, we can compare the standard cell potential (0.90 V) to 0 V. If the standard cell potential is positive, as it is in this case, it means the reaction will occur spontaneously without an external power source.
Therefore, the reaction will occur spontaneously.
c. Electrons flow from the anode (the electrode where oxidation occurs) to the cathode (the electrode where reduction occurs).
In the given half-cell reactions, Al(s) → Al3+(aq) + 3e− is the oxidation (anode) half-reaction, and Zn2+(aq) + 2e− → Zn(s) is the reduction (cathode) half-reaction.
Therefore, electrons will flow from the aluminum electrode (anode) to the zinc electrode (cathode).
d. The electrode at which oxidation occurs (anode) will lose mass. In this case, the aluminum electrode will lose mass as it is being oxidized to form aluminum ions.
The electrode at which reduction occurs (cathode) will gain mass. In this case, the zinc electrode will gain mass as it undergoes reduction to form solid zinc.
To answer these questions, we need to consider the standard reduction potentials of the half-cell reactions and use the Nernst equation to calculate the cell potential. Here's how you can solve each part:
a. Predict the potential of the cell under standard conditions:
The standard cell potential (E°cell) can be calculated by subtracting the standard reduction potential of the reduction half-reaction from the standard reduction potential of the oxidation half-reaction.
The standard reduction potential for the reaction Al3+(aq) + 3e− → Al(s) is -1.66 V (given), and for the reaction Zn2+(aq) + 2e− → Zn(s) is -0.76 V (also given).
To obtain the overall reaction, you need to reverse the reduction of zinc half-reaction and multiply it by a factor of 2:
2Zn(s) → 2Zn2+(aq) + 4e-
Now, you can calculate the standard cell potential:
E°cell = E°(Al3+(aq)/Al(s)) - E°(Zn2+(aq)/Zn(s))
= (-1.66 V) - (-0.76 V)
= -0.90 V
Therefore, the predicted potential of the cell under standard conditions is -0.90 V.
b. Predict whether the reaction will occur spontaneously or require a source of electricity:
A reaction will occur spontaneously if the cell potential is positive (E°cell > 0) and non-spontaneous if the cell potential is negative (E°cell < 0).
In this case, the calculated standard cell potential is -0.90 V, which is negative. Therefore, the reaction will not occur spontaneously. It will require a source of electricity to proceed.
c. Predict the direction electrons will flow in the reaction:
Electrons always flow from the anode (oxidation half-reaction) to the cathode (reduction half-reaction). In this case, the reduction half-reaction is the reduction of aluminum ions (Al3+(aq) + 3e- → Al(s)). Thus, electrons will flow from the anode to the cathode.
d. Predict which electrode will lose mass and which will gain mass:
In the oxidation half-reaction, aluminum (Al) is converted into Al3+ ions, losing electrons. This means that the aluminum electrode will lose mass as it oxidizes.
In the reduction half-reaction, zinc ions (Zn2+(aq)) are reduced to zinc (Zn) with the gain of electrons. Therefore, the zinc electrode will gain mass as it is reduced.
In summary, the aluminum electrode will lose mass, and the zinc electrode will gain mass in the reaction.