redox practice problems
indicate the oxidation states of each of the elements in the following chemical species:
1) naoh
2) hbr
3) vf₅
4) na₂so₄
5) no₂⁻¹
6) siof₂
determine which of the following reactions are redox reactions. for those that are, identify the element which has been oxidized, and the one that has been reduced:
7) 2 naoh + cacl₂ → 2 nacl + ca(oh)₂
8) 3 li + fecl₃ → 3 licl + fe
9) ccl₄ + 2 br₂ → cbr₄ + 2 i₂
10) hno₃ + nahco₃ → h₂o + co₂ + nano₃

Question

Accepted Answer

### Problem 1: Oxidation State of NaOH
# Explanation:
## Step 1: Recall Oxidation State Rules
- Group 1 metals (Na) have an oxidation state of +1.
- Oxygen (O) in compounds (except peroxides) has an oxidation state of -2.
- Hydrogen (H) in compounds (except hydrides) has an oxidation state of +1.
## Step 2: Calculate for NaOH
Let the oxidation state of Na be $ x $, O be $ y $, and H be $ z $.
For Na: $ x = +1 $ (Group 1 metal).
For O: $ y = -2 $ (standard rule for O in compounds).
For H: $ z = +1 $ (standard rule for H in compounds).
We can also verify using the overall charge (NaOH is neutral, so sum of oxidation states is 0):
$ (+1) + (-2) + (+1) = 0 $, which holds true.
# Answer:
Na: +1, O: -2, H: +1

### Problem 2: Oxidation State of HBr
# Explanation:
## Step 1: Recall Oxidation State Rules
- Hydrogen (H) in compounds (except hydrides) has an oxidation state of +1.
- Halogens (Br is a halogen) in compounds (except interhalogens or with O) have an oxidation state of -1.
## Step 2: Calculate for HBr
Let the oxidation state of H be $ x $ and Br be $ y $.
For H: $ x = +1 $ (standard rule for H in compounds).
For Br: $ y = -1 $ (halogen in compound, not with O or interhalogen).
Verify (HBr is neutral, sum of oxidation states is 0):
$ (+1) + (-1) = 0 $, which holds true.
# Answer:
H: +1, Br: -1

### Problem 3: Oxidation State of $ \text{VF}_5 $
# Explanation:
## Step 1: Recall Oxidation State Rules
- Fluorine (F) in compounds has an oxidation state of -1 (most electronegative element).
- Let the oxidation state of V be $ x $.
## Step 2: Calculate for $ \text{VF}_5 $
There are 5 F atoms, each with oxidation state -1. The compound is neutral (sum of oxidation states is 0).
So, $ x + 5(-1) = 0 $
$ x - 5 = 0 $
$ x = +5 $
# Answer:
V: +5, F: -1

### Problem 4: Oxidation State of $ \text{Na}_2\text{SO}_4 $
# Explanation:
## Step 1: Recall Oxidation State Rules
- Group 1 metals (Na) have an oxidation state of +1.
- Oxygen (O) in compounds (except peroxides) has an oxidation state of -2.
- Let the oxidation state of S be $ x $.
## Step 2: Calculate for $ \text{Na}_2\text{SO}_4 $
There are 2 Na atoms, 1 S atom, and 4 O atoms. The compound is neutral (sum of oxidation states is 0).
So, $ 2(+1) + x + 4(-2) = 0 $
$ 2 + x - 8 = 0 $
$ x - 6 = 0 $
$ x = +6 $
# Answer:
Na: +1, S: +6, O: -2

### Problem 5: Oxidation State of $ \text{NO}_2^{-1} $ (Nitrite Ion)
# Explanation:
## Step 1: Recall Oxidation State Rules
- Oxygen (O) in compounds (except peroxides) has an oxidation state of -2.
- Let the oxidation state of N be $ x $. The overall charge of the ion is -1.
## Step 2: Calculate for $ \text{NO}_2^{-1} $
There are 2 O atoms. So,
$ x + 2(-2) = -1 $
$ x - 4 = -1 $
$ x = -1 + 4 $
$ x = +3 $
# Answer:
N: +3, O: -2

