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pav-90 [236]
2 years ago
14

Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) +

4OH-(aq) 1. Which species is being oxidized? 2. Which species is being reduced? 3. Which species is the oxidizing agent? 4. Which species is the reducing agent? 5. How many electrons are being transferred? Hint: If you were to balance this equation how many electrons would be in each half-reaction? That is how many electrons are transferred.
Chemistry
1 answer:
frutty [35]2 years ago
5 0

Answer:

1. Chromium

2. Chlorine.

3. Chlorine.

4. Chromium.

5. 12 electrons.

Explanation:

Hello,

In this case, the given reaction with the appropriate oxidation states turns out:

2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)

In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:

2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-

On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:

6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)

Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.

Best regards.

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2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are
kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0
1 year ago
A solution is prepared by diluting 50.00 ml of 2.575 m solution of hno3 to 250.0 ml. what is the molarity of the resulting solut
larisa [96]
Thank you for posting your question here at brainly. Below are the choices that can be found elsewhere:

12.88 M 
<span>0.1278 M </span>
<span>0.2000 M </span>
<span>0.5150 M
</span>
Below is the answer:

<span>5 times diluted (250/50),so 2.575/5=0.515 M
</span>
I hope it helps. 
8 0
1 year ago
Read 2 more answers
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
2 years ago
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Convert 57.6 L to dm3 and divide it by 24
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