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3 years ago

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Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

in each of the following equations:
(a)Mg + NiCl2 ->MgCl2 + Ni
(b)PCl3 + Cl2 -> PCl5
(c)C2H4 + 3O2 -> 2CO2 + 2H2O
(d)Zn + H2SO4 -> H2 + ZnSO4
(e)2K2S2O3 + I2- > K2S4O6+2KI
(f)3Cu + 8HNO3 = 3Cu(NO3)2 + 4H2O + 2NO

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

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Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot

barxatty [35]

Ethanoic acid ionizes in aqueous solutions to form two ions which are $CH_3COO^-$ and $H^+$

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula $CH_3COOH$ .

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

$CH_3COOH ---- > CH_3COO^- + H^+$

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single $H^+$

More on ethanoic acid can be found here: brainly.com/question/9991017

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2 years ago

The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

MAVERICK [17]

Answer:

Approximately $2.46\; \rm mol$ .

Explanation:

Make use of the molar mass data ( $M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}$ ) to calculate the number of moles of molecules in that $64.0\; \rm g$ of $\rm C_2H_2$ :

$\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}$ .

Make sure that the equation for this reaction is balanced.

Coefficient of $\rm C_2H_2$ in this equation: $2$ .

Coefficient of $\rm H_2O$ in this equation: $2$ .

In other words, for every two moles of $\rm C_2H_2$ that this reaction consumes, two moles of $\rm H_2O$ would be produced.

Equivalently, for every mole of $\rm C_2H_2$ that this reaction consumes, one mole of $\rm H_2O$ would be produced.

Hence the ratio: $\displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1$ .

Apply this ratio to find the number of moles of $\rm H_2O$ that this reaction would have produced:

$\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}$ .

3 0

3 years ago

Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0

3 years ago

The spectrochemical series is I < Br< < Cl^- < F

vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0

3 years ago