In the following reaction,2I- + Cr2O72- + 14H+ → I2 + 2Cr3+ + 7H2OUnbalanced parts are
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a
H+, H2O
b
Cr2O72-, Cr3+
c
I-, I2
d
None of them are balanced
answer is C.
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Detailed Solution
(I) Cr2O72- + 6e- → 2Cr3+(II) 2I- → I2 + 2e-To balance electrons (II) is to be multiplied by (3). Thus,6I- + Cr2O72- + 14H+ → 3I2 + 2Cr3+ + 7H2OThus, I- and I2 are not balanced.