Chemistry, 03.12.2019 17:31 adriana145

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 ?

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Answer from: ItsNataly101

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:

$E=E^{o} -\frac{R*T}{n*F} *ln(Q)$

Where E refers to the electrode potential.

Eº = potential in standard conditions.

R = gas constant.

T = absolute temperature (in Kelvin degrees).

n = number of moles of electrons that have participation in the reaction.

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

Q = reaction ratio

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

$E=E^{o} -\frac{0.059}{n} *lnQ$

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:

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

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

Q=5.43

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:

E⁰=0.483 Vn=2Q=5.43

Replacing you get:

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

E=0.433 V

So, the value of the cell potential under the given conditions is 0.433 V

Answer from: Quest

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

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

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.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

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

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

The correct answer is b.hoped i you

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