Answer:
Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.
Answer:- 2.92 kJ of heat is released.
Solution:- We have water at 100 degree C and it's going to be cool to 15 degree C.
So, change in temperature,
= 15 - 100 = -85 degree C
mass of water, m = 8.2 g
specific heat of water, c = 
The equation used for solving this type of problems is:

Let's plug in the values in the equation and solve it for q which is the heat energy:
q = (8.2)(4.184)(-85)
q = -2916.248 J
They want answer in kJ. So, let's convert J to kJ and for this we divide by 1000.

q = -2.92 kJ
Negative sign indicates the heat is released. So, in the above process of coiling of water, 2.92 kJ of heat is released.
Answer:
the valence electrons of atoms in a pure metal can be modeled as a sea of electrons
When reversing a given reaction, we simply change the sign of the standard enthalpy change value. Therefore, the reaction will become:
H₂O → H₂ + 0.5O₂, ΔH = 286kJ
This is because if a certain amount of energy is released when a reaction occurs, the same amount of energy must be supplied for the reaction to occur in the reverse direction.