91影视

The cell reaction is the overall reaction that occurs in the cell, stated with the assumption that the right-hand electrode is the cathode, i.e., with the premise that the spontaneous reaction occurs in the right-hand compartment.

(a)

The given reaction is 鈥�

${\text{Zn(s) + MnO}}_{\text{2}}{\text{(s) +H}}_{\text{2}}\text{O(l)}鈫�{\text{ZnO(s) + Mn(OH)}}_{\text{2}}\text{(s)}$

The half reaction is given as 鈥�

$$\begin{gathered} {\text{Zn + OH}}^{\text{-}}鈫�{\text{ZnO +H}}_{\text{2}}{\text{O + 2e}}^{\text{-}}\text{oxidation} \\ {\text{MnO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}{\text{O + 2e}}^{\text{-}}鈫�{\text{Mn(OH)}}_{\text{2}}{\text{+ 2OH}}^{\text{-}}\text{reduction} \end{gathered}$$

Here it can be seen that for a mole of reaction, 2 moles of electron flow.

Therefore, the value for moles is obtained as 2 moles.

(b)

It is known that $4.50\mathrm{g}\mathrm{of}\mathrm{Zn}$is oxidized, hence find the mass of ${\mathrm{MnO}}_2\mathrm{and}{\mathrm{H}}_2\mathrm{O}$consumed.

The number of moles of $\mathrm{Zn}$oxidized is (molar mass of $\mathrm{Zn}\mathrm{is}65.38\mathrm{g}/\mathrm{mol}$) 鈥�

data-custom-editor="chemistry" $\begin{array}{rcl}& & \text{Moles Zn =}\frac{\text{4.50 g}}{\text{65.38 g/mol}}\\ & & \text{= 0.0688 mole Zn}\\ & & \end{array}$

For every $1\mathrm{mole}\mathrm{of}{\mathrm{Z}}_{\mathrm{n}}$that is oxidized, 1 mole of ${\mathrm{MnO}}_2\mathrm{and}{\mathrm{H}}_2\mathrm{O}$are consumed, so, the masses of ${\mathrm{MnO}}_2\mathrm{and}{\mathrm{H}}_2\mathrm{O}$consumed are (molar mass of ${\text{MnO}}_{\text{2}}$is $86.9368\mathrm{g}/\mathrm{mol}$, and molar mass of ${\text{H}}_{\text{2}}\text{O}$is $18.01528\mathrm{g}/\mathrm{mol}$)

Moles of ${\text{MnO}}_{\text{2}}$consumed $=0.688\mathrm{mol}路86.9368\mathrm{g}/\mathrm{mol}=5.98\mathrm{g}$

Moles of${\text{H}}_{\text{2}}\text{O}$consumed$=0.0688\mathrm{mol}路18.01528\mathrm{g}/\mathrm{mol}=1.24\mathrm{g}$

Therefore, the values of mass are obtained as$5.98\mathrm{g}\mathrm{and}1.24\mathrm{g}$.

(c)

Faraday constant: charge of 1 mole of electrons ($\text{F = 96485 C/mole}$)

The total mass of reactant consumed is 鈥�

Mass Consumed data-custom-editor="chemistry" $=4.50\mathrm{g}\mathrm{Zn}+5.98\mathrm{g}{\mathrm{MnO}}_2+1.24\mathrm{g}{\mathrm{H}}_2\mathrm{O}=11.72\mathrm{g}$

Therefore, the value for mass is obtained as 11.72 g.

(d)

For every mole of product, of electron flow, so for $0.0688\mathrm{mole}$of product, the moles of electrons that flow is 鈥�

$\begin{array}{rcl}\mathrm{Moles}{\mathrm{e}}^{-}& =& 0.0688\mathrm{mol}\mathrm{product}路\frac{2\mathrm{mol}{\mathrm{e}}^{-}}{1\mathrm{mol}\mathrm{product}}\\ & =& 0.1376\mathrm{mol}{\mathrm{e}}^{-}\end{array}$

Now calculate the charge using Faraday constant 鈥�

$\begin{array}{rcl}\mathrm{Charge}& =& \mathrm{Moles}{\mathrm{e}}^{-}路\mathrm{F}\\ & =& 0.1376\mathrm{mole}{\mathrm{e}}^{-}路96485\mathrm{C}/\mathrm{mol}{\mathrm{e}}^{-}\\ & =& 13276.336\mathrm{C}\end{array}$

Therefore, the value for charge is obtained as$13276.336C$.

(e)

Alkaline batteries do not consist only of cathode and anode, they consist of steel can, outer jacket, brass pin current collector,separator, steal vent, and metal base as well.

Therefore, that is the reason why the batteries are heavier than what is calculated.