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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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Kc at 298K is Kc = 1 × 10^84.

Calculation:

The relation between Kp and Kc is

      Kp = KcRT

The value of Kp is known

We can calculate the value of Kc with this

R =  8.3145 J mol^−1 K^−1 .

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      Kp = KcRT

      2.5×10⁸⁷ = Kc × 8.3145 × 298

      2.5×10⁸⁷ = Kc × 2477.572

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To learn more click the given link

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