Final answer:
For the given changes, the effect on the observed cell potential (E) can be predicted as: Increased E, Decreased E, No change in E, and No change in E, for increasing the amount of PbCl₂, diluting the solution, dissolving NaOH, and increasing the temperature, respectively.
Step-by-step explanation:
The electrochemical cell reaction shown is:
H₂(g) + PbCl₂(s) ⇋ Pb(s) + 2HCl(aq)
The reaction is an oxidation-reduction (redox) reaction where hydrogen gas (H₂) is oxidized to form protons (H⁺) and lead (Pb) is reduced to lead ions (Pb²⁺).
To predict the immediate effect on the observed cell potential (E) for the given changes, we can analyze each change:
- Increasing the amount of PbCl₂ will shift the reaction to the right, increasing the concentration of Pb²⁺ ions and HCl in the cell. This will increase the reaction rate and result in an increase in E. Therefore, the effect is Increased E.
- Diluting the solution by adding H₂O will decrease the concentration of reactants, leading to a decrease in the reaction rate and a decrease in E. Therefore, the effect is Decreased E.
- Dissolving NaOH into the solution will introduce hydroxide ions (OH⁻) which can react with H⁺ ions from the HCl, reducing their concentration. However, this will not affect PbCl₂ or the overall reaction. Consequently, there will be no change in E. Therefore, the effect is No change in E.
- Increasing the temperature will generally increase the reaction rate. However, since the reaction is non-spontaneous (ΔH⁰ > 0), increasing the temperature will not affect the sign of ΔH⁰ and only slightly increase the reaction rate. Therefore, the effect is No change in E.