Q21.112Cp - Chemistry: Molecular Nature Of Matter And Change

Question

The overall cell reaction occurring in an alkaline battery is 

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

(a) How many moles of electrons flow per mole of reaction? 

(b) If $\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{g}$ of zinc is oxidized, how many grams of manganese dioxide and of water are consumed? 

(c) What is the total mass of reactants consumed in part (b)? 

(d) How many coulombs are produced in part (b)? 

(e) In practice, voltaic cells of a given capacity (coulombs) are heavier than the calculation in part (c) indicates. Explain.

 

Step-by-Step Solution

Answer

(a) The number of moles of electrons that flow per mole of reaction is 2 moles. .

(b) The amount of manganese dioxide and water that are consumed is $5.98 \mathrm{g} \mathrm{and} 1.24 \mathrm{g}$ and respectively.

(c) The total mass of reactants consumed is $11.72 \mathrm{g}$.

(d) The charged produced is $13276.336 C$ .

(e) Batteries are heavier because they have a steel container, outer jacket, brass pin current collector, separator, steal vent, and metal base in addition to the cathode and anode.

 

Step 1: Concept Introduction

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.

Step 2: Calculation for Number of moles

(a)

The given reaction is –

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

The half reaction is given as –

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

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.

Step 3: Calculation for Mass

(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 $\mathrm{Zn} \mathrm{is} 65.38 \mathrm{g}/\mathrm{mol}$) –

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 $=0.688 \mathrm{mol}·86.9368 \mathrm{g}/\mathrm{mol}=5.98 \mathrm{g}$

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

 

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

 

Step 4: Calculation for Total Mass

(c)

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

The total mass of reactant consumed is –

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.

Step 5: Calculation for Charge

(d)

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

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

Now calculate the charge using Faraday constant –

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

 

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

Step 6: Charge is heavier than Mass

(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.