<u>Question:</u>
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K
<u>Answer:</u>
The mean activity coefficient for HCl solution is 0.78.
<u>Explanation:</u>
Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is
As we know that 
= 0.22 V and E = 0.342 V, the equation will become
So, the mean activity coefficient is 0.78.
Answer:
Explanation:
Fossil fuel is nonrenewable, whereas hydro, wind, solar, and wave power are generally ... depletes its fuel stock but at a relatively slow rate, and geothermal energy (Williamson, ... Future primary energy sources (e.g. nuclear, coal and solar) are primarily ... b. Distributed peak electrical generation, •. c. Distributed electricity ...
The building with incandescent light bulbs would have higher energy bills because less than 10% of the bulb is used for light while the rest is given off as heat. Fluorescent light bulbs use ¼ as much energy and provide the same amount of light.
The substances which allow the current to flow in the circuit are:
Lithium bromide in a solution and Graphite.
Lithium bromide can be pass the current in the circuit, due to the ability to form ions in the solution Li⁺ and Br⁻ which are mobile and can transport the electrical charge.
In the graphite because of the presence of delocalised electrons, each carbon atom forms three covalent bonds with three other carbon atoms, it conducts electrical current.
In the case of potassium chloride lattice there are in a solid form and the atoms are not mobile so they can not form ions. In the diamond all the carbon valences are satisfied so the electrical charge can not be transported.
Nitric oxide in a solution is isolated and can not transport the electrical current due to its inability of the to form ions
.
