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Black_prince [1.1K]
3 years ago
7

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.

Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) S (s) (nonspontaneous)? Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq) (nonspontaneous)? 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq) (spontaneous?) Are these correct?
Chemistry
2 answers:
sammy [17]3 years ago
8 0
A electrochemical reaction is said to be spontaneous, if E^{0} cell is positive. 

Answer 1:
Consider reaction: <span>Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s) 

The cell representation of above reaction is given by;
    </span>S^{2-}/S //  Ni^{2+}/Ni

Hence, E^{0}cell =  E^{0} Ni^{2+/Ni} -  E^{0} S/S^{2-}
we know that, {E^{0} Ni^{2+}/Ni  = -0.25 v
and {E^{0} S/ S^{2-}  = -0.47 v

Therefore, E^{0} cell = - 0.25 - (-0.47) = 0.22 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2: 
Consider reaction: <span>Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)
</span>
The cell representation of above reaction is given by;
    H_{2} /  H^{+} //  Pb^{2+} /Pb

Hence, E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}
we know that, {E^{0} Pb^{2+}/Pb = -0.126 v
and {E^{0} H_{2}/ H^{+} = -0 v

Therefore, E^{0} cell = - 0.126 - 0 = -0.126 v

Since,  E^{0} cell is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3: 
Consider reaction: <span>2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)
</span>
The cell representation of above reaction is given by;
    Cr/Cr^{2+} // Ag^{+}/Ag

Hence, E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}
we know that, {E^{0} Ag^{+}/Ag = -0.22 v
and {E^{0} Cr/ Cr^{2+} = -0.913 v

Therefore, E^{0} cell = - 0.22 - (-0.913) = 0.693 v

Since,  E^{0} cell is positive, hence cell reaction is spontaneous
solong [7]3 years ago
8 0

Answer: Ni^{2+}(aq)+S^{2-}(aq)\rightarrow Ni(s)+S(s)  : non spontaneous

Pb^{2+}(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)  : non spontaneous

2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)  : spontaneous

Explanation:

a) Ni^{2+}(aq)+S^{2-}(aq)\rightarrow Ni(s)+S(s)

Here S undergoes oxidation by loss of electrons, thus act as anode. Ni undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Ni^{2+}/Ni]}=-0.25V

E^0_{[S^{2-}/S]}=0.407VV

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[S^{2-}/S]}

E^0=-0.25-(0.407V)-0.657V

As value of E^0 is negative, the reaction is non spontaneous.

b) Pb^{2+}(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)

Here Hydrogen undergoes oxidation by loss of electrons, thus act as anode. Pb undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Pb^{2+}/Pb]}=-0.13

E^0_{[H^{+}/H_2]}=0V

E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[H^{+}/H_2]}

E^0=-0.13-(0V)=-0.13V

As value of E^0 is negative, the reaction is non spontaneous.

c) 2Ag^{+}(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^{2+}(aq)

Here Cr undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Ag^{+}/Ag]}=+0.80V

E^0_{[Cr^{2+}/Cr]}=-0.913V

E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Cr^{2+}/Cr]}

E^0=+0.80-(-0.913V)=1.713V

As value of E^0 is positive, the reaction is spontaneous.

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Salsk061 [2.6K]

The solubility KI is 50 g in 100 g of H₂O at 20 °C. if 110 grams of ki are added to 200 grams of H₂O <u>the </u><u>solution </u><u>will be </u><u>saturated</u><u>.</u>

<h3>What is solubility?</h3>

Solubility is a condition where the solute is fully dissolved in the solvent. When fully mixed with the solvent.

Given that 50 g of KI is added to 100 g of water at 20 °C it means 100 g of water can dissolve a maximum of 50 g of  KCl.

1 g of water will dissolve an quantity of 0.5 g of  KCl.

To assay for 200 g of water: 200 g of water can disintegrate a maximum of (0.5) x 200 g of  KCl.

The maximum amount of KCl that will dissolve is 100 g

Actualised amount dissolved = 110 g

when Amount dissolved > Maximum solubility limit

110 g > 100 g

Thus,  the solution is saturated.

To learn more about solubility, refer to the below link:

brainly.com/question/8591226

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6 0
1 year ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
nikdorinn [45]

Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

\text{Moles of }Al=\frac{55g}{27g/mol}=2mole

Now we have to calculate the moles of O_2.

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of O_2

So, 2 mole of Al react with \frac{3}{4}\times 2=0.75\times 2 moles of O_2

Now we have to calculate the volume of O_2 consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of O_2 occupies 22.4 liter volume of O_2 gas

So, 0.75\times 2 mole of O_2 occupies 0.75\times 2\times 22.4 liter volume of O_2 gas

Therefore, the volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

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Ethanol will evaporate first because it is more volatile or has a lower boiling point than water.

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