Thermodynamics - AP Chemistry

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Question

You want to prepare a cup of tea. To do so, you pour of tap water at $20^{0}C$ in a cup that does not absorb microwave radiation and heat it in a microwave oven at of power. If you assume a density of $1.00, \frac{g}{mL}$ for the water and know that its specific heat capacity is $4.184, \frac{J}{g^{o}C}$ , what time do you need to set in the microwave oven to heat the water to $60^{o}C$ ?

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Answer

$Q=m\cdot C\cdot\Delta T$

Since the density of water is $1.00, \frac{g}{mL}$ , the mass of is . Plug in known values to the equation and solve.

$100g\cdot\frac{4.184J}{g^{\text o}C}\cdot \left ( 60 - 20 \right )^{\text o}C=16736J$

Use the formula below to find the time needed to heat up the sample of water in the microwave:

$P=\frac{\Delta W}{\Delta t}$

$\Delta t=\frac{16736J}{700W}=23.9s$

Our answer must contain three significant figures.

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Question

How much heat is needed to raise grams of aluminum by $20^{o}:C$ ?

$C_{aluminum}=0.900: \frac{J}{g\cdot^{o}\textrm{C}}$ .

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Answer

To find the amount of heat needed to change the temperature of a given material by a certain amount, we'll need to use the equation for specific heat. The specific heat capacity of a compound represents the amount of energy necessary to raise gram of that substance by $1: ^{o}C$ .

$q=mC\Delta T$

$q=(5: grams)(0.900: \frac{J}{g\cdot ^{o}C})(20: ^{o}C)=90:J$

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Question

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

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Answer

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

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Question

Consider an electrochemical cell that has the following overall reaction:

2 H+(aq) + Sn (s) -> Sn2+ (aq) + H2 (aq)

Which of the following changes would alter the measured cell potential?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

All of these changes would change Q and thus change the measured cell potential.

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Question

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions:

\[Al3+\] = 2.0 M; \[Mn2+\] = 1.0 M.

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

Altering these conditions would increase Q, and thus result in a decrease in the measured cell potential.

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Question

Determine the Ecell for the following reaction at 25 C:

Zn (s) + 2 VO2+ (aq) + 4 H+ -> 2 VO2+ (aq) + Zn2+(aq) + 2 H2O (l)

Given that:

VO2+ (aq) + 2 H+ (aq) + e- -> VO2+(aq) + H2O (l) Eo = 1.00 V

Zn2+ (aq) + 2 e--> Zn (s) Eo = -0.76 V

And

\[ VO2+\] = 2.0 M; \[H+\] = 0.50 M; \[VO2+\] = 1.0 x 10-2M; \[Zn2+\] = 1.0 x 10-1M

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Answer

$E_{cell} = E^0_{cell} - \frac {0.0592}{n} : log : Q$

$E^o_{cell} = 1.00 - (-0.76) = 1.76 V$

$E_{cell} = 1.76 V - \frac{ 0.0592}{2} : log : \frac{[VO^{2+} ]^2 [Zn^{2+} ]}{[VO_2^+ ]^2 [H^+ ]^4 )}$

$= 1.76 V - \frac{0.0592}{2}: log : \frac{[1.0 x 10^{-2} ]^2 [1.0 x 10^{-1} ]}{[2.0]^2 [0.5]^4} = 1.89$

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Question

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions: \[Al3+\] = 1.0 M; \[Mn2+\] = 5.0 M.

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

At these concentrations Q will become smaller, and thus log Q will become smaller. This will give rise to a larger cell potential.

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Question

What is the cell potential of the following cell:

Zn (s) + 2 H+(aq) -> Zn2+ (aq) + H2 (g) Eo = 0.76 V

When the \[Zn2+\] = 1.0 M; PH2 = 1 atm, and the pH in the cathode is 5.2?

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Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log: Q$

$[H^+ ] = 10^{-pH}= 10^{-5.2} = 6.31 x 10^{-6}$

$E_{cell} = 0.76 V - \frac{0.0592}{2} : log : \frac{[Zn^{2+} ] P_{H_2} }{[H^+ ]^2 }$

$E_{cell} = 0.76 - \frac{0.0592}{2} : log: \frac{[1.0]1}{[6.31 x 10^{-6} ]^2} = 0.45 V$

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Question

Calculate the standard cell potential of the following reaction:

Cd(s) + MnO2 (s) + 4 H+ (aq) + -> Cd2+ (aq) + Mn2+ (aq) + 2 H2O (l)

Given:

MnO2 (s) + 4 H+ (aq) + 2e- -> Mn2+ (aq) + 2 H2O (l) Eo = 1.23 V

Cd2+ (aq) + 2 e- -> Cd (s) Eo = -0.40 V

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

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Question

Calculate the standard cell potential of the following reaction:

Zn (s) + Cu2+ (aq) -> Zn2+ (aq) + Cu (s)

Given:

Zn2+(aq)+ 2 e--> Zn (s) Eo = -0.76 V

Cu2+(aq)+ 2 e--> Cu (s) Eo = 0.34 V

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 V

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Question

Calculate the standard cell potential of the following reaction:

