Question

(a) What quantity of charge (in coulombs) is a fully charged 1.34-V zinc–mercuric oxide watch battery theoretically capable of furnishing if the mass of HgO in the battery is 0.50 g?

(b) What is the theoretical maximum amount of work (in joules) that can be obtained from this battery?

Short answer

a) The total quantity of charge of zinc-mercuric oxide battery is ${\text{4.4×10}}^{\text{2}}\text{C}$.

b) The maximum amount of work that can be obtained from the zinc-mercuric oxide battery is 590 J.

Step-by-step solution

Part a Step 1: a) Calculate the cell potential of the zinc-mercuric oxide battery. 

Given data:

${\text{E}}_{\text{Cell}}\text{=1.34V}$

Mass of cathode = 0.50 g.

Mass of mercuric oxide cathode = 0.50 g.

Molar mass of mercuric oxide (HgO) = 216.6 g/ mol.

Now, calculate the number of moles of mercuric oxide (HgO).

$\begin{array}{l}\text{=0.50g×}\frac{\text{1molHgO}}{\text{216.6g}}\\ {\text{=2.3×10}}^{\text{-3}}\text{molHgO}\end{array}$

So, the number of moles of mercuric oxide is${\text{2.3×10}}^{\text{-3}}\text{molHgO}$.

The anode half-cell reaction.

$\text{HgO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\to \text{Hg}\left(\text{l}\right){\text{+2OH}}^{\text{-}}\left(\text{aq}\right)$

Two moles of electronsare required toreduce one mole of mercuric oxide.

Here, calculate the number of moles of electrons required to reduce

.${\text{2.3×10}}^{\text{-3}}\text{molHgO}$

$\text{=}\left({\text{2.3×10}}^{\text{-3}}\text{molHgO}\right)\text{×}\left(\frac{{\text{2mole}}^{\text{-}}}{\text{1molHgO}}\right)$

${\text{=4.6×10}}^{\text{-3}}{\text{mole}}^{\text{-}}$

Hence, the number of moles of electrons required to oxidize${\text{2.3×10}}^{\text{-3}}\text{molHgO}$ is${\text{4.6×10}}^{\text{-3}}{\text{mole}}^{\text{-}}$.

Calculate the total quantity charge of the zinc-mercuric oxide battery.

The charge carried by one mole of electrons is

.${\text{=96385C/mole}}^{\text{-}}$

Calculate the total charge

$\left(\text{Q}\right){\text{=4.6×10}}^{\text{-3}}{\text{mole}}^{\text{-}}\text{×}\frac{\text{96485C}}{{\text{mole}}^{\text{-}}}$

$\left(\text{Q}\right){\text{=4.4×10}}^{\text{2}}\text{C}$

.

Therefore, the total number of charges of the zinc-mercuric oxide battery is.

${\text{4.4×10}}^{\text{2}}\text{C}$

Part b Step 4: b) Calculate the maximum amount of work obtained from the zinc-mercuric oxide battery.

The cell potential of the zinc-mercuric oxide battery is

.${\text{E}}_{\text{cell}}\text{=1.34V}$

The total quantity of charge present in a zinc-mercuric oxide battery is

.${\text{4.4×10}}^{\text{2}}\text{C}$

Here calculate the maximum amount of work obtained from the battery,

${\text{W}}_{\text{elec}}{\text{=itE}}_{\text{cell}}$

$\begin{array}{c}{\text{W}}_{\text{elec}}{\text{=QE}}_{\text{cell}}\\ \text{=}\left({\text{4.4×10}}^2\text{C}\right)\text{×}\left(\text{1.34V}\right)\\ \text{=}\left({\text{4.4×10}}^2\text{C}\right)\text{×}\left(\text{1.34}\frac{\text{J}}{\text{C}}\right)\end{array}$

$\begin{array}{l}{\text{=5.9×10}}^2\text{J}\\ \text{=590J}\end{array}$

Hence, the maximum amount of work obtained from the zinc-mercuric oxide battery is 590 J.