Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Answer:
false
Explanation:
a physical property can be observed and stay the same but a chemical property can't
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
