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.

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

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