The average current produced by the cell in figure 2, when it operates for 30.0 min and consumes 1.00 g of the anode is
4 years ago
Answered By Emily D
I'm hoping the question has a picture to go with it that gives more context because this is not enough for me to solve it. That said, I hope this can point you in the right direction:This sounds like a Faraday's Law problem, which will use this equation: $n_e=\frac{It}{F}$ne=ItF ne = moles of electrons transferred FIND FROM 100g OF ANODEI = current (C/s) THIS IS WHAT WE ARE LOOKING FORt = time (s) CONVERT 30 MIN TO SECONDS (30*60s/min = 1800s)F = 9.65 x 104 C/mol
You'll get ne from the number of moles of your anode that are in 100 grams. Then just solve for I
4 years ago
Answered By Emily D
I'm hoping the question has a picture to go with it that gives more context because this is not enough for me to solve it. That said, I hope this can point you in the right direction:This sounds like a Faraday's Law problem, which will use this equation: $n_e=\frac{It}{F}$ne=ItF ne = moles of electrons transferred FIND FROM 100g OF ANODEI = current (C/s) THIS IS WHAT WE ARE LOOKING FORt = time (s) CONVERT 30 MIN TO SECONDS (30*60s/min = 1800s)F = 9.65 x 104 C/mol
You'll get ne from the number of moles of your anode that are in 100 grams. Then just solve for I