Question

What happens when a bulb is connected in series with two electrolytic cells containing aqueous \(\mathrm{CuSO}_{4}\) and aqueous sucrose solution?

Short answer

Answer: The bulb will not light up because the aqueous sucrose solution is a non-electrolyte and cannot conduct electricity, preventing electrical current from flowing through the entire circuit.

Step-by-step solution

Understand the properties of the solutions

First, it's essential to be familiar with the properties of the two solutions. Aqueous \(\mathrm{CuSO}_{4}\) is an electrolyte, which means it can conduct electricity when dissolved in water. In the case of aqueous sucrose solution, sucrose does not dissociate into ions in an aqueous solution, and hence, it is a non-electrolyte.

Analyzing the effects of the solutions on the electrical circuit

Since the two electrolytic cells are connected in series, the electrical current will need to flow through both solutions in the circuit to light the bulb. The aqueous \(\mathrm{CuSO}_{4}\) can conduct electricity, while the sucrose solution cannot. Because the entire circuit comes to depend on the sucrose solution's conductivity, the electrical current cannot flow through the circuit.

Determine the impact on the bulb

As the sucrose solution is a non-conductor, it prevents electrical current from flowing through the circuit. This means that there will be no electrical current in the circuit, and the bulb will not light up.

Conclusion

When a bulb is connected in series with two electrolytic cells containing aqueous \(\mathrm{CuSO}_{4}\) and aqueous sucrose solution, the bulb will not light up. This is because the sucrose solution is a non-electrolyte and cannot conduct electricity, resulting in a lack of electrical current in the entire circuit.