The Nernst equation is:
E = E° - RTlnK/nF
where
E° is the standard potential voltage
R is the universal gas constant = 8.314 J/mol·K
K is the reaction quotient
n is the number of moles electrons transferred
F is Faraday's constant = 96,500 C/mol e⁻
Let's determine K first. The overall reaction is:
Fe(s) + Cd²⁺(aq) --> Fe²⁺(aq) + Cd(s)
Accounting for aqueous phases only,
K = [products]/[reactants] = [Fe²⁺]/[Cd²⁺] = 0.10 M/1.4 M = 1/14
From the reactions written, you can see that 2 electrons were transferred. So, n = 2.
Lastly, the value for E⁰ is the sum of individual E⁰ of the reactions.
E⁰ = E⁰cathode - E⁰anode
Cathode reaction: Fe(s) --> 2e + Fe²⁺ (E⁰cathode = 0.44 V)
Anode reaction: Cd²⁺ + 2e --> Cd(s) (E⁰anode = -0.4 V)
Thus,
E⁰ = 0.44 - -0.4 = 0.84 V
Substituting the values (assume T at room temperature = 298 K),
E = 0.84 - (8.314)(298 K)(ln 1/14)/(2)(96,500)
<em>E = 0.87 V</em>
<span>Polar covalent & Ionic are the two materials
</span>
Answer:
Rubidium and cesium
Explanation:
It is noteworthy to say here that larger cations have more stable superoxides. This goes a long way to show that large cations are stabilized by large cations.
Let us consider the main point of the question. We are told in the question that the reason why potassium reacts with oxygen to form a superoxide is because of its low value of first ionization energy.
The implication of this is that, the other two metals that can be examined to prove this point must have lower first ionization energy than potassium. Potassium has a first ionization energy of 419 KJmol-1, rubidium has a first ionization energy of 403 KJ mol-1 and ceasium has a first ionization energy of 376 KJmol-1.
Hence, if we want to validate the hypothesis that potassium's capacity to form a superoxide compound is related to a low value for the first ionization energy, we must also consider the elements rubidium and cesium whose first ionization energies are lower than that of potassium.
<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>