Question

How long will it take to plate out each of the following with a current of 100.0 A?
a. 1.0 kg of Al from aqueous Al³⁺
b. 1.0 g of Ni from aqueous Ni²⁺
c. 5.0 moles of Ag from aqueous Ag⁺

Short answer

(a) Time required to plate out 1 kg of Al is 30 hr

(b) Time required to plate out 1 g of Ni is 33 sec

(c) Time required to plate out 5.0 moles of Ag is 4820 s

Step-by-step solution

Time required to plate out 1 kg of Al

The reduction reaction can be denoted as,

${\mathrm{Al}}^{3+}3{\mathrm{e}}^{-}\to \mathrm{Al}$

The mole number of electron can be denoted as,

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{electron}& =& (1 \mathrm{kg} \mathrm{Al})\left(\frac{{10}^3 \mathrm{mol} \mathrm{Al}}{1 \mathrm{g} \mathrm{Al}}\right)\left(\frac{1 \mathrm{mol} \mathrm{Al}}{27 \mathrm{g} \mathrm{Al}}\right)\left(\frac{3 \mathrm{mol}}{1 \mathrm{mol} \mathrm{Al}}\right)\\ & =& 111.1 \mathrm{mol}\end{array}$

The requirement of charge can be calculated as,

$\begin{array}{rcl}\mathrm{Charge}& =& (111.1 \mathrm{mol})\left(\frac{96485 \mathrm{C}}{1 \mathrm{mol}}\right)\\ & =& 1.07\times {10}^7\mathrm{C}\end{array}$

The time required to charge release can be denoted as,

$\begin{array}{rcl}\mathrm{Time}& =& (1.07\times {10}^7\mathrm{C})\left(\frac{1 \mathrm{s}}{100 \mathrm{C}}\right)\left(\frac{1 \mathrm{hr}}{3600 \sec }\right)\\ & =& 30 \mathrm{hr}\end{array}$

Time required to plate out 1 g of Ni

The reduction reaction can be denoted as,

${\mathrm{Ni}}^{2+}2{\mathrm{e}}^{-}\to \mathrm{Ni}$

The mole number of electron can be denoted as,

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{electron}& =& (1 \mathrm{g} \mathrm{Ni})\left(\frac{1 \mathrm{mol} \mathrm{Ni}}{58.7 \mathrm{g} \mathrm{Ni}}\right)\left(\frac{2 \mathrm{mol}}{1 \mathrm{mol} \mathrm{Ni}}\right)\\ & =& 0.034 \mathrm{mol}\end{array}$

The requirement of charge can be calculated as,

$\begin{array}{rcl}\mathrm{Charge}& =& (0.034 \mathrm{mol})\left(\frac{96485 \mathrm{C}}{1 \mathrm{mol}}\right)\\ & =& 3.28\times {10}^3\mathrm{C}\end{array}$

The time required to charge release can be denoted as,

$\begin{array}{rcl}\mathrm{Time}& =& (3.28\times {10}^3\mathrm{C})\left(\frac{1 \mathrm{s}}{100 \mathrm{C}}\right)\\ & =& 33 \mathrm{s}\end{array}$

Time required to plat out 5.0 mole of Ag

The reduction reaction can be denoted as,

${\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}$

The mole number of electron can be denoted as,

$\begin{array}{rcl}\mathrm{Moles} \mathrm{of} \mathrm{electron}& =& (5 \mathrm{mol} \mathrm{Ag})\left(\frac{1 \mathrm{mol} \mathrm{e}}{1 \mathrm{mol} \mathrm{Ag}}\right)\\ & =& 5 \mathrm{mol}\end{array}$

The requirement of charge can be calculated as,

$\begin{array}{rcl}\mathrm{Charge}& =& (5 \mathrm{mol})\left(\frac{96485 \mathrm{C}}{1 \mathrm{mol}}\right)\\ & =& 4.82\times {10}^5\mathrm{C}\end{array}$

The time required to charge release can be denoted as,

$\begin{array}{rcl}\mathrm{Time}& =& (4.82\times {10}^5\mathrm{C})\left(\frac{1 \mathrm{s}}{100 \mathrm{C}}\right)\\ & =& 4820 \mathrm{s}\end{array}$