Zn (s) + 2 Ag1+ (aq) -> Zn2+ (aq) + 2 Ag (s)

Given:

Zn2+(aq)+ 2 e--> Zn (s) Eo = -0.76 V

Ag1+(aq)+ 1 e--> Ag (s) Eo = 0.80 V

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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Question

Calculate the standard cell potential of the following reaction:

3 F2 (g) + 2 Au (s) -> 6 F- (aq) + 2 Au3+

Given:

F2 (g) + 2 e- -> 2 F- (aq) Eo = 2.87 V

Au3+(aq)+ 3 e--> Au (s) Eo = 1.50 V

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Answer

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 V

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Question

How long would it take to electroplate 28.3g of silver at a constant current of 2.0 amps from a solution of silver nitrate?

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Answer

$Ag^+ + 1 e- \rightarrow Ag(s)$

$28.3g : Ag : x : \frac{1 mol Ag}{107.87g} : x : \frac{1 mol e-}{1 mol Ag^+ } : x : \frac{96485 C}{1 mol e^-} = 25313 C$

$time = \frac{25313 C}{2 A}: x : \frac{1 min}{60 sec} = 211 min$

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Question

How many grams of Cr can be obtained by the electrolysis of a Cr(NO3)3 if 10 amps are passed through the cell for 6 hours?

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Answer

$Cr^{3+} + 3 e- \rightarrow Cr(s)$

$mass = \frac{10 A , x, 6 hr , x , \frac{3600 s}{1 hr}}{ \frac{96485 C}{mol e-} } , x , \frac{1 mol , Cr}{3 mol e-} , x , \frac{51.996 g , Cr}{1 mol , Cr}= 38.8 g$

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Question

How much Al would be precipitated if an AlCl3 solution was electrolyzed for 1.00 hours with 5.00 amps?

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Answer

$Al^{3+} + 3 e- \rightarrow Al(s)$

$mass = \frac{5 A x 1 hr , x , \frac{3600 s}{1 hr} }{\frac{96485 C}{mol e-}} , x , \frac{1 mol , Al }{3 , mol , e-}, x, \frac{26.982 g Al}{1 mol Al} = 1.68 g$

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Question

What current must be past through a solution of AlCl3 for 1.19 hours to produce 40g of solid Al?

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Answer

$Al^{3+} + 3 e- \rightarrow Al(s)$

$40g , Al : x , \frac{1, mol , Al}{26.982g}, x, \frac{3 , mol , e-}{1 , mol , Al} , x , \frac{96485 C}{1 , mol , e^-} = 429108 , C$

$Current = \frac{429108 C}{1.19hr , x , \frac{3600 s}{1hr}} = 100 A$

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Question

In the electrolysis of CaCl2 a 1.15 amp current is passed through the cell for 5.0 hours. How much Ca is produced?

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Answer

$Ca^{2+} + 2 e- \rightarrow Ca(s)$

$mass = \frac{1.15 A , x , 5 , hr , x, \frac{ 3600 , s}{1 , hr} }{\frac{96485, C}{mol , e-}}, x , \frac{1 mol Ca }{2, mol , e-}, x , \frac{40.078 g , Al}{1 mol , Al}= 4.3 g$

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Question

Which of these conditions will result in a spontaneous reaction?

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Answer

To determine whether a reaction will occur spontaneously, one needs to look at whether the Gibbs free energy will decrease. In order for Gibbs free energy to decrease, the combination of the heat of the reaction and the change in entropy must give a negative Gibbs free energy. The equation is:

$\Delta G= \Delta H - T\Delta S$

If a reaction is exothermic, the heat of the reaction is negative. If the reactants go from liquid to gaseous state, the entropy increases (is positive). When the enthalpy is negative and the entropy is positive, the equation must always give a negative solution. This leads to a negative Gibbs free energy, so the reaction will be spontaneous.

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Question

Which of the following reaction conditions cannot result in a spontaneous reaction?

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Answer

To determine whether a reaction will occur spontaneously, one needs to look at whether the Gibbs free energy will decrease. In order for Gibbs free energy to decrease, the combination of the heat of the reaction and the change in entropy must give a negative Gibbs free energy. The equation is:

$\Delta G= \Delta H - T\Delta S$

If a reaction is endothermic, the heat of the reaction is positive. If the reactants go from gaseous to liquid state, the entropy decreases (is negative). When the enthalpy is positive and the entropy is negative, the equation must always give a positive solution. This leads to a positive Gibbs free energy, so the reaction can never be spontaneous.

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Question

Which of the following values must be negative in order for a reaction to be spontaneous?

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Answer

In order to determine if a reaction is spontaneous, one needs to look at $\Delta G$ , which is the change in Gibbs free energy. The formula to find change in Gibbs free energy is:

$\Delta G = \Delta H -T\Delta S$

$\Delta H$ and $\Delta S$ can have positive or negative values (under certain conditions), but $\Delta G$ must always be negative in order for the reaction to be spontaneous.

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