On balancing the given redox reaction, aCr₂O₇²⁻(aq) + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + c/2H₂O(l) the coefficients a, b and c are found to be, respectively:
On balancing the given redox reaction, aCr₂O₇²⁻(aq) + bSO₃²⁻(aq) + cH⁺(aq) → 2aCr³⁺(aq) + bSO₄²⁻(aq) + c/2H₂O(l) the coefficients a, b and c are found to be, respectively:
Option 1 -
8,1,3
Option 2 -
1,3,8
Option 3 -
3,8,1
Option 4 -
1,8,3
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1 Answer
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Correct Option - 2
Detailed Solution:Reduction half reaction.
Cr? O? ²? + 14H? + 6e? → 2Cr? ³ + 7H? O
Oxidation half reaction
SO? ²? + H? O → SO? ²? + 2e? ] × 3
Oxygen is balanced by adding water and hydrogen is balanced by adding H? and the charge is balanced by electrons.
Add ( eq. (i) + (3 × eq. (ii) )
Cr? O? ²? + 3SO? ²? + 8H? → 2Cr? ³ + 3SO? ²? + 4H? O
a = 1 b = 3 c = 8
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