What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10, according to the Nernst Equation?

• A. The cell voltage increases by 0.059 V
• B. The cell voltage decreases by 0.059 V
• C. The cell voltage decreases by 0.1 V
• D. The cell voltage remains unchanged

Answer: (B)

Verified step by step guidance

1. Identify the Nernst Equation: The Nernst Equation is used to calculate the cell potential under non-standard conditions. It is given by: $E=E{°}_{\mathrm{cell}}-\frac{0.059}{n}\log \frac{\mathrm{[Products]}}{\mathrm{[Reactants]}}$, where $E$ is the cell potential, $E{°}_{\mathrm{cell}}$ is the standard cell potential, and $n$ is the number of moles of electrons transferred in the reaction.
2. Determine the effect of increasing ion concentration: Increasing the concentration of ions in the cathode half-cell by a factor of 10 affects the reaction quotient $Q$ in the Nernst Equation. Specifically, the concentration of products in the reaction quotient increases.
3. Calculate the change in the reaction quotient: Since the concentration of ions in the cathode half-cell is increased by a factor of 10, the reaction quotient $Q$ becomes 10 times larger.
4. Apply the change to the Nernst Equation: Substitute the new value of $Q$ into the Nernst Equation. The logarithmic term $\log \frac{\mathrm{[Products]}}{\mathrm{[Reactants]}}$ will increase by $\log 10$, which is 1.
5. Determine the change in cell voltage: The increase in the logarithmic term by 1 results in a decrease in the cell voltage by $\frac{0.059}{n}$ volts. Since $n$ is typically 1 for a single electron transfer, the cell voltage decreases by 0.059 V.