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stira [4]
3 years ago
11

2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 Vwhat is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.695 M , [

Co2+]= 0.175 M , and [Cl−]= 0.315 M and the pressure of Cl2 is PCl2= 8.50 atm ?
Chemistry
1 answer:
ch4aika [34]3 years ago
6 0

Answer:

The value of the cell potential under the given conditions is 0.433 V

Explanation:

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

The Nernst equation is useful for finding the potential for reduction in electrodes under conditions other than standards. This is what happens in this case, since the concentrations are different than 1M and the gas pressure varies from the value 1 atm.

The Nernst equation is:

<em> E=E^{o} -\frac{R*T}{n*F} *ln(Q)</em>

<em>Where E refers to the electrode potential. </em>

<em>Eº = potential in standard conditions. </em>

<em>R = gas constant. </em>

<em>T = absolute temperature (in Kelvin degrees). </em>

<em>n = number of moles of electrons that have participation in the reaction. </em>

<em>F = Faraday constant (with a value of 96500 C / mol, approx.) </em>

<em>Q = reaction ratio </em>

<em> When the reaction occurs at 25 ° C, the numerical value of the constants is replaced by 0.059, the expression being as follows:</em>

<em>E=E^{o} -\frac{0.059}{n} *lnQ</em>

For the reaction aA + bB → cC + dD, Q adopts the expression:

Q=\frac{C^{c} *D^{d} }{A^{a} *B^{b} }

being for solutions the molar concentrations at any moment and for gases the pressure in atmospheres at any moment

In this case:

Q=\frac{[Co^{+2}] ^{2}*P_{Cl2}  }{[Co^{+3}] ^{2}*[Cl^{-}] ^{2}  }

Q=\frac{[0.175M] ^{2}*8.5  }{[0.695M] ^{2}*[0.315M] ^{2}  }

<em>Q=5.43</em>

On the other hand, by observing the following semi-reactions it is possible to see that the number of moles of electrons n involved is 2:

2 Cl- ⇒ Cl₂ + 2 e-

2* [Co³⁺ + e- ⇒ Co²⁺ ]

So, the data to be able to calculate the electrode potential in the requested conditions is:

  • <em>E⁰=0.483 V</em>
  • <em>n=2</em>
  • <em>Q=5.43</em>

Replacing you get:

E=0.483 -\frac{0.059}{2} *ln(5.43)

<em>E=0.433 V </em>

So, <u><em>the value of the cell potential under the given conditions is 0.433 V</em></u>

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