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scoray [572]
2 years ago
15

Write a balanced half-reaction for the reduction of permanganate ion mno−4 to solid manganese dioxide mno2 in acidic aqueous sol

ution. Be sure to add physical state symbols where appropriate
Chemistry
1 answer:
Svetlanka [38]2 years ago
8 0

The half-reaction includes either the reduction or the oxidation reaction of the redox reactions. In acidic solution permanganate ion will react with hydrogen ion to yield manganese ion and water.

<h3>What are Redox reactions?</h3>

Redox or oxidation-reduction reactions are the chemical reactions in which the oxidation and the reduction of the chemical species occur simultaneously.

Permanganate (VII) ion is a strong oxidizing agent and gets easily reduced to manganese ion in presence of the hydrogen ion in an acidic solution.

The balanced half-reaction for reduction is shown as,

\rm MnO_{4}^{-} \; (aq)+ 8H^{+} \; (aq)+ 5e^{-} \rightarrow Mn^{2+} \; (aq)+ 4H_{2}O \; (l)

Learn more about reduction reactions here:

brainly.com/question/10084275

#SPJ4

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2 years ago
A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts
Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

                    1 quarts = 0.95 liter

                   7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

        Given:

            1 liter of solution contains 61.2g of barium nitrate:

          7.125 liter will contain  7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

  In solution we have  Ba²⁺ + NO₃⁻

                Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

 

Number of moles of Ba(NO₃)₂ = \frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = \frac{436.05}{271} = 1.609mole

      1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

    therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions  x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0
3 years ago
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