Q21.109 P - Chemistry: Molecular Nature Of Matter And Change

Question

Zinc plating (galvanizing) is an important means of corrosion protection. Although the process is done customarily by dipping the object into molten zinc, the metal can also be electroplated from aqueous solutions. How many grams of zinc can be deposited on a steel tank from a  ${\mathbf{\text{ZnSO}}}_{\mathbf{\text{4}}}$ solution when a $\mathbf{\text{0.855 - A}}$  current flows for $\mathbf{\text{2.50}}$  days?

Step-by-Step Solution

Answer

The amount of Zinc that can be deposited on a steel tank from a $\mathrm{ZnS}{O}_4$  solution is $\text{62.57 g}$

Step 1: Concept Introduction

The act of adding a protective zinc coating to steel or iron to prevent rusting is known as galvanization or galvanizing (sometimes spelled galvanization or galvanizing). The most popular process is hot-dip galvanizing, which involves immersing the pieces in molten zinc.

Step 2: Information Provided

• A current of $\text{0.855 A(A = C/s)}$ is passed through a   ${\text{ZnSO}}_{\text{4}}$solution, and it flows for  2.5 days.
• Faraday constant:  $\text{F = 96485 C/mole}$
• Find the mass of  Zn (in grams) that can be deposited.

 

Step 3: Calculation for Charge

The reaction is –

 ${\text{ZnSO}}_{\text{4}}\to {\text{Zn}}^{\text{2 + }}{\text{ + SO}}_{\text{4}}^{\text{2 - }}$

The half-reaction for   reduction is –

 ${\text{Zn}}^{\text{2 + }}{\text{ + 2e}}^{\text{ - }}\to \text{Zn}$

Convert   days into seconds –

 $$\begin{gathered} \text{Time = 2.5d.}\frac{24h}{1d}\text{.}\frac{3600s}{1h} \\ {\text{=2.16.10}}^{\text{5}}\text{ s} \end{gathered}$$

Since the value of the current and the time is known, calculate the charge –

 $\begin{array}{rcl}& & \text{Current = }\frac{\text{ Charge }}{\text{ Time }}\\ & & \text{Charge = Current . Time}\\ & & {\text{ = 0.855 C/s . 2.16×10}}^{\text{5}}\text{S}\\ & & {\text{ = 1.847. 10}}^{\text{5}}\text{ C}\end{array}$

Step 4: Calculation for Mass of Zinc

Now, calculate the number of moles of electrons –

 $$\begin{gathered} {\text{Charge = Moles}}^{\text{ - }}·{\text{F Moles}}^{\text{ - }} \\ \text{ = }\frac{\text{ Charge }}{\text{F}} \\ \text{ = }\frac{\text{1.847}·{\text{10}}^{\text{5}}\text{C}}{{\text{96485 C/mole}}^{\text{ - }}} \\ {\text{ = 1.914 mole}}^{\text{- }} \end{gathered}$$

Since  of electrons can deposit  $\text{1 mole}$ of Zn , the number of moles of  Zn deposited is –

Moles Zn deposited $=1.914{\text{mole}}^{\text{-}}\text{ . }\frac{1 mol Zn}{2 mol{e}^{-}}\text{=0.957 mol Zn}$

The molar mass of  Zn is 65.38 g/ mol, so, the mass of  Zn deposited is –

Mass of Zn deposited $0.957 mol Zn. 65.38 g/mol=62.57 g$

 

Therefore, the value for mass is obtained as  $\text{62.57 g}$.