Here we consider instead dissolving guns (which we will assume are pure iron) with sulfuric acid.
Fe+H2SO4->FeSO4+H2
The characters realize that their hideout has been discovered by the police and they still have a last handgun that weighs 0.25 kg to dissolve. The cops will get there in an hour, so they have to speed up the reaction, by driving it at a higher current. What is the minimum total voltage in volts they'll need to drive the reaction at to get rid of the gun in time? Consider excess potential because of activation losses only and the exchange current I0 to be 1 A for the reaction over the surface of the entire tank (not a current density). £\ is 0.5 and everything is being done at room temperature. Assume standard electrochemical potentials.
To determine the minimum total voltage required to drive the reaction at a higher current, we need to consider the electrochemical aspects of the dissolution process.
First, let's understand the concept of standard electrochemical potentials. Standard potentials (E°) are measures of the thermodynamic driving force of a redox reaction. They are specific to each half-reaction and provide information about the relative ease of that reaction.
In this case, we need to calculate the voltage required to drive the dissolution of iron (Fe) in sulfuric acid (H2SO4) at a higher current. The half-reaction for the dissolution of iron can be represented as:
Fe(s) → Fe2+(aq) + 2e-
The standard potential for this reaction can be found in tables or references. For iron dissolution, the standard potential (E°) is approximately -0.44 V.
Now, let's consider the concept of exchange current (I0) and the overpotential (£\). The exchange current is a measure of the rate at which the redox reaction proceeds at equilibrium. It represents the current when the activation losses are negligible. In this case, the exchange current (I0) is given as 1 A.
The overpotential (£\) accounts for the excess potential required to drive the reaction at a different current than the exchange current. In this situation, we can use the value of £\, which is stated to be 0.5 V.
To calculate the minimum total voltage needed, we can use the Butler-Volmer equation:
I = I0 [exp((£\) / (RT)) - exp((E° - £\) / (RT))]
In this equation, I is the current, T is the temperature in Kelvin, R is the universal gas constant (8.314 J/(mol·K)), and £\ and E° are in volts.
Given that the current (I) needed is the same as the exchange current (I0) of 1 A, we can rearrange the Butler-Volmer equation and solve for £\:
I0 = I = I0 [exp((£\) / (RT)) - exp((E° - £\) / (RT))]
exp((£\) / (RT)) - exp((E° - £\) / (RT)) = 1
Using the given values and assuming room temperature (around 298 K), we can solve for £\:
exp((0.5 V) / (8.314 J/(mol·K) × 298 K)) - exp(((-0.44 V) - 0.5 V) / (8.314 J/(mol·K) × 298 K)) = 1
Solving this equation will give us the value of £\.
Once we have the value of £\, we can determine the minimum total voltage required by summing the standard potential (E°) and the overpotential (£\):
Minimum Total Voltage = E° + £\
Using the calculated value of £\ and the standard potential for iron dissolution, we can find the minimum total voltage needed to dissolve the gun in the given time frame.