CuCl2<span> + Fe ⟶ Cu + FeCI</span><span>2</span>
Answer:
Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²
Explanation:
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²
Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48
Ca. #Substrate e⁻ = 6F = 6(8) = 48
#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs
#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs
BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²
XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)
Answer:
1.The electrode on the right is positive
2. 0.058V
Explanation:
The above cell is a concentration cell.
A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).
Part 2: Please, see the attachment below for the calculations.