Calculate the Ecell for this reaction
3HClO2(aq) +2Cr3+(aq) + 4H2O(l) -> 3HClO(aq) + (Cr2O7)2–(aq) + 8H+(aq)
I thought Cr + e -> Cr2O7 is the cathode
and 3HClO2 -> 3HClO + e- is the anode.
Using Ecell = Cathode - Anode
I get -0.31, but the answer I checked is supposed to be positive.
If someone can explain me the steps again that'll be great, thanks
See your other post above.
To calculate the Ecell for a redox reaction, you have correctly identified the cathode and anode half-reactions. However, it seems there might be a misunderstanding regarding the direction of electron flow.
In a redox reaction, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). Therefore, the reduced species should be at the cathode, and the oxidized species should be at the anode.
Let's reassign the half-reactions as follows:
Cathode half-reaction: 2Cr3+(aq) + 7H2O(l) → (Cr2O7)2-(aq) + 14H+(aq) + 6e-
Anode half-reaction: 3HClO2(aq) → 3HClO(aq) + 2e-
Now, let's find the standard reduction potentials (E°) for each half-reaction. Look up the standard reduction potentials for Cr3+/Cr2O7^2- and HClO2/HClO half-reactions in a reference table or textbook. For the purposes of this explanation, let's assume the values are as follows:
E°(2Cr3+/Cr2O7^2-) = +1.33 V
E°(HClO2/HClO) = +1.20 V
To calculate the standard cell potential (E°cell), you subtract the anode potential from the cathode potential:
E°cell = E°(cathode) - E°(anode)
= (+1.33 V) - (+1.20 V)
= +0.13 V
Since the calculated E°cell value is positive, it indicates that the reaction is spontaneous under standard conditions. Therefore, the answer you checked is correct.
Keep in mind that the E°cell value represents the cell potential under standard conditions. If you want to calculate the actual Ecell for a given concentration of reactants, you will need to use the Nernst equation.