Question

    Consider a galvanic cell based upon the following half reactions:

Ag⁺ + e^- → Ag; 0.80 V
Cu²⁺ + 2 e^- → Cu; 0.34 V

Which of the following changes will decrease the potential of the cell?

	Adding Cu2+ ions to the copper half reaction (assuming no volume change).
	Adding equal amounts of water to both half reactions.
	Adding Ag+ ions to the silver half reaction (assume no volume change)
	Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Answer 1

We have, Nernst equation for the cell.

Ecell = E°cell– (2.303 x RT / 6F) log [Cu²⁺] / [Ag+] = E°cell– (2.303 x RT / 6F) logQ

Now Let us consider the options one by one

1)       Adding Cu²⁺ ions to the copper half reaction (assuming no volume change).

This will increases the Q value and results in decrease of of Ecell.

2)       Adding equal amounts of water to both half reactions.

Q value remain unaffected. Therefore, cell potential remains the same

3)        Removing Cu²⁺ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

This will produce no effect. Because based on Le Chatelier's principle the system attempts to generate more Cu²⁺ ions to counter the affect the effect

4)       Adding Ag⁺ ions to the silver half reaction (assume no volume change)

This will decrease the Q value. Therefore, Ecell will increase

Answer: 1)   Adding Cu²⁺ ions to the copper half reaction (assuming no volume change).