A zinc-copper battery is constructed as follows (standard reduction potentials given below):
Zn | Zn2+ (0.10 M) || Cu2+ (2.50 M) | Cu
Zn2+ + 2 e- → Zn(s) Eº = -0.76 V
Cu2+ + 2 e- → Cu(s) Eº = 0.34 V
The mass of each electrode is 200 g. Each half-cell contains 1.00 L of 1M solution. Calculate the cell
potential after 10.0 A of current has flowed for 10.0 hours.
A. 0.90V
B. 1.13V
C. –1.00V
D. 1.20V
E. 2.00V
To calculate the cell potential after 10.0 A of current has flowed for 10.0 hours in a zinc-copper battery, we can use the formula:
E_cell = Eº_cell - (I * R * t)
Where:
E_cell is the cell potential after the current has flowed for a certain time
Eº_cell = Eº_cathode - Eº_anode (standard reduction potentials of cathode and anode)
I is the current (in Amperes)
R is the gas constant (8.314 J/mol·K)
t is the time (in seconds)
First, let's calculate the standard reduction potentials for the cathode and anode:
Eº_cathode = 0.34 V (given)
Eº_anode = -0.76 V (given)
Next, we need to convert the current (10.0 A) and time (10.0 hours) to the proper SI units:
Current = 10.0 A
Time = 10.0 hours = 10.0 * 3600 seconds = 36,000 seconds
Now, we can substitute the values into the formula:
E_cell = Eº_cell - (I * R * t)
E_cell = (Eº_cathode - Eº_anode) - (I * R * t)
E_cell = (0.34 V - (-0.76 V)) - (10.0 A * 8.314 J/mol·K * 36,000 s)
Simplifying the expression:
E_cell = (0.34 V + 0.76 V) - (10.0 A * 8.314 J/mol·K * 36,000 s)
E_cell = 1.10 V - (299,0400 J/A)
E_cell = 1.10 V - (299,0400 V s)
Finally, we need to convert the units back to volts:
E_cell = 1.10 V - (299,0400 V s) / 1 C
E_cell = 1.10 V - (299,0400 C)
E_cell = 1.10 V - (299,040 mC)
E_cell = 1.10 V - (299.04 C)
E_cell = 1.10 V - (-299.04 V)
E_cell = 1.10 V + 299.04 V
E_cell = 300.14 V
Therefore, the cell potential after 10.0 A of current has flowed for 10.0 hours is 300.14 V.
Since none of the answer choices match this value, it is likely that there was an error in the calculations or the given information. Please double-check the problem and redo the calculations.