Complete Question
Consider the rechargeable battery: [tex]Zn(s)||ZnCl _{(aq)}||Cl_2_{(aq)}|Cl_2 (l)|C_{(s)}[/tex]
(a) Write reduction half-reactions for each electrode. From which electrode will electrons flow from the battery into a circuit if the electrode potentials are not too different from [tex]E^o[/tex] values
(b)
if the battery delivers a constant current of [tex]1 .00 *10^{3} \ A[/tex] for 1.00 h , how many kg of [tex]Cl_2[/tex] will be consumed
Answer:
a
At the anode
[tex]Zn^{2+} + 2e^{-} \to Zn_{s} \ \ \ E^o = -0.762 V[/tex]
At the cathode
[tex]Cl_2 _{(l)} + 2e^{-} \to 2Cl^{-} _{aq} \ \ \ \ E^o = 1.396 V[/tex]
b
The value is [tex]mCl = 1.32583 \ kg [/tex]
Explanation:
Generally the half-reactions for each electrode is mathematically represented as
At the anode
[tex]Zn^{2+} + 2e^{-} \to Zn_{s} \ \ \ E^o = -0.762 V[/tex]
At the cathode
[tex]Cl_2 _{(l)} + 2e^{-} \to 2Cl^{-} _{aq} \ \ \ \ E^o = 1.396 V[/tex]
Generally from the question we are told that
The current is [tex]I = 1.00 *10^{3} \ A[/tex]
The time is [tex]t = 1.00\ h= 3600[/tex]
Generally the quantity of charge consumed is
[tex]Q = It[/tex]
=> [tex]Q = 1.00 *10^{3} * 3600 [/tex]
=> [tex]Q = 3.6 *10^{6} \ C [/tex]
Generally the number of moles of electron consumed is
[tex]n = \frac{Q}{F}[/tex]
Here F is the faradays constant with value [tex]F = (96500C/mole \ e-)[/tex]
So
[tex]n = \frac{ 3.6 *10^{6} }{96500}[/tex]
=> [tex]n = 37.3 \ moles [/tex]
Generally the number of moles of [tex]Cl_2[/tex] consumed is mathematically represented as
[tex]nCl = \frac{1}{2} * n[/tex]
=> [tex]nCl = \frac{1}{2} * 37.3 [/tex]
=> [tex]nCl = 18.7 \ moles [/tex]
Generally the mass of [tex]Cl_2[/tex] consumed is mathematically represented as
[tex]mCl = nCl * Z[/tex]
Here Z is the molar mass of [tex]Cl_2[/tex] is Z = 70.9 g/mole
So
[tex]mCl = 18.7 * 70.9 [/tex]
=> [tex]mCl = 18.7 * 70.9 [/tex]
=> [tex]mCl = 1325.83 \ g [/tex]
=> [tex]mCl = 1.32583 \ kg [/tex]
