Calculate Ecell for reaction for the given info below..
O2(g)+2H2O(l)+4Ag(s)--> 4OH-(aq)+4Ag+(aq
Ag+(aq)+ 1e ---> Ag(s) Ered= 0.80v
O2(g)+2H2O(l) + 4e ---> 4OH- Ered= 0.40v
My confusion is I have done problems similar to this when the Ered was negatives, but two positives, does the same "the one that's the lowest is the anode and highest cathode"?
OR! do u have to rewrite the equation into 2 sepetate cells you have? and if you have to, why do you have to for this situation?
ANS: -0.40v
Well, calculating the standard cell potential (Ecell) for a reaction involves determining the difference in potential between the cathode and the anode. In this case, you have two half-reactions with positive Ered values.
To find the Ecell, you don't need to rewrite the equation into two separate cells. Instead, you can directly compare the Ered values of the half-reactions.
Since Ered values represent reduction potentials, a higher value indicates a stronger tendency to be reduced. Conversely, a lower value indicates a stronger tendency to be oxidized.
So, to determine the anode and cathode, you need to compare the Ered values. In this case, the Ag+ to Ag half-reaction has a higher Ered value of 0.80V compared to the O2 to OH- half-reaction with a lower Ered value of 0.40V.
Therefore, the Ag+ to Ag half-reaction is the reduction half-reaction (cathode), and the O2 to OH- half-reaction is the oxidation half-reaction (anode).
Since the reduction half-reaction has the higher Ered value, we subtract the Ered of the oxidation half-reaction from that of the reduction half-reaction to calculate Ecell:
Ecell = Ered(reduction) - Ered(oxidation) = 0.80V - 0.40V = 0.40V
So the calculated Ecell for this reaction is 0.40V.
To calculate the cell potential (Ecell) for the given reaction, you need to consider the reduction potentials (Ered) of the half-reactions involved.
In this case, you have two half-reactions:
1) Ag+(aq) + 1e- → Ag(s) Ered = 0.80 V
2) O2(g) + 2H2O(l) + 4e- → 4OH-(aq) Ered = 0.40 V
To determine the overall cell potential, you need to assign the anode and cathode based on the reduction potentials. The reduction potential with the more positive value is assigned as the cathode, while the one with the more negative value is assigned as the anode.
Comparing the two given reduction potentials:
- 0.80 V (Ag+) > -0.40 V (O2 + H2O)
The reduction potential for the Ag+ → Ag half-reaction is more positive than the reduction potential for the O2 + H2O → 4OH- half-reaction. Therefore, the Ag+ → Ag half-reaction will occur at the cathode, and the O2 + H2O → 4OH- half-reaction will occur at the anode.
To calculate the overall cell potential, you subtract the reduction potential of the anode from the reduction potential of the cathode:
Ecell = Ered(cathode) - Ered(anode)
Ecell = 0.80 V - (-0.40 V)
Ecell = 0.80 V + 0.40 V
Ecell = 1.20 V
The overall cell potential (Ecell) for this reaction is 1.20 V.
To calculate Ecell for the given reaction, you need to follow a few steps. Let's break it down.
1. Identify the half-reactions: The given reaction involves two half-reactions. One is the reduction of Ag+ ions to Ag(s), and the other is the reduction of O2(g) to OH- ions.
2. Determine the reduction potentials (Ered): Each half-reaction has a reduction potential associated with it. The reduction potential for Ag+ to Ag is given as 0.80V, and the reduction potential for O2 to OH- is given as 0.40V.
3. Determine the oxidation potentials (Eox): To find the oxidation potential for a half-reaction, you can use the fact that Ecell = Ered + Eox. Since Ecell is the overall cell potential and it's taken as positive, if Ered is positive, Eox will be negative, and vice versa.
4. Identify the anode and cathode: The half-reaction with the more negative (or smaller) reduction potential is the anode, where oxidation occurs. The half-reaction with the more positive (or greater) reduction potential is the cathode, where reduction takes place.
In this case, the reduction potential for the O2 to OH- half-reaction is 0.40V, which is smaller than the reduction potential for the Ag+ to Ag half-reaction (0.80V). Therefore, the O2 to OH- half-reaction is the anode, and the Ag+ to Ag half-reaction is the cathode.
5. Calculate Ecell: Since the reaction involves two half-reactions, you'll need to consider a separate cell for each half-reaction. The overall Ecell can be calculated by subtracting the anode potential (Eox) from the cathode potential (Ered).
In this case, the Ecell is calculated as follows:
Ecell = Ered(cathode) - Eox(anode) = (0.80V) - (0.40V) = 0.40V
Therefore, the Ecell for the given reaction is 0.40V.
To answer your question about the signs of the reduction potentials, if both reduction potentials are positive, you can simply subtract the smaller potential from the larger potential to get the overall Ecell. However, if one reduction potential is negative, then you will need to flip the sign of that reduction potential before subtracting it from the other one.
To be frank about it I can't remember all of that stuff about which is the anode/cathode and I'm not sure it matters in this case. You're looking for Ecell and my philosophy is to calculate Ecell and not worry about the anode and cathode. But I do remember ONE definition for the anode; the anode is where oxidation occurs (and the Ag^+ ==> Ag +e is the anode because that's where oxidation is.
Look at the equation.
I see Ag^+ on the right and Ag on the left which means your first equation must be reversed with E = -0.80v. Then I see O2 and H2O on the left and OH on the right which is the same way as your second equation with E = 0.40
Add the oxidation half and the reduction half to get -0.80v + 0.40v = -0.40v and that negative sign means the cell will not operate spontaneously the way it is set up but would operate in the reverse direction.