In a particular redox reaction, mno2 is oxidized to mno4– and cu2 is reduced to cu . complete and balance the equation for this

In a particular redox reaction, mno2 is oxidized to mno4– and cu2 is reduced to cu . complete and balance the equation for this reaction in acidic solution. phases are optional.

Answers

First, formulate the half-reactions. Let's take the reaction involving Mn. Determine the oxidation number of Mn to determine the number of electrons in the reaction.

For MnO₂: x + 2(-2) = 0 --> x = +4
For MnO₄⁻: x + 4(-2) = -1 --> x = +7
Thus,
MnO₂ --> MnO₄⁻ + 3e⁻

For Cu₂: 0
For Cu²⁺: 2⁺
Thus,
Cu²⁺ + 2e⁻ --> Cu

Now, to eliminate the electrons, multiply the first reaction with 2 and the second half-reaction with 3.

2 MnO₂ --> 2 MnO₄⁻ + 6e⁻
3Cu²⁺ + 6e⁻ --> 3 Cu

Add the reactions:

2 MnO₂ + 3Cu²⁺ + 6e⁻ --> 2 MnO₄⁻ + 6e⁻ + 3 Cu

Eliminate like terms in the reactant and product side:

2 MnO₂ + 3Cu²⁺ --> 2 MnO₄⁻ + 3 Cu

2H2O + MnO2 + 3Ag^+> MnO4^- + 4H^+ + 3Ag

Explanation:

Oxidation half equation:

2H2O + MnO2 > MnO4^- + 3e + 4H^+

Reduction half equation:

3Ag^+ + 3e > 3Ag

Overall redox reaction equation:

2H2O + MnO2 + 3Ag^+> MnO4^- + 4H^+ + 3Ag

The hydrogen ions comes from water.

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)>3Sb^5+(aq) + Br^-(aq) 3H2O(l)

Explanation:

When we want to balance redox reaction equations, we must ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation.

After we have done this, we can now write the overall balanced reaction equation without including the number of electrons lost or gained. Hence;

Oxidation half equation;

3Sb^3+(aq) > 3Sb^5+(aq) +6e

Reduction half equation;

BrO3^-(aq) + 6H^+(aq) + 6e > Br^-(aq) 3H2O(l)

Overall balanced reaction equation;

3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)>3Sb^5+(aq) + Br^-(aq) 3H2O(l)

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺

Explanation:

In order to balance a redox reaction, we will use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Ag⁺ → Ag

Oxidation: BrO⁻ → BrO₃⁻

Step 2: Perform the mass balance adding H⁺ and H₂O where necessary

Ag⁺ → Ag

2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺

Step 3: Perform the electrical balance adding electrons where necessary

1 e⁻ + Ag⁺ → Ag

2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺ + 4 e⁻

Step 4: Multiply both half-reactions by numbers that secure that the number of electrons gained and lost are equal

4 × (1 e⁻ + Ag⁺ → Ag)

1 × (2 H₂O + BrO⁻ → BrO₃⁻ + 4 H⁺ + 4 e⁻)

Step 5: Add both half-reactions

4 e⁻ + 4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺ + 4 e⁻

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO₃⁻ + 4 H⁺

6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺

Explanation:

Step 1: Write the unbalanced reaction

BrO₃⁻ + Sb³⁺ ⟶ Br⁻ + Sb⁵⁺

Step 2: Identify both half-reactions

Reduction: BrO₃⁻ ⟶ Br⁻

Oxidation: Sb³⁺ ⟶ Sb⁵⁺

Step 3: Perform the mass balance, adding H⁺ and H₂O where appropriate

6 H⁺ + BrO₃⁻ ⟶ Br⁻ + 3 H₂O

Sb³⁺ ⟶ Sb⁵⁺

Step 4: Perform the charge balance, adding electrons where appropriate

6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O

Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻

Step 5: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost is the same

1 × (6 H⁺ + BrO₃⁻ + 6 e⁻ ⟶ Br⁻ + 3 H₂O)

3 × (Sb³⁺ ⟶ Sb⁵⁺ + 2 e⁻)

Step 6: Add both half-reactions and cancel what is repeated in both sides

6 H⁺ + BrO₃⁻ + 3 Sb³⁺ ⟶ Br⁻ + 3 H₂O + 3 Sb⁵⁺

M=0.120M

Explanation:

Hello,

In this case, the undergone chemical reaction is:

MnO_4^-(aq)+H_2C_2O_4(aq)rightarrow Mn^{+2}+CO_2

In such a way, the acidic redox balance turns out:

(Mn^{+7}O_4)^-+5e^-+8H^+rightarrow Mn^{+2}+4H_2O\H_2C_2O_4rightarrow2CO_2+2H^++2e^-

Which leads to the total balanced equation as follows:

2(MnO_4)^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)rightarrow2Mn^{+2}(aq)+8H_2O(l)+10CO_2(g)

Thus, as the mass of oxalic acid is not given, one could suppose a value of 1 g (which you can modify based on the actual statement) in order to compute the oxalic acid moles as shwon below:

1gH_2C_2O_4*frac{1molH_2C_2O_4}{90.04gH_2C_2O_4} *frac{2mol(MnO_4)^-}{5molH_2C_2O_4} =0.00444mol(MnO_4)^-

Whereby the molality results:

M=frac{0.00444mol(MnO_4)^-}{0.03702L} =0.120M

Remember you can modify the oxalic acid mass as you desire.

Best regards.

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

Explanation:

A redox reaction is a reaction that occurs with oxidation and a reduction. The compound or the element that oxides are called reducted agent because it promotes the reduction of the other one. The compound or the element that reduces is called the oxidation agent, for the same reason.

In the oxidation, the substance increases its oxidation number: it loses electrons. The opposite occurs in reduction: there is a decrease in the oxidation number and the substance gain electrons.

So, the oxidation reaction is:

NO₂⁻(aq) ⇄ NO₃⁻(aq) + 2e⁻

And the reduction:

Cu²⁺(aq) + 2e⁻ ⇄ Cu⁰(aq)

Note that in the first one, the oxidation number of O is always -2, and the oxidation number of N was from +3 to +5.

The balanced reaction will be:

Cu²⁺(aq) + 3NO₂⁻(aq) ⇄ Cu⁰(aq) + 2NO₃⁻(aq)

The number of atoms of an element must be the same on each side of the equation.

Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).


Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.  

Balanced chemical reaction:

Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).

Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).

The complete and balance the equation for this reaction in acidic solution is given by:

Cr+4H_2O+6Ag^+rightarrow CrO_4^{2-}+8H^++6Ag

Explanation:

Oxidation reaction of chromium to chromate in acidic medium:

Crrightarrow CrO_4^{2-}

Add 4 molecules of water on the left hand side to balance the oxygen:

Cr+4H_2Orightarrow CrO_4^{2-}

Now balance hydrogen atoms by adding hydrogen ions on opposite to the side where water molecule are present .

Cr+4H_2Orightarrow CrO_4^{2-}+8H^+6e^-....[1]

In last, to balance the charge on the both sides add electrons to sides where positive charge is more.

Reduction reaction of silver ions to silver in acidic medium:

Ag^++e^-rightarrow Ag..[2]

By [1] + 6 × [2] , we will get the balance equation for this reaction in acidic solution:

Cr+4H_2O+6Ag^++6e^-rightarrow CrO_4^{2-}+8H^+6e^-+6Ag

Cr+4H_2O+6Ag^+rightarrow CrO_4^{2-}+8H^++6Ag

Cro42ag What is the percent for the fraction nine twentieths? 1. 9% 2. 20% 3. 45% 4. 90%

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