Q - 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)
Predict the potential of the cell under standard conditions.
Predict whether the reaction will occur spontaneously, or whether a source of electricity will be required for the reaction. Justify your answer.
In terms of the metals involved, predict the direction electrons will flow in the reaction.
Predict which electrode will lose mass and which will gain mass.
DrBob222 please could you help me with this? Can you explain to me how to answer it?
Can anyone explain how to find potential of the cell?
To predict the cell potential under standard conditions, we'll need to look at the standard reduction potentials of the half-cell reactions. The standard reduction potential (E°) is a measure of the tendency of a species to gain electrons.
The reduction potential for the Al3+(aq) + 3e− → Al(s) half-reaction is +1.66 V (given in tables).
The reduction potential for the Zn2+(aq) + 2e− → Zn(s) half-reaction is -0.76 V (given in tables).
To calculate the cell potential, we subtract the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs):
E°cell = E°cathode - E°anode
E°cell = -0.76 V - (+1.66 V)
E°cell = -0.76 V - 1.66 V
E°cell = -2.42 V
Since the calculated cell potential is negative, the reaction will not occur spontaneously under standard conditions. In order for the reaction to occur, an external source of electricity will be required to provide the necessary energy.
In terms of the metals involved, aluminium (Al) has a higher tendency to lose electrons, and zinc (Zn) has a lower tendency to gain electrons. Therefore, electrons will flow from the Al electrode to the Zn electrode.
In terms of mass changes, during the reaction, the Al electrode will lose mass as it oxidizes to form Al3+ ions, while the Zn electrode will gain mass as it reduces to form Zn metal.
To predict the potential of the cell under standard conditions, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n)log(Q)
Where:
Ecell = cell potential
E°cell = standard cell potential
n = number of electrons transferred in the balanced equation
Q = reaction quotient
In this case, the balanced equation for the cell reaction is:
Al(s) + Zn2+(aq) → Al3+(aq) + Zn(s)
From the given half-cell reactions, we see that Al(s) is being oxidized (losing electrons) and Zn2+(aq) is being reduced (gaining electrons).
The standard cell potential, E°cell, can be calculated by subtracting the reduction potential of the anode (Al3+(aq) + 3e- → Al(s)) from the reduction potential of the cathode (Zn2+(aq) + 2e- → Zn(s)).
Using standard reduction potentials from a reference table, we find:
E°cell = E°cathode - E°anode
E°cell = 0.00 V - (-1.66 V)
E°cell = 1.66 V
The positive value of 1.66 V indicates a spontaneous reaction under standard conditions. Therefore, no external source of electricity is required for the reaction to proceed.
Since Al is being oxidized (losing electrons) at the anode, electrons will flow from Al to Zn.
Regarding the mass changes of the electrodes:
- The anode (Al) will lose mass because it is being oxidized and transformed into Al3+ ions.
- The cathode (Zn) will gain mass because it is being reduced and deposited as solid metallic Zn.
So, the anode (Al electrode) will lose mass, while the cathode (Zn electrode) will gain mass.