Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
A pure solid is heated and turns into a pure liquid.
Explanation:
No colour change recorded, only change of state, hence this is a physical change - physical changes I.e. change of state and temperature are not chemical reactions.
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
2.41 molecules, should be it
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
mol = mass ÷ molar mass
If mass of hydrazine (N₂H₄) = 5.29 g
then mol of hydrazine = 5.29 g ÷ ((14 ×2) + (1 × 4))
= 0.165 mol
mole ratio of hydrazine to Nitogen is 1 : 1
∴ if moles of hydrazine = 0.165 mol
then moles of nitrogen = 0.165 mol
Mass = mol × molar mass
Since mol of nitrogen (N₂) = 0.165
then mass of hydrazine = 0.165 × (14 × 2)
= 4.62 g
