The reduction potentials for the species involved in an electrochemical cell are as follows:
M+ + e– --> M is 0.30 V
J+ + e– --> J is 0.40 V.
All solutions are 1.00 M and J+ / J is connected to M+ / M.
What is Ecell (in V)?
Which species is easier to oxidize?
The electrons flow from ? (M or J) to ? (M or J).
Take the most negative reduction potential and turn it around. Leave the other as is.
J + e ==> J E = 0.40 v
M ==> M^+ + e E = -0.30 v
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now add the equation and you will obtain a positive voltage for Ecell.
The cell notation is as follows:
M/M^+(1 M)||J^+(1 M)/J
So M is the anode (-) and J is the cathode (+). Electrons flow from - to +.
Write them both as oxidations. The easier one to oxidize will have the larger E value.
its unlimited tries, so if you are just looking for the answers, for the first part, just take the larger minus the smaller value, the species that is easier to oxidize should be the smaller value V, and the flow goes from less V to more V if that made any sense
ANSWER:
+.1
M
M
J
Jenn is correct!
ANSWER:
+.1
M
M
J
To determine the potential difference (Ecell) of an electrochemical cell, you need to calculate the difference between the reduction potentials of the species involved. The reduction potential is a measure of the tendency of a species to gain electrons, which is also an indicator of its ease of reduction.
Given the reduction potentials:
- For the half-reaction M+ + e- -> M, the reduction potential (E°) is 0.30 V.
- For the half-reaction J+ + e- -> J, the reduction potential (E°) is 0.40 V.
To find Ecell, subtract the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs). In this case, the anode is M+ / M, and the cathode is J+ / J.
Ecell = E°cathode - E°anode
Ecell = 0.40 V - 0.30 V
Ecell = 0.10 V
Therefore, the value of Ecell is 0.10 V.
To determine which species is easier to oxidize, we should compare their reduction potentials. The species with the lower reduction potential is easier to oxidize, as it has a stronger tendency to lose electrons.
In this case, given that the reduction potential of M+ / M is lower (0.30 V) compared to J+ / J (0.40 V), M+ / M is easier to oxidize.
The flow of electrons in the cell is from the anode to the cathode. The species being oxidized (losing electrons) is present at the anode, and the species being reduced (gaining electrons) is at the cathode.
Therefore, in this electrochemical cell, the electrons flow from M to J.