Chemistry - Redox Reactions Question with Solution | TestHub

To refine the solution, we first balance the redox reactions and calculate the molarity of K₂Cr₂O₇, then H₂O₂.

1. Reaction between K₂Cr₂O₇ and H₂C₂O₄:

Cr₂O₇²⁻ + 8H⁺ + 3H₂C₂O₄ -> 2Cr³⁺ + 6CO₂ + 7H₂O

Here, n-factor for K₂Cr₂O₇ is 6 (Cr goes from +6 to +3) and for H₂C₂O₄ is 2 (C goes from +3 to +4).

Using M₁V₁n₁ = M₂V₂n₂:

$M_{{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}\times 2mL\times 6=1.0M\times 5mL\times 2$

$M_{{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}=\frac{1\times 5\times 2}{2\times 6}=\frac{10}{12}=\frac{5}{6}M$

2. Reaction between K₂Cr₂O₇ and H₂O₂:

Cr₂O₇²⁻ + 8H⁺ + 3H₂O₂ -> 2Cr³⁺ + 3O₂ + 7H₂O

Here, n-factor for K₂Cr₂O₇ is 6 and for H₂O₂ is 2 (O goes from -1 to 0).

Using M₁V₁n₁ = M₂V₂n₂:

$M_{{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}\times V_{{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}\times n_{{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}=M_{{\mathrm{H}}_2{\mathrm{O}}_2}\times V_{{\mathrm{H}}_2{\mathrm{O}}_2}\times n_{{\mathrm{H}}_2{\mathrm{O}}_2}$

$\frac{5}{6}M\times 40mL\times 6=M_{{\mathrm{H}}_2{\mathrm{O}}_2}\times 20mL\times 2$

$200=M_{{\mathrm{H}}_2{\mathrm{O}}_2}\times 40$

$M_{{\mathrm{H}}_2{\mathrm{O}}_2}=\frac{200}{40}=5M$

So, statement A is correct.

3. Volume strength of H₂O₂:

Volume strength = Molarity $\times$ 11.2 (at STP)

Volume strength = $5M\times 11.2=56V$

So, statements B and C are incorrect.

4. If 40 mL of $5/8M{H}_2{O}_2$ is added to 10 mL of the above H₂O₂ solution (which is 5 M):

Total moles of H₂O₂ = $(5M\times 10mL)+(\frac{5}{8}M\times 40mL)$

Total moles = $50+25=75mmol$

Total volume = $10mL+40mL=50mL$

Final molarity ($M_f$) = $\frac{75mmol}{50mL}=1.5M$

Volume strength of resulting solution = $1.5M\times 11.2=16.8V$

So, statement D is incorrect.

The final answer is $\boxed{A}$.