Question

In an electrochemical cell, the following reaction takes place : $2\text{Cu}^{2+} (\text{aq}) + \text{Zn} (\text{s}) \rightarrow 2\text{Cu} (\text{s}) + \text{Zn}^{2+} (\text{aq})$
$E^\circ_{\text{cell}} = 1.28 \, \text{V}$
As the reaction progresses, what will happen to the overall voltage of the cell?

• A. Voltage will remain constant.
• B. It will decrease as $[\text{Zn}^{2+}]$ increases.
• C. It will increase as $[\text{Cu}^+]$ increases.
• D. It will increase as $[\text{Zn}^{2+}]$ increases.

Hint:

Remember to apply the Nernst equation to understand the relationship between concentration and cell potential in electrochemical reactions.

Answer 1

The Correct Option is B

As the reaction progresses, the concentration of $[\text{Zn}^{2+}]$ increases. According to the Nernst equation, as the concentration of a reactant increases, the cell potential decreases. Therefore, the overall voltage of the cell will decrease as $[\text{Zn}^{2+}]$ increases.