Q21.110CP - Chemistry: Molecular Nature Of Matter And Change

Question

The ${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}$ used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is 

${\mathbf{\text{Mn}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + 2H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}\mathbf{\to }{\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) + 4H}}}^{\mathbf{\text{ + }}}{\mathbf{\text{(aq) + 2e}}}^{\mathbf{\text{ - }}}$

If a current of is $\mathbf{\text{25.0 A}}$ used, how many hours are needed to produce $\mathbf{\text{1.00 kg}}$ of ${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}$? At which electrode is the ${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}$formed?

Step-by-Step Solution

Answer

The number of hours needed to produce 1.00kg of ${\text{MnO}}_{\text{2}}$  is 24.66h .

${\text{MnO}}_{\text{2}}$ is formed at anode.

Step 1: Concept Introduction

Any process that is either produced or accompanied by the passage of an electric current and involves the transfer of electrons between two substances—one solid and the other liquid—is known as an electrochemical reaction.

Step 2: Information Provided

• A current of $\text{0.855 A(A = C/s)}$  is used to produce ${\text{MnO}}_{\text{2}}$ .
• Find the time (in hours) it takes to produce $\text{1.00 kg}$ of ${\text{MnO}}_{\text{2}}\text{ (1000 g)}$ .
• Faraday constant: $\text{F = 96485 C/mole}$

Step 3: Calculation for Charge

The given balanced half-reaction –

${\text{Mn}}^{\text{2 + }}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\to {\text{MnO}}_{\text{2}}{\text{(s) + 4H}}^{\text{ + }}{\text{(aq) + 2e}}^{\text{ - }}$

It can be seen that for every mole of produced, of electrons are involved.

The number of moles of produced is (the molar mass of is $\text{86.9368 g/mol}$) –

Moles produced $=\frac{1000\mathrm{g}}{86.9368\mathrm{g}/\mathrm{mol}}=11.50 \mathrm{mol} {\mathrm{MnO}}_2$

Hence, the number of moles of electrons involved is –

$\begin{array}{rcl}\mathrm{Moles} {\mathrm{e}}^{-}& =& 11.50 \mathrm{mol} {\mathrm{MnO}}_2.\frac{2 \mathrm{mol} {\mathrm{e}}^{-}}{1 \mathrm{mol} {\mathrm{MnO}}_2}\\ & =& 23 \mathrm{mol} {\mathrm{e}}^{-}\end{array}$

Now, calculate the charge, using Faraday constant –

$\begin{array}{rcl}\mathrm{Charge} & =& {\mathrm{mol}}^{-}·\mathrm{F}\\ & =& 23 \mathrm{mol} {\mathrm{e}}^{-}·96485\frac{\mathrm{C}}{\mathrm{mol} {\mathrm{e}}^{-}}\\ & =& 2219.155\mathrm{C}\end{array}$

Step 4: Calculation for Time Taken

Finally, calculate the time it takes to produce 1.00 kg of $Mn{O}_2$.

$\begin{array}{rcl}\mathrm{Current}& =& \frac{\mathrm{Charge}}{\mathrm{Time}}\\ \mathrm{Time}& =& \frac{\mathrm{Charge}}{\mathrm{Current}}\\ & =& \frac{2219.155\mathrm{C}}{25.0 {\displaystyle \frac{\mathrm{C}}{\mathrm{s}}}}\\ & =& 88766.2\mathrm{s}·\frac{1 \mathrm{h}}{3600 \mathrm{s}}\\ & =& 24.66 \mathrm{h}\end{array}$

Since given half-reaction is oxidation half-reaction, ${\text{MnO}}_{\text{2}}$ forms at anode.

 

Therefore, the value for time is obtained as $24.66 \mathrm{h}$.