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?

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.

• 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.

The reaction is –

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

The half-reaction forreduction is –

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

Convertdays 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 –

$\begin{array}{rcl}& & \text{Current =}\frac{\text{Charge}}{\text{Time}}\\ & & \text{Charge = Current.Time}\\ & & {\text{= 0.855 C/s.2.16×10}}^{\text{5}}\text{S}\\ & & {\text{= 1.847.10}}^{\text{5}}\text{C}\end{array}$

Now, calculate the number of moles of electrons –

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

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

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

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

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

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