Question

A voltaic cell using Cu/ ${\text{Cu}}^{\text{2 +}}$and Sn/${\text{Sn}}^{\text{2 +}}$half-cells is set up at standard conditions, and each compartment has a volume of345mL. The cell delivers $\text{0.17}$A for$\text{48.0}$ h. (a) How many grams of Cu(s) are deposited? (b) What is the [ ${\text{Cu}}^{\text{2 +}}$] remaining?

Short answer

1. The amount of Cu(s) are deposited as: $\text{9.67 g Cu}$
2. The remaining [ ${\text{Cu}}^{\text{2 +}}$] are obtained as: ${\text{[Cu}}^{\text{2 +}}\text{]=0.51 M}$

Step-by-step solution

Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

How many grams of Cu(s) are deposited?

a. Multiply the current by the given time to get the charge.

$$\begin{gathered} \text{Charge =}\frac{\text{0.17 C}}{\text{s}}\text{×48 hr×}\frac{\text{60 min}}{\text{1 hr}}\text{×}\frac{\text{60 sec}}{\text{1 min}} \\ \text{= 29,376 C} \end{gathered}$$

Divide the charge by Faraday's constant to get the moles of electrons reacting. Convert the electron moles to Cu reactive moles. In the reduction of${\text{Cu}}^{\text{2 +}}$, two electrons are involved.

$\text{29,376 C×}\frac{{\text{1 mole}}^{\text{-}}}{\text{96,500 C}}\text{×}\frac{\text{1 mol Cu}}{{\text{2 mole}}^{\text{-}}}\text{= 0.152 mol Cu}$

Multiply the moles of Cu by their molar mass to get mass.

$\text{0.152 mol Cu×}\frac{\text{63.55 g}}{\text{1 mol Cu}}\text{= 9.67 g Cu}$

Therefore, the value is: $\text{9.67 g Cu}$

Step 3: What is the [ Cu2 + ] remaining?

The moles of copper ion in the solution are:

$\text{0.345 L×}\frac{\text{1 mol}}{\text{1 L}}{\text{= 0.345 molCu}}^{\text{2 +}}$

Now, subtracting the moles of copper being deposited is:

${\text{0.345 mol Cu}}^{\text{2 +}}{\text{- 0.152 mol Cu}}^{\text{2 +}}{\text{= 0.193 mol Cu}}^{\text{2 +}}$

Then, evaluating the new molarity of the solution as:

$$\begin{gathered} \text{M =}\frac{{\text{0.193 mol Cu}}^{\text{2 +}}}{\text{0.375 L}} \\ \text{= 0.51 M} \end{gathered}$$

Therefore, the value is: ${\text{[Cu}}^{\text{2 +}}\text{]=0.51 M}$