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Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

0.0027432 using scientific notation

alexira [117]

Answer: 2.7423 × 10^-3

Explanation:

4 0

3 years ago

Calculate the formula mass of magnesium chloride, MgCl2?

viktelen [127]

Mg = 24.3
Cl = 35.5

24.3 + 35.5 x 2 = 95.3 ~ 95.21 ( all periodic tabes have different accuracies)

Let me know if you have any questions and please give brainliest if you like my answer:)

8 0

3 years ago

Why is a quantitative observation more useful than a non- quantitative one? Which of the following are quantitative?

daser333 [38]

A quantitative observation is not necessarily more useful than a non-quantitative one. However, quantitative observations do allow one to ﬁnd trends.

(a), the sun rising is a non-quantitative observation.

(b), knowledge of the numerical relationship between the weight on the Moon and on Earth, is a quantitative observation.

(c), watching ice ﬂoat on water does not involve a measurement; therefore, it must be a qualitative observation.

(d) the fact that we know that the water pump won’t work for depths more than 34 feet makes it quantitative. Again, seeing numbers is a giveaway that it’s a quantitative observation. Quantitative is where you deal with numbers.

7 0

3 years ago

PLEASE ANSWER THIS I AM DOING SCIENCE AND IT SUCKS

Aleks [24]

The scientific question is “what is the growth rate of the fingerlings in 5% and 10% saline solutions?”

the other is a hypothesis

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3 years ago

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