Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
1
Explanation:
1 hydrogen is displaced from H2so4 in the reaction
We add up all the various atoms:
C: 55
H: 72
Mg: 1
N: 4
O: 5
55 + 72 + 1 + 4 + 5
= 137
The answer is B.
Answer:
dS= 1.79*169.504
j/k = 303.41 j/k
Explanation:
Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)
dS(Fe3O4) =146.4 j/k
dS(H2) =130.684
dS(Fe) =27.78
dS(H2O) =188.825
dSrxn = dS[product]-dS[reactants]
= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]
= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k
This is the dS for 1mole Fe3O4
for 1.79 mols Fe3O4
dS= 1.79*169.504 j/k = 303.41 j/k
