assign oxidation states to each atom in each of the following species

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Assign oxidation states to each atom in each element, ion, or compound. Ag, Ca2+, BaO, H2S, NO3-, CrO42-

Master Calculate Oxidation Numbers with a bite sized video explanation from Jules Bruno

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4.3: Oxidation Numbers and Redox Reactions

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Redox reactions may involve proton transfers, electron transfers, and other bond-breaking and bond-making processes. As a result, the equations involved are can be more difficult to interpret and balance than those describing acid-base reactions. To recognize redox reactions, we must be able to identify when a species is oxidized or reduced; to do this we assign assign  oxidation numbers to each atom in a reaction. Then, we compare the oxidation states of eac atom on the reactants side and to the atoms on the products side. When changes occur, we know a readox reaction has taken place.

In the reaction above, the nitrogen in the NO 3 –  ion is assigned an oxidation number of +5, and each oxygen is assigned an oxidation number of –2. This assignment corresponds to the nitrogen having lost five valence electrons to the more electronegative oxygen atoms. The nitrogen in the NO 2  ion has an oxidation number of + 4, which corresponds to the loss of four valence electrons to the more electronegative oxygen atoms. The nitrogen atom may be thought of as having one valence electron for itself, that is, one more electron than it had in NO 3 – . Through this reaction, nitrogen gains one electron in going from NO 3 – to NO 2 , and we say nitrogen is reduced by the reaction. 

In the same reaction, the oxidation number of copper increased from 0 to +2; we say that copper is oxidized by the reaction and lost two electrons. The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agent. Copper was oxidized because its electrons were accepted by an oxidizing agent, nitrogen (or nitrate ion).

A word of caution: Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO 3 – does not really have a +5 charge which can be reduced to +4 in NO 2 . Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atoms in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid.

The general rules for oxidation numbers are seen below, taken from the following page in the Analytical Chemistry Core Textbook: Oxidation States

Rules for Assigning Oxidation Numbers

• The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl 2 , S 8 , and large structures of carbon or silicon each have an oxidation state of zero. ( Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl 2 .)
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. (W hen an electron is lost by one atom (+1 contribution to oxidation number), the same electron must be gained by another atom (–1 contribution to oxidation number).)
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
• Some elements almost always have the same oxidation states in their compounds:

Exceptions:

Hydrogen in the metal hydrides : Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H - . The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.

Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).

Oxygen in peroxides : Peroxides include hydrogen peroxide, H 2 O 2 . This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F 2 O : The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen : Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference.

Example \(\PageIndex{1}\): Chromium

What is the oxidation state of chromium in Cr 2 + ?

For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)

What is the oxidation state of chromium in CrCl 3 ?

This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:

n + 3(-1) = 0

The oxidation state of chromium is +3.

Example \(\PageIndex{2}\): Chromium

What is the oxidation state of chromium in Cr(H 2 O) 6 3+ ?

This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.

The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr 3 + . The oxidation state is +3.

What is the oxidation state of chromium in the dichromate ion, Cr 2 O 7 2 - ?

The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

Example \(\PageIndex{3}\): Copper

What is the oxidation state of copper in CuSO 4 ?

Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states.

In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem:

• Recognize CuSO 4 as an ionic compound containing a copper ion and a sulfate ion, SO 4 2 - . To form an electrically neutral compound, the copper must be present as a Cu 2 + ion. The oxidation state is therefore +2.
• Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below). 

Exercise \(\PageIndex{1}\)

Assign oxidation states to each atom in the reaction. Determine which species are the oxidizing and reducing agents.

\(\ce{2MnO4^{–} + 5SO2 + 6H2O -> 5SO4^{2–} + 2Mn^{2+} + 4H3O^{+}}\)

The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO 2 has been oxidized by MnO 4 – , and so MnO 4 – is the oxidizing agent. MnO 4 – has been reduced by SO 2 , and so SO 2 is the reducing agent.

