Question

Question:A zinc-copper battery is constructed as follows:

Zn | Zn²⁺(0.10 M) | | Cu²⁺(2.50 M) | Cu

The mass of each electrode is 200. g.

a. Calculate the cell potential when this battery is first connected.

b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.)

c. Calculate the mass of each electrode after 10.0 h.

d. How long can this battery deliver a current of 10.0 A before it goes dead?

Short answer

1. $E^{\circ }=1.10\hspace{0.17em}V$
   
2. $1.09V$
   
3. $39\mathrm{g}\mathrm{Cu},78\mathrm{g}\mathrm{Zn}$
   
4. role="math" localid="1663742780218" $t=16.9\mathrm{hours}$
   

Step-by-step solution

Determining the cell potential when this battery is first connected

$\overline{)\mathrm{Zn}}{\mathrm{Zn}}^{+2}\left(0.1\mathrm{m}\right)\left|\right|\overline{){\mathrm{Cu}}^{+2}\left(2.5\mathrm{m}\right)}\mathrm{Cu}$

Mass of each electrode as$200\mathrm{g}$

localid="1663743373544" $\begin{array}{l}Zn\to Z{n}^2+2e^{-}\left(E^{\circ }=-0.76V\right)\\ C{u}^{+2}+2e^{-}\to Cu\left(E^{\circ }=0.34V\right)\end{array}$

Overall reaction;

$Zn+C{u}^{2+}\underset{}{\to }Cu+Z{n}^{2+}$

$\begin{array}{rcl}{E}_{cell}^{\circ }& =& E_{reduction}-E_{oxidation}\\ & =& 0.34V-\left(-0.76V\right)\\ & =& 1.10V\end{array}$

Hence, the cell potential of the battery is $1.10\mathrm{V}$

Calculating the cell potential when 10.0 A current has flowed for 10.0 h. 

Rate at which metal is being reduced or produced;
$\frac{60\min }{1\mathrm{hour}}\times \frac{60\mathrm{s}}{1\min }\times \frac{10\mathrm{C}}{1\mathrm{s}}\times \frac{1\mathrm{mol}}{96485\mathrm{C}}\times \frac{\mathrm{mole}}{2\mathrm{mol}{\mathrm{e}}^{-1}}=0.187\mathrm{moles}/\mathrm{hour}$

Concentration of $\left[Z{n}^{+2}\right]$after $10.0\mathrm{hour}$$10.0\mathrm{hours}$

$\left[{\mathrm{Zn}}^{2+}\right]=0.10\mathrm{M}+10\mathrm{hour}\times 0.187\mathrm{M}/\mathrm{hour}=1.97\mathrm{M}$

Concentration of role="math" localid="1663755903924" $\left[{\mathrm{Cu}}^{+2}\right]$after $10.0\mathrm{hour}$

$\left[{\mathrm{Cu}}^{2+}\right]=2.5\mathrm{M}-10\mathrm{hour}\times 0.187\mathrm{M}/\mathrm{hour}=0.634\mathrm{M}$

$\begin{array}{l}{\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.0591}{\mathrm{n}}log\frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{+2}\right]}\\ {\mathrm{E}}_{\mathrm{cell}}=1.10-\frac{0.0591}{2}\log \frac{1.97}{0.634}\\ {\mathrm{E}}_{\mathrm{cell}}=1.09\mathrm{V}\end{array}$

Hence, the cell potential is$1.09\mathrm{V}$.

Calculating the mass of electrode after 10.0 hr

Coulombs produced

$\frac{10\mathrm{C}}{1\mathrm{s}}\times \frac{60\min }{1\mathrm{hr}}\times \frac{60\mathrm{s}}{1\min }\times 10\mathrm{hr}=36000\mathrm{C}$

We know that anode is giving away zinc metal ions, then mass of anode shall decrease

$36000\mathrm{C}\times \frac{1\mathrm{mol}}{96485\mathrm{C}}\times \frac{1\mathrm{mol}\mathrm{Zn}}{2\mathrm{mol}{\mathrm{e}}^{-1}}\times 63.5\mathrm{g}/\mathrm{mol}=112\mathrm{g}$

Mass of anode$=200\mathrm{g}-112\mathrm{g}=78\mathrm{g}$

We know that at the cathode, more and more Cu is being deposited. Hence, the mass of cathode shall increase;

$36000\mathrm{C}\times \frac{1\mathrm{mol}}{96485\mathrm{C}}\times \frac{1\mathrm{mol}\mathrm{Zn}}{2\mathrm{mol}{\mathrm{e}}^{-1}}\times 65.39\mathrm{g}/\mathrm{mol}=119\mathrm{g}$

Mass of cathode $=200\mathrm{g}+119\mathrm{g}=319\mathrm{g}$

Hence, the mass of the cathode is $319\mathrm{g}$

Determining the time of the battery passing current after it got dead

The battery will die when the amount of zinc will exhaust

$\begin{array}{rcl}\mathrm{t}& =& 200\mathrm{g}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}}\times \frac{1\mathrm{s}}{10\mathrm{C}}\times \frac{96485\mathrm{C}}{1\mathrm{mol}}\times \frac{2\mathrm{mol}\mathrm{Zn}}{63.5\mathrm{g}/\mathrm{mol}}\\ & =& 16.9\mathrm{hr}\end{array}$

Hence, the time of the battery passing current after it got dead is$16.9\mathrm{hr}$.