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In a zinc manganese dioxide dry cell, the anode is made up of zinc and the cathode of a carbon rod surrounded by a mixture of$\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$, carbon, $\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }$ and $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ in aqueous base. The cathodic reaction may be represented as:
$\text{ 2Mn}{{\text{O}}_{\text{2}}}\text{(s)+Z}{{\text{n}}^{\text{2+}}}\text{+2}{{\text{e}}^{\text{-}}}\to \text{ZnM}{{\text{n}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ (s)}$
 Let there be $\text{ 8 g }$ $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$ in the cathodic compartment. How many days will the dry cell continue to give a current of$\text{ }4\times {{10}^{-3}}~\text{ampere }$?

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Last updated date: 20th Jun 2024
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Answer
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Hint: The faraday's law is a combination of faraday's first law and second law. The relation is given as follows:
$\text{ W = }\dfrac{\text{ I}\times \text{t}\times \text{E}}{\text{F}}\text{ }$
Where W is the weight of the substance given, I is the current flowing through the cell, t is the time required in seconds, E is the equivalent mass of the substance and F is the faraday's constant $\text{ }96500\text{ }{{\text{C}}^{-1}}\text{ }$.

Complete step by step answer:
A zinc-carbon battery is a dry cell. It provides the direct current produced from the electrochemical reaction between zinc and manganese oxide $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$. In the cell, the zinc is a negatively charged electrode that is oxidized by charge carriers. The chemical half-reaction which takes place at the cathode is given as follows:
$\text{ 2Mn}{{\text{O}}_{\text{2}}}\text{(s)+Z}{{\text{n}}^{\text{2+}}}\text{+2}{{\text{e}}^{\text{-}}}\to \text{ZnM}{{\text{n}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ (s)}$
We observe that the current is produced when zinc reacts with the $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$. Thus, the net current produced by the reaction will depend on the amount of reactant.

Here, when all the $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$ is used up in the cathodic process, then the dry cell will stop producing current.
$\text{ 2}\overset{+4}{\mathop{\text{Mn}}}\,{{\text{O}}_{\text{2}}}\text{(s)+Z}{{\text{n}}^{\text{2+}}}\text{+2}{{\text{e}}^{\text{-}}}\to \text{Zn}{{\overset{+3}{\mathop{\text{Mn}}}\,}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ (s)}$
During the cathodic process, the oxidation state of manganese $\text{ Mn }$ changes from $\text{ +4 }$to$\text{ +3 }$.

We are provided with the following data,
The amount of manganese is $\text{ 8 g }$
The current produced by the dry cell is $\text{ }4\times {{10}^{-3}}~\text{ampere }$
We have to find the number of days this dry cell will continue to give $\text{ }4\times {{10}^{-3}}~\text{ampere }$ current.
Here, Faraday first and second law. The relation obtained by combining faraday laws is as follows,
$\text{ W = }\dfrac{\text{ I}\times \text{t}\times \text{E}}{96500}\text{ }$ (1)
Where,
W is the weight of the compound
I is the current produced
t is time in seconds
E is the equivalent mass

Faraday's constant $\text{ }96500\text{ }{{\text{C}}^{-1}}\text{ }$
Let’s first calculate the equivalent mass of $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$.
$\text{ Equivalent mass of Mn}{{\text{O}}_{\text{2}}}\text{ = }\dfrac{\text{Molecular mass}}{\text{Change in oxidation state}}\text{ = }\dfrac{87}{1}\text{ = 87 }$
Thus, the equivalent mass $\text{ Mn}{{\text{O}}_{\text{2}}}\text{ }$is 87.
Substitute these values in equation (1). we have,
$\begin{align}
  & \text{ 8 = }\dfrac{\text{ 4}\times {{10}^{-3}}\times \text{t}\times 87}{96500}\text{ } \\
 & \Rightarrow \text{t = }\dfrac{\text{8}\times 96500}{\text{4}\times {{10}^{-3}}\times 87} \\
 & \therefore \text{t = 2218390}\text{.805 sec} \\
\end{align}$
The time of the dry cell is equal to $\text{2218390}\text{.805 sec}$.
The time in days is,
$\text{ t = 2218390}\text{.805 sec = }\dfrac{\text{2218390}\text{.805 sec}}{3600\times 24}\text{ = 25}\text{.675 days }$
Thus, the zinc –manganese oxide will continue up to $\text{25}\text{.675 days }$ giving out the current of $\text{ }4\times {{10}^{-3}}~\text{ampere }$

Note: Note that, the time required was asked in terms of days. But Faraday's law only considers time in terms of seconds. Thus, convert the time from seconds to the days. We know that each day has 24 hours, each hour contains 60minute and 1 minute is equal to 60 seconds. Thus seconds in the 24 hours or 1 day can be written as,
$\begin{align}
  & \text{ 1 min = 60 sec } \\
 & \text{ 1 hr = 60 min } \\
 & \text{ 24 hr = 1 day } \\
 & \therefore \text{1 day = 24 }\times \text{ 60 min }\times \text{ 60 sec = 24 }\times \text{ 3600 sec } \\
\end{align}$