The equilibrium constant, K, for a redox reaction is related to the standard potential, E∘, by the equation

lnK=nFE∘RT

where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e−) , R (the gas constant) is equal to 8.314 J/(mol⋅K) , and T is the Kelvin temperature.

Standard reduction potentials

Reduction half-reaction E∘ (V)
Ag+(aq)+e−→Ag(s) 0.80
Cu2+(aq)+2e−→Cu(s) 0.34
Sn4+(aq)+4e−→Sn(s) 0.15
2H+(aq)+2e−→H2(g) 0
Ni2+(aq)+2e−→Ni(s) −0.26
Fe2+(aq)+2e−→Fe(s) −0.45
Zn2+(aq)+2e−→Zn(s) −0.76
Al3+(aq)+3e−→Al(s) −1.66
Mg2+(aq)+2e−→Mg(s) −2.37
Part A

Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction:

Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)

Express your answer numerically.

K =
Part B

Calculate the standard cell potential (E∘) for the reaction

X(s)+Y+(aq)→X+(aq)+Y(s)

if K = 4.88×10−3.

Express your answer to three significant figures and include the appropriate units.

E∘ =

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.