Balance the reaction between CIO, and Cr to form CIO3 and Cr3+ in acidic solutio
ID: 1045929 • Letter: B
Question
Balance the reaction between CIO, and Cr to form CIO3 and Cr3+ in acidic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. Cr CiO3 (reactant, product, neither) with a coefficient of (Enter 0 for Water appears in the balanced equation as a neither.) How many electrons are transferred in this reaction? Balance the reaction between Ag and Mn2+ to form Ag and MnO2 in acidic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. Mn2+ Mno2 in the balanced equation as a (reactant, product, neither) with a coefficient of (Enter 0 for Water appears neither.) How many electrons are transferred in this reaction?Explanation / Answer
1.
Seprate the redox reaction into oxidation half and reduction half reacions.
Oxidation half reaction Redcution half reaction
Cr --------> Cr3+ ClO4- ----------> ClO3-
2. Balance all aotms other than O and H
Cr --------> Cr3+ ClO4- ----------> ClO3-
3. To balance oxygen atoms in acidic medium add H2O molecules.
Cr --------> Cr3+ ClO4- ----------> ClO3- + H2O
4. To balance hydrogen atoms in acidic medium add H+ ions.
Cr --------> Cr3+ ClO4- +2 H+ (aq.)----------> ClO3- + H2O
5. To balance charges add electrons at more positive charge side.
Cr --------> Cr3+ + 3 e- ClO4- +2 H+ (aq.) + 2 e-----------> ClO3- + H2O
6. To equalise electrons in both half reactions, do OHR * 2 and RHR * 3
2 Cr --------> 2 Cr3+ + 6 e- 3 ClO4- +6 H+ (aq.) + 6 e-----------> 3 ClO3- + 3 H2O
7. Now add the both balanced half reacitons to get balanced redox reaction.
2 Cr --------> 2 Cr3+ + 6 e-
3 ClO4- +6 H+ (aq.) + 6 e-----------> 3 ClO3- + 3 H2O
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2 Cr + 3 ClO4- + 6 H+ ------------> 2 Cr3+ + 3 ClO3- + 3 H2O
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Verify
Atoms Left Right
Cr 2 2
Cl 3 3
O 12 12
H 6 6
Charge +3 +3
Therefore,
Water appears in the balanced equation as Product with coefficient 3
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