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givi [52]
3 years ago
13

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

um and cadmium:        In(s)|In3 (aq)||Cd2 (aq)|Cd(s)
Chemistry
1 answer:
AlexFokin [52]3 years ago
4 0

Explanation:

The given cell reaction is as follows.

       In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : In(s) \rightarrow In^{3+}(aq) + 3e^{-} ...... (1)

At cathode; Reduction-half reaction : Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s) ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

      2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}

      3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)

Net reaction: 2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

Thus, we can conclude that the overall cell reaction is as follows.

        2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

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Use the periodic table to determine the electron configuration for Ca and Pm in noble-gas notation Ca: [Ar]4s2 [Ar]4s1 [Ar]3s2 [
sveta [45]
Answer:

1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵

Explanation:

1) Ca:

Its atomic number is 20. So it has 20 protons and 20 electrons.

Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.

So, you have two more electrons (20 - 18 = 2) to distribute.

Those two electrons go the the orbital 4s.

Finally, the electron configuration is [Ar] 4s².

2) Pm

The atomic number of Pm is 61, so it has 61 protons and 61 electrons.

Pm is in the row (period) 6. So, the noble gas before Pm is Xe.

The atomic number of Xe is 54.

Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.

The resultant distribution for Pm is: [Xe]6s² 4f⁵.
5 0
3 years ago
Read 2 more answers
when a substance undergoes combustion and carbon completely it produces carbon monoxide and water true or false ​
eimsori [14]

False

complete combustion produces carbon dioxide + water

6 0
3 years ago
What problem would arise if the excess acetyl‑CoA were not converted to ketone bodies?
HACTEHA [7]

Answer:

The correct answer is C fatty acid oxidation would stop when all of the CoA is bound as acetyl CoA.

Explanation:

Acetyl CoA is the principle end product of beta oxidation of even chain fatty acid such as palmitic acid.

    When the cellelar label of actyl CoA increases at that time the excess acetyl CoA is converted to ketone bodies by the process called ketogenesis.

 According to the question if the excess acetyl CoA is not converted to ketone bodies then it will interfere with the oxidation of fatty acid because fatty acid molecules will not get any CoA SH molecule to activate themselves to initiate a new round of beta oxidation.

As a result fatty acid oxidation will stop.

6 0
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How many moles of aluminum do 4.40×1024 aluminum atoms represent?
zhannawk [14.2K]
For this problem we use the Avogadro's number which is an empirical value that relates the number of particles to the number of moles. Its approximated value is 6.022×10²³ atoms/mole. The solution is as follows:

4.40×10²⁴ aluminum atoms * 1 moles/6.022×10²³ atoms = 7.306

Thus, there are 7.306 moles of aluminum.
3 0
4 years ago
How many grams of ammonia are produced when 1.0 mole of nitrogen reacts
Varvara68 [4.7K]

Answer: N2(g) + 3H2-> 2NH3(g)  This is the balanced equation

Note the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important.

 

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

 

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced  

grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

 

NOTE:  The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

Explanation: here you go!! good luck! hope this helped

7 0
3 years ago
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