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⁻
I believe you have to label out the positive metal ion and the delocalized electrons. They're the 2 things that makes up a metal structure.
In the diagram, the circles with the + symbol are the positive metal ions, since + represents positive. And the remaining - circles are the delocalized electrons, as electrons are negative.
And for how a metal conducts electricity, since they're delocalized mobile electrons present in any metal structures, they're able to move away from the metal to the positive side of the battery and more electrons can replace their place flowing from the negative side.
Its d its the only one that makes sense
Answer:
why did you post a link to brainly when we are on brainly already??????
Explanation:
Answer: B
Explanation:
According to Ohm's Law, the answer is B.
Ohm's Law states that power is equal to volume x current.
If volume x current equals power, that means they are both 50% of power.
Ohm's Law:
power = voltage x current
current = voltage x power
voltage = power x current
I hope this answer helped.