### Problem 6: Oxidation State of $ \text{SiOF}_2 $
# Explanation:
## Step 1: Recall Oxidation State Rules
- Fluorine (F) in compounds has an oxidation state of -1.
- Oxygen (O) in compounds (except peroxides) has an oxidation state of -2.
- Let the oxidation state of Si be $ x $. The compound is neutral (sum of oxidation states is 0).
## Step 2: Calculate for $ \text{SiOF}_2 $
There are 1 Si, 1 O, and 2 F atoms.
So, $ x + (-2) + 2(-1) = 0 $
$ x - 2 - 2 = 0 $
$ x - 4 = 0 $
$ x = +4 $
# Answer:
Si: +4, O: -2, F: -1

### Problem 7: Reaction $ 2 \text{NaOH} + \text{CaCl}_2
ightarrow 2 \text{NaCl} + \text{Ca(OH)}_2 $
# Explanation:
## Step 1: Identify Reaction Type
This is a double - displacement reaction (ions exchange places: $ \text{Na}^+ $ and $ \text{Ca}^{2+} $ exchange with $ \text{OH}^- $ and $ \text{Cl}^- $).
## Step 2: Check Oxidation States
- For Na: In $ \text{NaOH} $ and $ \text{NaCl} $, oxidation state is +1.
- For Ca: In $ \text{CaCl}_2 $ and $ \text{Ca(OH)}_2 $, oxidation state is +2.
- For O: In $ \text{NaOH} $ and $ \text{Ca(OH)}_2 $, oxidation state is -2.
- For H: In $ \text{NaOH} $ and $ \text{Ca(OH)}_2 $, oxidation state is +1.
- For Cl: In $ \text{CaCl}_2 $ and $ \text{NaCl} $, oxidation state is -1.

Since no element changes its oxidation state, this is not a redox reaction.
# Answer:
Not a redox reaction (no change in oxidation states of elements).

### Problem 8: Reaction $ 3 \text{Li} + \text{FeCl}_3
ightarrow 3 \text{LiCl} + \text{Fe} $
# Explanation:
## Step 1: Determine Oxidation States
- For Li: In elemental form (Li), oxidation state is 0; in $ \text{LiCl} $, oxidation state is +1.
- For Fe: In $ \text{FeCl}_3 $, oxidation state is +3; in elemental form (Fe), oxidation state is 0.
- For Cl: In $ \text{FeCl}_3 $ and $ \text{LiCl} $, oxidation state is -1.
## Step 2: Identify Oxidation and Reduction
- Oxidation: Li is oxidized (oxidation state increases from 0 to +1).
- Reduction: Fe is reduced (oxidation state decreases from +3 to 0).
So, this is a redox reaction.
# Answer:
Redox reaction. Oxidized: Li (oxidation state 0→+1); Reduced: Fe (oxidation state +3→0)

