Chapter 17, Problem 17.115CHP

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{Zn}}_{\left(\text{s}\right)}$ in grams formed from $\text{Ni}{\left(\text{OH}\right)}_{\text{2}}$ has to be given.

Concept Introduction:

The mass of a substance is calculated from the constant current and the time duration of the reaction.  Conversion factors are used along with the atomic weight of the substance.  The general formula for obtaining the mass using the conversion factor is given as,

$\text{Mass}\text{=}\text{Current×time(in min)×}\frac{\text{60s}}{\text{1min}}\text{×}\frac{\text{1}\text{mole}{\text{e}}^{\text{-}}}{\text{96,500C}}\text{×}\frac{\text{1}\text{mole}\text{of}\text{substance}}{\text{n}\text{mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{Atomic}\text{weight}}{\text{1}\text{mole}\text{of}\text{substance}}$

The flow chart below is useful in the calculation of grams or litres of the product.

Figure 1

(b)

Interpretation Introduction

Interpretation:

The time in minutes that is required to fully recharge a dead battery has to be given.

Concept Introduction:

Constant Current:

The number of coulombs of charge passed through the cell is equal to the product of current in amperes (coulombs per second) and time in seconds.  The constant current is calculated using the formula,

$\text{Charge(C)}\text{=}\text{Current(A)×Time(s)}$

Moles of Electrons:

The charge on one mole of electrons is $\text{96,500}\text{C,}$ and the number of moles of electrons passed through the cell is calculated as,

$\text{Moles}\text{of}{\text{e}}^{\text{-}}\text{=}\text{Charge(C)×}\frac{\text{1}\text{mole}{\text{e}}^{\text{-}}}{\text{96,500}\text{C}}$