The typical battery used in a standard flashlight produced approximately 1.5 V. If the value for ΔG° of the reaction were −289,500 J, what would you calculate as the moles of electrons exchanged in the balanced redox reaction?
I got 1 mols for this...but i think it's wrong, can explain what am I suppose to do?
DGo = -nEF
-289,500 = -n(1.5)(96,485)
n = ?
okay I ended up getting 2 instead of one? XD I made a mistake in my half reactions...thanks!
To calculate the moles of electrons exchanged in a redox reaction, you can use Faraday's constant, which is 96,485 C/mol e-. This constant relates the amount of charge (in Coulombs) to the number of moles of electrons exchanged.
First, convert the value of ΔG° from J to C (Coulombs). Since 1 J = 1 C, you can skip this step.
Next, use the equation:
ΔG° = -nFΔE°
Where:
ΔG° = standard Gibbs free energy change (in J)
n = moles of electrons exchanged
F = Faraday's constant (in C/mol e-)
ΔE° = standard cell potential (in V)
Rearrange the equation to solve for n:
n = -ΔG° / (FΔE°)
Given that the standard cell potential (ΔE°) for a standard flashlight battery is 1.5 V and ΔG° is -289,500 J, we can substitute these values into the equation:
n = -(-289,500 J) / (96,485 C/mol e- * 1.5 V)
Simplifying the equation:
n = 289,500 J / (96,485 C/mol * 1.5 V)
n ≈ 1.99 mol e-
Therefore, the number of moles of electrons exchanged in the balanced redox reaction is approximately 1.99 mol e-.