An elctrochemical cell is made up of Zn&Co
with their standard reduction potential -0.76V,-0.28V respectively.
1. Select the anode&cathode?
2. Write down the cell reactions?
3. Compute the e.m.f of the cell?
Select the larger value of reduction potential, reverse it and change the sign. That will be Zn^2+ + 2e ==> Zn Eored = -0.76
Zn ==> Zn^2++ 2e Eo ox = 0.76v
Co^2+ + 2e ==> Co Eo red = -.28v
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Zn +Co^2+ ==>Zn2+ + Co is cell rxn.
Eocell = Eoox + Eored = +0.76 + (-0.28) = ? is cell voltage.
That anode is where oxidation occurs. Which of the two half cells I've written above is oxidation(loss of electrons)? The cathode of course is the other one.
To answer these questions, we need to understand the concept of standard reduction potentials and their relation to electrochemical cells.
1. Selecting the anode and cathode:
In an electrochemical cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. To determine which element will be the anode and which will be the cathode, we compare the standard reduction potentials of the elements involved. The element with the more negative (more readily reduced) standard reduction potential will be the cathode, while the element with the less negative (less readily reduced) standard reduction potential will be the anode.
In this case, zinc (Zn) has a standard reduction potential of -0.76V, and cobalt (Co) has a standard reduction potential of -0.28V. Since the standard reduction potential of zinc is more negative than that of cobalt, zinc will be the cathode, and cobalt will be the anode.
2. Writing down the cell reactions:
The anode will undergo oxidation, releasing electrons, while the cathode will undergo reduction, accepting those electrons. The overall cell reaction can be determined by combining the two half-reactions:
Anode (oxidation):
Co(s) -> Co2+(aq) + 2e-
Cathode (reduction):
Zn2+(aq) + 2e- -> Zn(s)
To balance the number of electrons transferred, we need to multiply the reduction half-reaction by 2:
2Zn2+(aq) + 4e- -> 2Zn(s)
Now, we can write the overall cell reaction by combining the two half-reactions:
Co(s) + 2Zn2+(aq) -> Co2+(aq) + 2Zn(s)
3. Computing the e.m.f of the cell:
The electromotive force (e.m.f) of a cell is a measure of the cell's voltage or potential difference between the anode and cathode. The e.m.f of a cell can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.
In this case, the standard reduction potential of the anode (cobalt): -0.28V
The standard reduction potential of the cathode (zinc): -0.76V
e.m.f of the cell = E(cathode) - E(anode)
= -0.76V - (-0.28V)
= -0.48V
Therefore, the e.m.f of the cell is -0.48V.