Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
Answer:
4
10
Explanation:
The reaction equation is given as;
Ca(OH)₂ → Ca²⁺ + 2OH⁻
Concentration of Ca(OH)₂ = 5 x 10⁻⁵M
Unknown:
pOH of the solution = ?
pH of the solution = ?
Solution:
Solve for the pOH of this solution using the expression below obtained from the ionic product of water;
pOH = ⁻log₁₀[OH⁻]
Ca(OH)₂ → Ca²⁺ + 2OH⁻
1moldm⁻³ 1moldm⁻³ 2 x 1moldm⁻³
5 x 10⁻⁵moldm⁻³ 5 x 10⁻⁵moldm⁻³ 2( 5 x 10⁻⁵moldm⁻³ )
1 x 10⁻⁴moldm⁻³
Therefore;
pOH = -log₁₀ 1 x 10⁻⁴ = 4
Since
pOH + pH = 14
pH = 14 - 4 = 10
1-4
2-2
3-3
4-1
Hope you are m right
Peter because he’s going faster using more energy