Silver chloride is an insoluble salt. What will be the effect on the cell potential when a small amount of silver nitrate solution is added to half-cell A?
When a small amount of silver nitrate solution is added to half-cell A, silver ions (Ag+) from the silver nitrate will react with chloride ions (Cl-) in the silver chloride to form silver chloride precipitate (AgCl). Since silver chloride is an insoluble salt, it will not dissociate in solution and will remain as a solid.
The presence of the silver chloride precipitate will decrease the concentration of silver ions in half-cell A. This decrease in concentration of reactant ions will lower the concentration of reactants in the anode half-cell, leading to a decrease in the oxidation reaction at the anode.
This reduction in the oxidation reaction at the anode will in turn decrease the overall cell potential. Essentially, the addition of the silver nitrate solution to half-cell A will reduce the number of reactants available for the electrochemical reaction and therefore decrease the overall cell potential.
When a small amount of silver nitrate solution is added to half-cell A, the following reactions occur:
1. In half-cell A (which contains silver chloride), the silver nitrate reacts with the chloride ions to form silver chloride:
AgNO3 + Cl- → AgCl + NO3-
2. In half-cell B (which contains a standard reference electrode), no reaction occurs as it serves as a reference to measure the potential.
As a result, the concentration of silver ions (Ag+) in half-cell A increases, which causes an increase in the concentration gradient between the two half-cells. This increase in concentration gradient leads to an increase in cell potential.
In summary, the addition of silver nitrate solution to half-cell A increases the concentration of silver ions, which increases the cell potential.