Exercise \(\PageIndex{2}\)

\(\ce{NH4^+ + PO4^{3–} -> NH3 + PO4^{2–}}\)

Oxidation numbers show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred.

Exercise \(\PageIndex{3}\)

\(\ce{HClO + H2S -> H3O^+ + Cl^{–} + S}\)

H 2 S has been oxidized, losing two electrons to form elemental S. Since H 2 S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl – . Since it accepts electrons, HClO is the oxidizing agent.

Exercise \(\PageIndex{4}\)

Determine the Oxidation States of each element in the following reactions:

• \(\ce{Fe(s) + O2(g) -> Fe2O3(g)}\)
• \(\ce{Fe^{2+}(aq)}\)
• \(\ce{Ag(s) + H2S -> Ag2S(g) + H2(g)}\)
• \(\ce{Fe}\) and \(\ce{O2}\) are free elements; therefore, they each have an oxidation state of 0 according to Rule #1. The product has a total oxidation state equal to 0, and following Rule #6, \(\ce{O}\) has an oxidation state of -2, which means \(\ce{Fe}\) has an oxidation state of +3. \
• The oxidation state of \(\ce{Fe}\) ions just corresponds to its charge since it is a single element species; therefore, the oxidation state is +2.
• \(\ce{Ag}\) has an oxidation state of 0, \(\ce{H}\) has an oxidation state of +1 according to Rule #5, \(\ce{H2}\) has an oxidation state of 0, \(\ce{S}\) has an oxidation state of -2 according to Rule #7, and hence \(\ce{Ag}\) in \(\ce{Ag2S}\) has an oxidation state of +1.

Exercise \(\PageIndex{5}\)

Determine the oxidation states of the phosphorus atom bold element in each of the following species:

• \(\ce{Na3PO3}\)
• \(\ce{H2PO4^{-}}\)
• The oxidation numbers of \(\ce{Na}\) and \(\ce{O}\) are +1 and -2. Because sodium phosphite is neutral species, the sum of the oxidation numbers must be zero. Letting x be the oxidation number of phosphorus, 0= 3(+1) + x + 3(-2). x =oxidation number of P= +3.
• Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-2), y = oxidation number of P= +5.

Exercise \(\PageIndex{6}\)

Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each):

• \(\ce{Zn + 2H^{+} -> Zn^{2+} + H2}\)
• \(\ce{2Al + 3Cu^{2+} -> 2Al^{3+} +3Cu}\)
• \(\ce{CO3^{2-} + 2H^{+} -> CO2 + H2O}\)
• Zn is oxidized (Oxidation number: 0 → +2); H + is reduced (Oxidation number: +1 → 0)
• Al is oxidized (Oxidation number: 0 → +3); Cu 2 + is reduced (+2 → 0)
• This is not a redox reaction because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.

Exercise \(\PageIndex{7}\)

Determine what is the oxidizing and reducing agents in the following reaction.

\[\ce{Zn + 2H^{+} -> Zn^{2+} + H2} \nonumber \]

The oxidation state of \(\ce{H}\) changes from +1 to 0, and the oxidation state of \(\ce{Zn}\) changes from 0 to +2. Hence, \(\ce{Zn}\) is oxidized and acts as the reducing agent . \(\ce{H^{+}}\) ion is reduced and acts as the oxidizing agent .

Assign oxidation number to the underlined elements in each of the following species. (i) N a H 2 P – – O 4 (ii) N a H S – – O 4 (iii) H 4 P – – 2 O 7 (iv) H 2 S – – 2 O 7 (v) C a O – – 2 (vi) N a B – – H 4 (vii) K A l ( S O – –– – 4 ) 2 .12 H 2 O (viii) C r 2 – –– – O 2 − 7

