Determine the emf for the cell noted below. (For Co3+(aq) + e- → Co2+(aq), E0 = 1.80 V, and for Na+(aq) + e- → Na(s), E0 = -2.71 V.)
Na(s) | Na+(aq) || Co3+(aq), Co2+(aq) | Co(s)
What is being oxidized? What is being reduced? What has the more positive reduction potential? Is this a voltaic cell? What is that absolute difference between the reduction potentials? What is the emf?
To determine the emf for the given cell, we need to identify what is being oxidized and what is being reduced, determine the more positive reduction potential, and calculate the absolute difference in reduction potentials. Let's go through these steps one by one.
1. What is being oxidized?:
In this cell notation, the left side of the double bar represents the anode (oxidation half-reaction). The anode is where oxidation occurs. From the notation, we can see that Na(s) is being oxidized, as it is on the left side of the double bar.
2. What is being reduced?:
The right side of the double bar in the cell notation represents the cathode (reduction half-reaction). The cathode is where reduction occurs. From the notation, we can see that Co3+(aq) is being reduced, as it is on the right side of the double bar.
3. Which has the more positive reduction potential?:
To determine the more positive reduction potential, we compare the values of E0 associated with the reduction half-reactions.
For Co3+(aq) + e- → Co2+(aq), E0 = 1.80 V, and for Na+(aq) + e- → Na(s), E0 = -2.71 V.
Since a more positive reduction potential indicates a stronger tendency to be reduced, in this case, Co3+(aq) has the more positive reduction potential with E0 = 1.80 V.
4. Is this a voltaic cell?:
Yes, this is a voltaic cell. Voltaic cells, also known as galvanic cells, are spontaneous electrochemical cells in which chemical energy is converted into electrical energy.
5. What is the absolute difference between the reduction potentials?:
The absolute difference between the reduction potentials can be calculated by subtracting the less positive reduction potential from the more positive reduction potential:
|1.80 V - (-2.71 V)| = |4.51 V| = 4.51 V
6. What is the emf?:
The emf (electromotive force) of a cell is the difference in electrical potential between the cathode and the anode. Since Co3+(aq) has the more positive reduction potential, the emf can be determined by subtracting the reduction potential of Na(s) from the reduction potential of Co3+(aq):
Emf = E0 (Co3+(aq)) - E0 (Na(s))
Emf = 1.80 V - (-2.71 V)
Emf = 1.80 V + 2.71 V
Emf = 4.51 V
Therefore, the emf for the given cell is 4.51 V.
To determine the emf for the cell noted, we can follow these steps:
1. Identify half-reactions: The cell notation indicates two half-reactions in the following format:
Na(s) | Na+(aq) || Co3+(aq), Co2+(aq) | Co(s)
The half-reactions are:
Oxidation: Na(s) → Na+(aq) + e-
Reduction: Co3+(aq) + e- → Co2+(aq)
2. Determine what is being oxidized: In this case, Na(s) is being oxidized to Na+(aq). This is evident from the fact that Na(s) is on the left side of the double line (||), indicating that it is the anode.
3. Determine what is being reduced: Co3+(aq) is being reduced to Co2+(aq). This is evident from the fact that Co3+(aq) is on the right side of the double line (||), indicating that it is the cathode.
4. Determine which half-reaction has the more positive reduction potential: The reduction potential for Co3+(aq) + e- → Co2+(aq) is given as E0 = 1.80 V. The reduction potential for Na+(aq) + e- → Na(s) is given as E0 = -2.71 V.
Since a more positive reduction potential indicates a stronger reduction, the half-reaction with Co3+(aq) has the more positive reduction potential.
5. Determine if this is a voltaic cell: A voltaic cell is an electrochemical cell that generates electrical energy from a spontaneous redox reaction. In this case, the given reduction potentials indicate that both half-reactions are spontaneous, so this is indeed a voltaic cell.
6. Calculate the absolute difference between the reduction potentials: The absolute difference between the reduction potentials is given by |E0(cathode) - E0(anode)|.
|1.80 V - (-2.71 V)| = |1.80 V + 2.71 V| = 4.51 V
7. Calculate the emf: The emf (electromotive force) of the cell is given by the reduction potential difference between the cathode and the anode in a voltaic cell. Since the reduction potential of the cathode is higher, the emf can be calculated as:
Emf = E0(cathode) - E0(anode)
Emf = 1.80 V - (-2.71 V)
Emf = 4.51 V
So, to summarize:
- Na(s) is being oxidized.
- Co3+(aq) is being reduced.
- Co3+(aq) has the more positive reduction potential.
- This is a voltaic cell.
- The absolute difference between the reduction potentials is 4.51 V.
- The emf of the cell is 4.51 V.