Chapter 14, Problem 14.AE

Interpretation Introduction

Interpretation:

• Cell voltage has to be calculated.
• How many kilo grams of $\text{HgO}$ will be consumed if power required to operate the pacemaker is $\text{0}\text{.0100}\text{W}$ in 365days has to be calculated and this is how many pounds of $\text{HgO}$ has to be calculated.

Answer to Problem 14.AE

• Cell voltage is found to be $\text{1}\text{.35V}$.
• $\text{0}\text{.262 kg}$ $\text{HgO}$ has consumed that is equal to $\text{0}\text{.578 Ib}$.

Explanation of Solution

Given data:

${\text{E}}^{\text{o}}=\text{1}\text{.35 V}$

To calculate: cell voltage

The cell voltage = 1.35V.  Because of all activity are unity.

To calculate: hoe many kilograms of $\text{HgO}$ will be consumed when power required to operate the pacemaker is $\text{0}\text{.0100}\text{W}$ in 365days

Faraday constant $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$

$\begin{array}{l}\text{I =}\raisebox{1ex}{$\text{P}$}\!\left/ \!\raisebox{-1ex}{$\text{E}$}\right.\\ \\ \text{=}\raisebox{1ex}{$\text{ 0}\text{.0100}\text{W}$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.35V}$}\right.\\ \\ \text{= 7}{\text{.41×10}}^{\text{-3}}\text{C/s}\\ \\ \text{mol}{\text{e}}^{\text{-}}\text{/s =}\left(\text{7}{\text{.41×10}}^{\text{-3}}\text{C/s}\right)\left(\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}\right)\\ \\ \text{= 7}{\text{.68×10}}^{\text{-8}}\text{mol}{\text{e}}^{\text{-}}\text{/365d}\end{array}$

2 electrons are accepted by $\text{HgO}$ when it is reduced to $\text{Hg}\left(\text{II}\right)$ from $\text{Hg}\left(0\right)$, therefore rate of the consumption is,

$\left(\text{2}{\text{.42mol e}}^{\text{-}}\text{/365d}\right)\left(\text{1mol HgO/2mol}{\text{e}}^{\text{-}}\right)\text{=1}\text{.21mol HgO/365d = 0}\text{.262kg HgO }$

One pound is equal to $\text{453}\text{.6}\text{g}$

Therefore, $\text{0}\text{.262kg HgO}$ is equal to $\text{0}\text{.578}\text{Ib}$.

Conclusion

• Cell voltage was found to be $\text{1}\text{.35V}$.
• It is found to be $\text{0}\text{.262 kg}$ $\text{HgO}$ has consumed that is equal to $\text{0}\text{.578 Ib}$.