Answer:
An input of heat energy Yr don't you know stupied
Answer:
The reaction combines the sodium with the hydrogen and oxygen in water to form sodium hydroxide and hydrogen gas, and you get a lot of energy released as heat as well. This heat actually melts any remaining sodium that has not reacted yet, and ignites the hydrogen gas, so you get the bang and the flash.
Explanation:
zoom in a bit more thanks
<span>Separate this redox reaction into its component half-reactions.
Cl2 + 2Na ----> 2NaCl
reduction: Cl2 + 2 e- ----> 2Cl-1
oxidation: 2Na ----> 2Na+ & 2 e-
2) Write a balanced overall reaction from these unbalanced half-reactions:
oxidation: Sn ----> Sn^2+ & 2 e-
reduction: 2Ag^+ & 2e- ----> 2Ag
giving us
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
</span>
Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)

![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)

Now we have to calculate the standard Gibbs free energy.
Formula used :

where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.

Therefore, the standard Gibbs free energy is +91 kJ/mole