### Problem 9: Reaction $ \text{CCl}_4 + 2 \text{Br}_2
ightarrow \text{CBr}_4 + 2 \text{I}_2 $ (Note: There is a typo, it should be $ 2\text{I}^- $ or some other iodine - containing species instead of $ 2\text{I}_2 $ as written. Assuming it is a reaction like $ \text{CCl}_4 + 2 \text{Br}^-
ightarrow \text{CBr}_4 + 2 \text{Cl}^- $ or a similar halogen - exchange reaction with a reducing agent for iodine. But if we consider the given reaction as is, there is an error. However, if we assume it is a reaction where Br and I are involved in redox:
## Step 1: Check Oxidation States (assuming the reaction is a redox reaction with I being oxidized and Br being reduced, but the given reaction as $ \text{CCl}_4 + 2 \text{Br}_2
ightarrow \text{CBr}_4 + 2 \text{I}_2 $ has an incorrect reactant - product combination as $ \text{Br}_2 $ and $ \text{I}_2 $ are both halogens in elemental form. If we correct it to a reaction like $ \text{CCl}_4 + 2 \text{NaBr}
ightarrow \text{CBr}_4 + 2 \text{NaCl} $, it is a non - redox double - displacement. But if we consider a redox reaction involving Br and I, for example, if the reaction is $ \text{CCl}_4 + 2 \text{KI}
ightarrow \text{CBr}_4 + 2 \text{KCl}+ \text{I}_2 $ (correcting Br to I in reactant or product):
- For I: In $ \text{KI} $ (oxidation state - 1) and $ \text{I}_2 $ (oxidation state 0), so I is oxidized.
- For Cl: In $ \text{CCl}_4 $ (oxidation state - 1) and $ \text{KCl} $ (oxidation state - 1), no change.
- For C: In $ \text{CCl}_4 $ and $ \text{CBr}_4 $, oxidation state of C is +4 (since Cl and Br are - 1, and 4*(- 1)+x = 0→x = +4), no change.
- For Br: If in a corrected reaction, but in the given reaction, since it has a typo, we can say that the given reaction as written is incorrect. But if we assume a redox reaction where, for example, Br is reduced and I is oxidized (in a proper reaction), the oxidized element would be I and reduced element would be Br (depending on the correct reaction). However, the given reaction $ \text{CCl}_4 + 2 \text{Br}_2
ightarrow \text{CBr}_4 + 2 \text{I}_2 $ has an incorrect stoichiometry and reactant - product combination as Br and I are both in elemental form and there is no source of I in the reactants.

Since the reaction as written has an error, we can state that the reaction is incorrectly written. But if we consider a possible intended redox reaction (e.g., with a metal halide as reactant), the key is to check oxidation states.
# Answer:
The reaction as written ($ \text{CCl}_4 + 2 \text{Br}_2
ightarrow \text{CBr}_4 + 2 \text{I}_2 $) is incorrect (no source of iodine in reactants and incorrect halogen combination). If it was a redox reaction involving halide ions, for example, in a reaction like $ \text{CCl}_4+2\text{KI}
ightarrow\text{CCl}_4 + 2\text{KCl}+\text{I}_2 $ (correcting Br to I in reactant), I would be oxidized (from - 1 to 0) and no element is reduced (Cl remains - 1), but this is just a speculation due to the typo in the given reaction.

### Problem 10: Reaction $ \text{HNO}_3 + \text{NaHCO}_3
ightarrow \text{H}_2\text{O} + \text{CO}_2 + \text{NaNO}_3 $
# Explanation:
## Step 1: Identify Reaction Type
This is an acid - base (neutralization) and decomposition (of carbonic acid $ \text{H}_2\text{CO}_3 $ formed from $ \text{H}^+ $ and $ \text{HCO}_3^- $ into $ \text{H}_2\text{O} $ and $ \text{CO}_2 $) reaction.
## Step 2: Check Oxidation States
- For H: In $ \text{HNO}_3 $, $ \text{NaHCO}_3 $, $ \text{H}_2\text{O} $, oxidation state is +1.
- For N: In $ \text{HNO}_3 $ and $ \text{NaNO}_3 $, oxidation state is +5.
- For O: In all compounds, oxidation state is -2 (except in $ \text{O}_2 $, but there is no $ \text{O}_2 $ here).
- For C: In $ \text{NaHCO}_3 $ (oxidation state: $ \text{Na}^+1, \text{H}^+1, \text{O}^-2 $, so $ +1+(+1)+x + 3*(-2)=0
ightarrow x = +4 $) and in $ \text{CO}_2 $ (oxidation state of C is +4, since $ 2*(-2)+x = 0
ightarrow x = +4 $), no change.
- For Na: In $ \text{NaHCO}_3 $ and $ \text{NaNO}_3 $, oxidation state is +1.

Since no element changes its oxidation state, this is not a redox reaction.
# Answer:
Not a redox reaction (no change in oxidation states of elements).

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QUESTION IMAGE

Question

Explanation:

Problem 1: Oxidation State of NaOH

Step 1: Recall Oxidation State Rules

Step 2: Calculate for NaOH

Step 1: Recall Oxidation State Rules

Step 2: Calculate for HBr

Step 1: Recall Oxidation State Rules

Step 2: Calculate for \( \text{VF}_5 \)

Answer:

Problem 2: Oxidation State of HBr