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USPshnik [31]
3 years ago
14

Reduction and oxidation must occur together. The electrons from the oxidized species are transferred to the reduced species. Che

mists often break these two processes apart and write what are referred to as "half reactions." Consider these two half reactions: Zn2+ (aq) + 2 e- → Zn (s) Cu2+ (aq) + 2 e- à Cu (s) a. Are these oxidation or reduction half reactions? b. Calculate ΔG° for each of these reactions. Note that ΔG°f for the electron is 0 kJ/mol. c. Based on your answer to (b), does Zn2+ or Cu2+ more strongly favor reduction?
Chemistry
1 answer:
Aneli [31]3 years ago
7 0

Explanation:

When a specie tends to lose electrons in a chemical reaction then it means oxidation has taken place. When a specie tends to gain electrons then it means reduction has taken place.

So, for the given reactions the value of E_{o} are as follows.

    Zn^{2+} + 2e^{-} \rightarow Zn,      E_{o} = -0.76 V

    Cu^{2+} + 2e^{-} \rightarow Cu,      E_{o} = 0.34 V

   \Delta G^{o} = -nFE^{o}_{cell}

(a)   Therefore, according to the given reactions the process is reduction.

(b)   Here,   n = 2

So, value of Zn^{2+} is as follows.

     \Delta G^{o}_{Zn^{2+}} = -2 \times (-0.76) \times 96500

                   = +146680 V

  \Delta G^{o}_{Cu^{2+}} = -2 \times (-0.34) \times 96500

                   = -65620 V

(c)  As per the calculated values, Zn^{2+} is the best reducing agent and Cu^{2+} is the best oxidizing agent.

Also, more positive is the E_{o} value more good it acts as an oxidizing agent. Hence, it is able to favor reduction.

Therefore, Cu^{2+} will more strongly favor reduction.

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Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
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Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

<em />

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A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

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It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

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Regards!

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