Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺ | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺ + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺ + 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺ + 2e⁻⇄ Cu²⁺ + 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺ ⇄ Cu²⁺ + 2Ag
amount of product formed or amount of reactants used / time
Answer:
See explanation
Explanation:
The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.
As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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It is 79 - + 3 = 76 electrons.
