The concentration of hydrogen ions in a solution is a measure of its acidity. So the correct option is (b) false.
When an Arrhenius acid is dissolved in water, hydrogen ions are produced:
H+(aq) + A- = HA + H2O (aq)
Here, H+ is the hydrogen cation, A- is the solvated anion, also known as the conjugate base, and HA is the non-dissociated acid. When an Arrhenius base is dissolved in water, hydroxide ions are produced:
BOH + H2O → B+(aq) + OH-(aq)
Is a material with at least one hydrogen atom that has the ability to split apart in an aqueous solution to produce an anion and an H + ion (a proton), creating an acidic solution. Bases are substances that, when dissolved in water, create hydroxide ions (OH) and a cation, resulting in a basic solution.
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Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Most transition metal form more than one cation but aluminum forms the Al3+ cation only.
27.6 - 22.3 = 5.3; you simply subtract the old density from the new density.
