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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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3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen​
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Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

Thus, the total number of molecules turns out:

Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec

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6 0
3 years ago
What is the pH of 1.0M HI solution?
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Answer:

0

Explanation:

Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.

7 0
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Read 2 more answers
What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen?
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Answer:

= C3H4N

Explanation:

We are given; 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen.

We first calculate the number of moles of each element.

Carbon = 90g/12 g/mol

            = 7.5 moles

Hydrogen = 11 g/ 1 g/mol

                 = 11 moles

Nitrogen = 35 g/ 14 g/mol

               = 2.5 moles

The we get the mole ratio of the elements;

= 7.5/2.5 : 11/2.5 : 2.5 /2.5

= 3 : 4.4 : 1

= 3 : 4 : 1

Therefore;

The empirical formula will be; C3H4N

6 0
3 years ago
Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur
yuradex [85]

<u>Answer:</u> The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For magnesium:</u>

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol

  • <u>For iron(III) chloride:</u>

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = \frac{2}{3}\times 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0
3 years ago
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