Answer:
Keq = 1.17 × 10²⁰
Explanation:
Let's consider the following redox reaction.
Cu²⁺(aq) + Ni(s) → Cu(s) + Ni²⁺(aq)
We can identify 2 half-reactions.
Cathode (reduction): Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red = 0.337 V
Anode (oxidation): Ni(s) → Ni²⁺(aq) + 2 e⁻ E°red = -0.257 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.337 V - (-0.257V) = 0.594 V
We can calculate the equilibrium constant (Keq) using the following expression.
where,
n are the moles of electrons transferred
It's likely what's wanted is
<span><span>Li</span>→<span><span>Li</span><span>2+</span></span>+2<span>e−
</span></span>
The reason is because IEs are usually reported from the neutral atom, that is, IE2 is the energy required to remove two electrons from a neutral Li atom, as above, rather than the additional energy required to remove one more electron from an Li+ cation.
Answer:
2
Explanation:
The sample has halved twice.
1/2 of 1/2 is 1/4, and 1/2 of 1 is 1/2.
Answer:
Electrons on atoms have different amounts of energy proportional to the distance of their orbital from the nucleus. So in the flame, electrons get excited and pushed to higher energy levels by the heat energy. When they fall back down, they give off photons of light of different colors, based upon how far they fall.
Answer:
Oxidation occurs when a reactant loses electrons during the reaction. Reduction occurs when a reactant gains electrons during the reaction.