Ask question

Ask question

All categories

kakasveta [241]

2 years ago

9

Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) Fe(CN)₆³⁻(aq) + Re(s) →\mathrm{Fe}(\

mathrm{CN})_{6}^{4-}(a q)+\mathrm{ReO}_{4}^{-}(a q) [basic]

1 answer:

photoshop1234 [79]2 years ago

6 0

The given chemical equation is:

Fe(CN)63-(aq) + Re(s)-> Fe(CN)64-(aq) + ReO4-(aq)

Consider oxidation half reaction and balance it first in acidic conditions:

Re- > ReO4-

Add water on the left side to balance the O-atoms:

Re- +4H20- > ReO4-

Add protons on the right side to balance H-atoms:

Re- +4H20- > ReO4-+8H+

To balance the charge add electrons:

Re- +4H20- > ReO4-+8H+ 7e-------------(1)

Reduction half reaction: Fe(CN)63-(aq) -> Fe(CN)64-(aq)

Add electrons to balance the charge:

Fe(CN)63-(aq) - +e- > Fe(CN)64- (aq)---------------(2)

Multiply equation(2) with seven :

7{Fe(CN)63-(aq) - +7e-} > 7{Fe(CN)64- (aq)} ------(3)

Add (1) and (3)

7Fe(CN)63-(aq) - +7e-} > 7{Fe(CN)64- (aq)} + Re- +4H20- > ReO4-+8H+ 7e-

Add 8OH- on both sides:7Fe(CN)63-(aq) - +7e-} > 7{Fe(CN)64- (aq)} + Re- +4H20- + OH- > ReO4-+8H+ 7e- +OH-

It becomes:

7Fe(CN)63-(aq) + rE +8OH-->OH- >7{Fe(CN)64- + ReO4-+ 4H20

This is the final equation in the basic medium.

Re(s) is oxidised. So it is the reducing agent.

Fe(CN)63- is reduced.It is the oxidising agent.

To know more about skeleton reactions questions

brainly.com/question/24225616

#SPJ4

You might be interested in

The process, when an atom gains electrons is called

max2010maxim [7]

Answer:

When an atom gains/loses an electron, the atom becomes charged, and is called an ion. Gaining an electron results in a negative charge, so the atom is an anion. Losing an electron results in a positive charge, so atom ion is a cation

Explanation:

Answerd the question : ion

8 0

3 years ago

3. If x electrons are needed to displace 108 g silver from a solution which contains Ag ions,

ss7ja [257]

Answer:

Choice A. $x$ electrons would be required for displacing $9\; \rm g$ of aluminum from a solution of $\rm Al^{3+}$ ions.

Assumption: by " $\rm Ag$ ions" the question meant $\rm Ag^{+}$ with a charge of $+1$ on each ion.

Explanation:

The question states that the relative atomic mass of $\rm Ag$ is $108$ . In other words, each mole of

Therefore, that $108\; \rm g\!$ of silver that were formed would contain $1\; \rm mol$ of silver atoms.

Metallic silver would precipitate out of this $\rm Ag^{+}$ solution only after these ions are turned into $\rm Ag$ atoms.

One $\rm Ag^{+}$ ion carries one unit of positive electrical charge. On the other hand, each $e^{-}$ carries one unit of negative electrical charge.

Therefore, each $\rm Ag^{+}\!$ ion will need to gain one electron to form a neutral $\rm Ag$ atom.

${\rm Ag^{+}}\; (aq) + e^{-} \to {\rm Ag}\; (s)$ .

At least $1\; \rm mol$ of electrons would be required to turn $1\; \rm mol\!$ of $\rm Ag^{+}$ ions into that $1\; \rm mol\!\!$ of silver atoms (which have a mass of $108\; \rm g\!$ .)

Hence, $x = 1\; \rm mol$ .

Unlike $\rm Ag^{+}$ ions, each aluminum ion $\rm Al^{3+}$ carries three units of positive electrical charge. That is three times the amount of charge on one $\rm Ag^{+}\!$ ion. Therefore, three electrons will be required to turn one $\rm Al^{3+}\!$ ion to an $\rm Al$ atom.

${\rm Al^{3+}}\; (aq) + 3\, e^{-} \to {\rm Al}\; (s)$

The question states that the relative atomic mass of $\rm Al$ is $27$ . Therefore, each mole of $\rm Al\!$ atoms would have a mass $27\; \rm g$ . There would be $\displaystyle \frac{9\; \rm g}{27\; \rm g \cdot mol^{-1}} = \frac{1}{3} \; \rm mol$ of atoms in that $9\; \rm g$ of $\rm Al\!\!$ .

It takes $3\; \rm mol$ of electrons to turn one mole of $\rm Al^{3+}$ ions to one mole of $\rm Al$ atoms. Hence, $\displaystyle \frac{1}{3}\times 3\; \rm mol = 1\; \rm mol$ of electrons would be required to produce that $\displaystyle \frac{1}{3}\; \rm mol$ of $\rm Al\!$ atoms (which has a mass of $9\; \rm g$ ) from $\rm Al^{3+}\!$ ions.

That corresponds to the first choice, $x$ electrons.

7 0

3 years ago

How many grams are in 8.23 moles of sodium oxide?

konstantin123 [22]

Answer:

So, the mole ratio between sodium and sodium oxide is 2:1=2

Explanation:

6 0

2 years ago

Read 2 more answers

How many grams of carbon dioxide are created from the complete combustion of 21.3 L of butane at STP?

Bond [772]

Answer:

165.5 g of CO2

Explanation:

We must first put down the balanced reaction equation:

C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)

From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:

22.4 L of butane created 174 g of CO2

Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2

3 0

4 years ago

You've decided to try to make your own wine by adding yeast to a sweet grape juice. After allowing plenty of time for the yeast

aalyn [17]

Answer:

possibly he did not stir it or did not store it at the right temperature

Explanation:

4 0

3 years ago

Read 2 more answers