Question

If you start with the cell at the standard conditions and 298K and add a few milliliters of 1m NaOH to the half-cell that contains the silver electrode, what can you say about the value of Eₑₗₗ?

a. Eₑₗₗ = 0
b. Eₑₗₗ = E∘
c. Eₑₗₗ > E∘ₑₗₗ
d. Eₑₗₗ < E∘ₑₗₗ

Answer 1

Adding 1M NaOH to the half-cell with the silver electrode decreases the concentration of Ag+ ions due to the formation of Ag2O, resulting in an increase in the cell potential compared to the standard cell potential, leading to Ecell > E°cell.

Step-by-step explanation:

If you start with a cell at standard conditions and 298K and add a few milliliters of 1M NaOH to the half-cell that contains the silver electrode, the concentration of Ag+ ions near the electrode will decrease as the OH- ions from the NaOH react with the Ag+ ions to form Ag2O, a precipitate. This change in concentration will cause a shift in the cell potential according to the Nernst equation, which takes into account the concentrations of the ionic species. Therefore, the addition of NaOH will alter the half-cell potential and the overall cell potential (Ecell) will become different from the standard cell potential (E°cell). Particularly, Ecell will be greater than E°cell.

So, the addition of NaOH will result in Ecell > E°cell as the local concentration of Ag+ decreases. It's important to note that this example assumes a significant concentration change enough to affect the potential of the half-cell.