Consider the galvanic cell constructed from the following metals and their corresponding metal ions:
E degrees (V)
M^3+^ + 3e^-^ ----> M 0.29
N^2+^ + 2e^-^ ----> N 0.61
To analyze the galvanic cell constructed from the given metals and metal ions, we need to compare their standard reduction potentials (E degrees) and determine the overall cell reaction and its potential.
1. Write the half-reactions:
a) M^3+^(aq) + 3e^-^ → M(s) (reduction half-reaction)
b) N^2+^(aq) + 2e^-^ → N(s) (reduction half-reaction)
2. Determine the standard reduction potentials:
The standard reduction potential (E degrees) for the reduction half-reactions are given:
E deg (M^3+/M) = 0.29 V
E deg (N^2+/N) = 0.61 V
3. Identify the oxidation half-reactions:
Whenever we have reduction half-reactions, the corresponding oxidation half-reactions can be obtained by reversing the reduction reactions:
a) M(s) → M^3+^(aq) + 3e^-^ (oxidation half-reaction)
b) N(s) → N^2+^(aq) + 2e^-^ (oxidation half-reaction)
4. Calculate the overall cell potential:
The overall cell potential (E cell) of the galvanic cell can be calculated by taking the difference between the reduction potentials of the cathode and the anode (reduction and oxidation half-reactions):
E cell = E deg (cathode) - E deg (anode)
= E deg (reduction) - E deg (oxidation)
E cell = (E deg reduction-M^3+/M) - (E deg reduction-N^2+/N)
= 0.29 V - 0.61 V
5. Determine the overall reaction:
The overall cell reaction is obtained by combining the reduction half-reactions and canceling out any common species, leaving only the net reaction:
a) M^3+^(aq) + 3e^-^ → M(s) (reduction half-reaction)
b) N(s) → N^2+^(aq) + 2e^-^ (oxidation half-reaction)
Multiply the oxidation half-reaction by 3 to balance the electrons:
3N(s) → 3N^2+^(aq) + 6e^-^ (oxidation half-reaction)
Add the two half-reactions together:
3N(s) + M^3+^(aq) → 3N^2+^(aq) + M(s)
Therefore, the overall cell reaction is:
3N(s) + M^3+^(aq) → 3N^2+^(aq) + M(s)
6. Conclusion:
The galvanic cell constructed using the given metals and metal ions has an overall reaction of 3N(s) + M^3+^(aq) → 3N^2+^(aq) + M(s) with a calculated cell potential of -0.32 V (0.29 V - 0.61 V).
To determine the overall cell potential (E°) of the galvanic cell constructed from these metals and their corresponding metal ions, you need to calculate the difference in standard reduction potentials (also known as standard electrode potentials or standard half-cell potentials) between the two half-reactions involved.
The half-reactions given are:
M^3+ + 3e^- -> M (with E° = 0.29 V)
N^2+ + 2e^- -> N (with E° = 0.61 V)
In a galvanic cell, oxidation (loss of electrons) occurs at the anode, while reduction (gain of electrons) occurs at the cathode. To construct the galvanic cell, you need to determine which metal/ion pair will undergo reduction (cathode) and which will undergo oxidation (anode).
Based on the values of the standard reduction potentials, the species with the higher reduction potential (more positive E°) will be reduced, and the species with the lower reduction potential (less positive E°) will be oxidized.
In this case, N^2+ (E° = 0.61 V) has a higher reduction potential than M^3+ (E° = 0.29 V). Therefore, N^2+ will be reduced at the cathode, and M^3+ will be oxidized at the anode.
The overall cell reaction is obtained by adding the two half-reactions together. However, we need to multiply each half-reaction by a whole number to balance the number of electrons transferred. In this case, the number of electrons transferred in the first half-reaction (3) must be equal to the number of electrons transferred in the second half-reaction (2). To achieve this, the first half-reaction must be multiplied by 2, and the second half-reaction must be multiplied by 3.
The balanced half-reactions are:
2M^3+ + 6e^- -> 2M
3N^2+ + 6e^- -> 3N
Now we can add the two balanced half-reactions together to obtain the overall balanced cell reaction:
2M^3+ + 6e^- + 3N^2+ + 6e^- -> 2M + 3N
Simplifying the equation:
2M^3+ + 3N^2+ -> 2M + 3N
The overall cell potential (E°cell) can be calculated by subtracting the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).
E°cell = E°cathode - E°anode
E°cell = 0.61 V - 0.29 V
E°cell = 0.32 V
Therefore, the overall cell potential of the galvanic cell constructed from these metals and their corresponding metal ions is 0.32 V.
I'm confused.
1. Apparently you want the cell construction so as to generate a voltage.
2. However, you don't say what you want.
3. I assume you want to know the voltage generated.
M ==> M^3+ + 3e........Eo = -0.29 v
N^2+ + 2e ==> N.......Eo = 0.61 v
--------------------------------------------
2M + 3N^2+ + 6e==>2M^3+ + 3N + 6e
Eo cell = -0.29 + 0.61 = ?
(NOTE) Remember you do not multiply the voltages when you balance the equation.