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Step-by-Step Solution
Step 1: Identify the Redox Reaction
In acidic medium, the balanced reaction for the oxidation of Fe2+ to Fe3+ using K2Cr2O7 (dichromate) is:
\mathrm{Cr_2O_7^{2-}} + 6\,\mathrm{Fe^{2+}} + 14\,\mathrm{H^+} \rightarrow 2\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + 7\,\mathrm{H_2O}
From this balanced reaction, 1 mole of dichromate ( \mathrm{Cr_2O_7^{2-}} ) oxidizes 6 moles of Fe2+.
Step 2: Express the Equivalence Relation
Using the concept of milli-equivalents (meq) to track the reaction:
- Let M = molarity of Fe2+ solution (unknown).
- Volume of Fe2+ solution used = 10 mL.
- Molarity of K2Cr2O7 = 0.02 M.
- Volume of K2Cr2O7 solution used = 15 mL.
- Each mole of K2Cr2O7 is equivalent to 6 moles of Fe2+.
Milli-equivalents of Fe2+ = milli-equivalents of dichromate
For Fe2+, millimoles = M \times \text{(Volume in mL)} . When thinking in terms of equivalence, Fe2+ has an n -factor of 1 for this oxidation step (since it goes from +2 to +3).
For dichromate, millimoles = (0.02)(15) and its n -factor is 6 because 1 mole of dichromate requires 6 moles of Fe2+.
Step 3: Set up the Equation
Milli-equivalents of Fe2+:
M \times 10 \times 1
Milli-equivalents of K2Cr2O7 = (moles of dichromate) \times (n-factor) in millimoles:
(0.02 \times 15) \times 6
Equating both sides:
M \times 10 = (0.02 \times 15) \times 6
Step 4: Solve for M
Calculate the right-hand side first:
0.02 \times 15 = 0.3
0.3 \times 6 = 1.8
Therefore,
M \times 10 = 1.8
M = \frac{1.8}{10} = 0.18 \text{ M}
Since the question states that the molarity is in the form x \times 10^{-2} M, we rewrite:
0.18 = 18 \times 10^{-2}
Thus, x = 18 .