(i) let assume oxidation number of p is x. we know that, oxidation number of na = +1 oxidation number of h = +1 oxidation number of o = -2 then we have 1(+1) + 2(+1) + 1 (x) + 4(-2) = 0 ⇒ 1 + 2 + x - 8 = 0 ⇒ x - 5 = 0 ⇒ x = + 5 hence the oxidation number of p is +5 (ii) let assume oxidation number of s is x. oxidation number of na = +1 oxidation number of h = +1 oxidation number of o = -2 then we have: 1(+1) + 1(+1) + 1 (x) + 4(-2) = 0 ⇒ 1 + 1 + x - 8 = 0 ⇒ x-6 = 0 ⇒ x = +6 hence the oxidation number of s is +6 (iii) let assume oxidation number of p is x. oxidation number of h = +1 oxidation number of o = -2 then we have: 4(+1) + 2(x) + 7 (-2) = 0 ⇒ 4 + 2x - 14 = 0 ⇒ 2x - 10 = 0 ⇒ 2x = +10 ⇒ x = +5 hence, oxidation number of p is +5 (iv) let assume oxidation number of s is x. oxidation number of o = -2 oxidation number of h = +1 then we have: 2(+1) + 2(x) + 7(-2) = 0 ⇒ 2 + 2x - 14 = 0 ⇒ 2x - 12 = 0 ⇒ x = +6 hence, oxidation number of s is +6. (v) let assume oxidation number of o is x. oxidation number of ca = +2 then we have: 1(+2) + 2(x) = 0 ⇒ 2 + 2x = 0 ⇒ 2x = -2 ⇒ x = -1 hence, oxidation number of o is -1 (vi) let assume oxidation number of b is x. oxidation number of na = +1 oxidation number of h = -1 then we have: 1(+1) + 1(x) + 4(-1) = 0 ⇒ 1 + x -4 = 0 ⇒ x - 3 = 0 ⇒ x = +3 hence, oxidation number of b is +3. (vii) let assume oxidation number of s is x. oxidation number of k = +1 oxidation number of al = +3 oxidation number of o = -2 oxidation number of h = +1 then we have: 1(+1) + 1 (+3) + 2(x) + 8(-2) + 24(+1) + 12 (-2) = 0 ⇒ 1 + 3 + 2x -16 +24 -24 = 0 ⇒ 2x - 12 = 0 ⇒ 2x = +12 ⇒ x = +6 hence, oxidation number of s is +6. (viii) let us assume oxidation number of cr is x. oxidation number of o = -2 then we have : 2(x) -2(7) = -2 x = +6 hence, oxidation number of cr is +6..

Which of the following represent the correct Element and Symbol Pair?

(i) Cobalt –CO

(ii) Copper-Cu

(iii)Potassium- K

(iv) Calcium-C

(v) Silicon-Sl

(vi) Silver- Ag

(vii)Barium-Ba

(viii)Iron-Ir

Assign oxidation numbers to the underlined elements in each of the following species: N a H 2 P O 4

N a H S – – O 4

H 4 P – – 2 O 7

K 2 M n – –– – O 4

C a O – – 2

N a B – – H 4

H 2 S – – 2 O 7

K A l S – – O 4 ) 2 .12 H 2 O

Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH 2 P O 4 (b) NaH S O 4 (c) H 4 P 2 O 7 (d) K 2 Mn O 4

(e) Ca O 2 (f) Na B H 4 (g) H 2 S 2 O 7 (h) KAl( S O 4 ) 2 .12 H 2 O

Round off each of the following numbers to nearest hundreds: (i) 3,985 (ii) 7,289 (iii) 8,074 (iv) 14,627 (v) 28,826 (vi) 4,20,387 (vii) 43,68,973 (viii) 7,42,898

Represent each of the following numbers on the number line: (i) - 1 3 (ii) - 3 4 (iii) - 1 2 3 (iv) - 3 1 7 (v) - 4 3 5 (vi) - 2 5 6 (vii) −3 (viii) - 2 7 8

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• 5. A sample of gas in a cylinder of volume 4.05 L at 342 K and 3.1 atm expands to 6.83 L by two differe
• 6. How much work is done by the system when 2.00 moles of O2 expand from a volume of 3.00 liters to a v
• 7. gas used in welding
• Programming
• Engineering

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