One more problem with 1/2 reactions
Pb(s) + PbO2(s) + 2H2SO4(aq)->2PbSO4(aq)+2H2O(l).
How do you split them up with two Pb solids on the same side of the equation?
Again you need to identify what is being reduced and what is being oxidised.
I think I read the question wrong Im not sure I need to split the reaction. I am trying to calculate delta G given an Ecell of 2.04V. Using Delta G = -nFEcell. I know F and Ecell but I need to find the value of n.
Pb goes from 0 on the left to +2 on the right as PbSO4.
Pb in PbO2 goes from +4 on the left to +2 on the right as PbSO4.
So change in electrons is 2 for n.
To calculate the value of 'n' in the equation ΔG = -nFEcell, you need to determine the change in the number of electrons (n) involved in the redox reaction.
In the given equation, you correctly identified the oxidation states of Pb. Pb on the left side of the equation (as a solid) has an oxidation state of 0, whereas on the right side (as part of PbSO4), it has an oxidation state of +2. This indicates that Pb is being reduced, meaning it gains electrons.
For PbO2, it starts with an oxidation state of +4 on the left side and becomes +2 in PbSO4 on the right side. Therefore, the oxidation state of Pb is decreasing, indicating that it is being reduced. Reduction involves gaining electrons.
Now, to find the change in electrons (n), subtract the initial oxidation state from the final oxidation state. For Pb, it goes from 0 to +2, which is a change of +2. For PbO2, the change is +4 - (+2) = +2.
Thus, the total change in electrons (n) for the overall reaction is 2 + 2 = 4 electrons.
Therefore, when calculating ΔG = -nFEcell, the value of 'n' is 4, assuming the reaction is happening under standard conditions.