Answer:
The smallest whole-number coefficient for OH⁻ is 2
Explanation:
Step 1: The equation redox reaction is divided into two half equations
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 5 : addition of the two half equations to yield a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole number coefficient for OH⁻ is 2
A redox reaction is divided into two half equations which are shown below:
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The two half equations are then added and written together to form a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole-number coefficient for OH⁻ is therefore 2.
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