A battery is constructed based on the oxidation of magnesium and the reduction of Cu^2+. The initial concentrations of Mg^2+ and Cu^2+ are 1.2*10^-4M and 1.5M, respectively, in 1.0-L half-cells.
The initial voltage of the battery is 2.83V with the standard voltage at 2.71V.
QUESTION: What is the voltage of the battery after delivering 5.0A for 7.0hr?
bad explanation
To find the voltage of the battery after delivering 5.0A for 7.0hr, we need to calculate the change in the concentrations of Mg^2+ and Cu^2+ and then determine the change in voltage.
1. Calculate the moles of electrons transferred:
The balanced chemical equation for the oxidation of magnesium and reduction of Cu^2+ is:
Mg(s) + Cu^2+(aq) → Mg^2+(aq) + Cu(s)
This equation shows that 2 moles of electrons are transferred for every 1 mole of Cu^2+ ions.
Given that we delivered a current of 5.0A for 7.0hr, the total charge passed through the circuit can be calculated using the equation:
Q = I * t
where Q is the charge, I is the current, and t is the time.
Q = 5.0A * 7.0hr * 3600s/hr (to convert hours to seconds)
2. Calculate the moles of Cu^2+ reacted:
Since 2 moles of electrons are transferred for every 1 mole of Cu^2+ ions, we can calculate the moles of Cu^2+ reacted as follows:
moles of Cu^2+ reacted = (Q / Faraday's constant) / 2
where Faraday's constant is 96485 C/mol e^-
3. Calculate the change in concentration of Cu^2+:
The initial concentration of Cu^2+ is 1.5M. To find the final concentration, we subtract the moles of Cu^2+ reacted from the initial moles:
final concentration = initial concentration - (moles of Cu^2+ reacted / volume)
volume = 1.0L (given)
4. Calculate the change in concentration of Mg^2+:
Since the moles of Mg^2+ and Cu^2+ are in a 1:1 ratio, the change in concentration of Mg^2+ will be the same as the change in concentration of Cu^2+.
5. Calculate the new voltage of the battery:
To calculate the new voltage, we can use the Nernst equation:
V = V0 - (RT / (nF)) * ln([Mg^2+]/[Cu^2+])
where V is the new voltage, V0 is the standard voltage (2.71V), R is the gas constant (8.314 J/(mol K)), T is the temperature (in Kelvin), n is the number of electrons transferred, F is Faraday's constant, and [Mg^2+] and [Cu^2+] are the concentrations of Mg^2+ and Cu^2+, respectively.
Now we can plug in the values and calculate the new voltage.
To determine the voltage of the battery after delivering a current for a given time, we need to use the equation:
Voltage = Initial Voltage - (Current × Time) / Charge
First, we need to calculate the charge that has been delivered by the battery. Charge is given by the equation:
Charge = Current × Time
In this case, the current is 5.0 A, and the time is 7.0 hours. So, the charge can be calculated as:
Charge = 5.0 A × 7.0 hr
Once we have the value of the charge, we can substitute it into the voltage equation along with the other known values:
Voltage = 2.83 V - (5.0 A × 7.0 hr) / Charge
However, we still need to calculate the charge. To do this, we'll use Faraday's law of electrolysis, which states that:
Charge = Current × Time × (1 Faraday / 96500 C)
Here, 96500 C represents the charge of 1 mole of electrons (Faraday constant). We need to convert the charge to coulombs, so the charge equation becomes:
Charge = 5.0 A × 7.0 hr × (1 Faraday / 96500 C)
Now we can substitute this value for charge in the voltage equation:
Voltage = 2.83 V - (5.0 A × 7.0 hr) / (5.0 A × 7.0 hr × (1 Faraday / 96500 C))
Simplifying the equation further:
Voltage = 2.83 V - 1 / (1 Faraday / 96500 C)
However, we still need to calculate the value of 1 Faraday / 96500 C. The Faraday constant is defined as the charge of 1 mole of electrons, which is 96485.34 C/mol. Therefore:
1 Faraday / 96500 C = (1 mol / 96485.34 C) × (96500 C)
Simplifying further:
1 Faraday / 96500 C = 1.000057471
Now we can substitute this value back into the voltage equation:
Voltage = 2.83 V - 1 / 1.000057471
Simplifying the equation:
Voltage = 2.83 V - 0.999942544 V
Finally, simplifying further:
Voltage = 1.057456 V
So, the voltage of the battery after delivering 5.0 A for 7.0 hours is approximately 1.057 V.
E = Ecello - (0.0592/n)*log Q. [See the last line for the equation from which you can get the expression for Q).
You know Ecello = 2.71 v from the problem.
What you must do is calculate the new concns of Cu^+2 (which is being used up) and Mg^+2 (which is being increased) after the cell has run 5.0 amperes for 7 hours. This is how you do it.
Coulombs = amperes x time = 5.0 x 7 hrs x (3600 sec/hr) = ??
You know that Cu^+2 will be removed. How much? 96,485 coulombs will remove 1/2 mol Cu^+2 which is (63.54g/2). How much was there originally. 1.5 M, convert to grams. Subtract what you started with minus the final and that is the grams remaining (and it was in 1 L) so convert that to moles and that is the molarity. Do the same process for Mg. Now that you have the NEW concns of Cu^+2 and Mg^+2, substitute those into Q and calculate the new E for the cell.
By the way, the cell you are talking about is
Mg(s) + Cu^+2 ==<> Cu(s) + Mg^+2