What makes this an oxidation-reaction? 3Ag2S+2Al(s) -> Al2S3+6 Ag(s)?
Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reason.
What is oxidized in the reaction? What is reduced?
In this simple electrochemical cell, what functions as the anode? What is the cathode?
Is this a galvanic cell or electrolytic cell?
What would the overall potential for this cell be?
What makes this an oxidation-reaction? 3Ag2S+2Al(s) -> Al2S3+6 Ag(s)?
Loss and gain of electrons
Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reason.
oxidation half cell is
Al ==> Al^3+ + 3e
Reduction half cell is
Ag^+ + e ==> Ag
I have omitted the spectator ions You can add those if you wish.
What is oxidized in the reaction? What is reduced?
answered above.
In this simple electrochemical cell, what functions as the anode? What is the cathode?
The electrode losing electrons is the anode; the electrode gaining electrons is the cathode.
Is this a galvanic cell or electrolytic cell?
This is a galvanic cell since the cell voltage is positive
What would the overall potential for this cell be?
I will leave this for you
To determine if a reaction is an oxidation-reduction (redox) reaction, we need to identify the species that is being oxidized and the one that is being reduced. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.
In the given reaction:
3Ag2S + 2Al(s) -> Al2S3 + 6Ag(s)
Let's identify the half-reactions:
Oxidation half-reaction: 2Al(s) -> Al3+(aq) + 3e^-
Reduction half-reaction: 3Ag2S + 6e^- -> Al2S3 + 6Ag(s)
In the oxidation half-reaction, aluminum (Al) is oxidized, as it loses electrons and is converted into Al3+ ions. Therefore, the oxidation reaction is Al(s) -> Al3+(aq) + 3e^-.
In the reduction half-reaction, the compound 3Ag2S gains electrons and is converted into Al2S3 and Ag(s). Thus, the reduction reaction is 3Ag2S + 6e^- -> Al2S3 + 6Ag(s). Here, Ag2S is reduced to Ag(s).
In this reaction:
- Aluminum (Al) is oxidized.
- Silver sulfide (Ag2S) is reduced.
In an electrochemical cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Therefore, in this reaction, aluminum (Al) functions as the anode, while silver sulfide (Ag2S) functions as the cathode.
This reaction represents a galvanic cell because it occurs spontaneously and produces an electric current.
To calculate the overall potential (cell potential) for this cell, we need to find the difference between the reduction potentials of the cathode and anode half-reactions.
To determine if a reaction is an oxidation-reduction (redox) reaction, we need to examine the change in oxidation numbers for the elements involved. In this reaction, we have silver sulfide (Ag2S) reacting with aluminum (Al) to form aluminum sulfide (Al2S3) and silver (Ag).
To identify the oxidation and reduction reactions, we can write the half-reactions for each species involved:
Oxidation half-reaction: Ag2S -> 2 Ag (reduction in oxidation number of silver from +1 to 0)
Reduction half-reaction: 2 Al -> Al2S3 (increase in oxidation number of aluminum from 0 to +3)
In this reaction, silver (Ag) is being reduced, and aluminum (Al) is being oxidized. Therefore, the reduction reaction is the reduction of silver, and the oxidation reaction is the oxidation of aluminum.
Regarding the functions of the anode and cathode in the simple electrochemical cell, the anode is the electrode where oxidation takes place, and the cathode is where reduction occurs. In this case, aluminum (Al) is oxidized, so it functions as the anode, while silver (Ag) is reduced and acts as the cathode.
Next, we need to determine whether this is a galvanic cell or an electrolytic cell. A galvanic cell, also known as a voltaic cell, generates electricity from a spontaneous redox reaction, while an electrolytic cell requires an external power source to drive a non-spontaneous redox reaction.
Since the reaction given is not part of an external circuit and is spontaneous, it is a galvanic cell (voltaic cell).
Finally, the overall potential for a cell can be determined by calculating the difference in standard reduction potentials between the two half-reactions. We can then use the Nernst equation to account for the concentrations and temperatures specific to the given cell. Without the concentrations and temperatures, we cannot determine the exact value for the overall potential in this case.