Question

The following equations are balanced atomwise and chargewise.
$\text{(i)}C{r}_2{O}_7^{2-}+8H^{+}+3H_2{O}_2\to 2C{r}^{3+}+7H_{2}O+3O_2\text{(ii)}C{r}_2{O}_7^{2-}+8H^{+}+5H_2{O}_2\to 2C{r}^{3+}+9H_{2}O+4O_2\text{(iii)}C{r}_2{O}_7^{2-}+8H^{+}+7H_2{O}_2\to 2C{r}^{3+}+11H_{2}O+5O_2$
The precise equation/equations representing the oxidation of $H_2{O}_2$ is/are:

• A. (i) only
• B. (ii) only
• C. (iii) only
• D. all the three

Answer 1

The correct option is A (i) only

n-factor for dichromate ion under acidic condition = 6
n-factor for hydrogen peroxide = 2
3 moles of hydrogen peroxide reduces 1 mole of dichromate.
The balanced chemical equation is given below:

$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O(H_2{O}_2\to O_2+2H^{+}+2e^{-})\times 3$
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$C{r}_2{O}_7^{2-}+8H^{+}+3H_2{O}_2\to 2C{r}^{3+}+7H_{2}O+3O_2$
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The reaction practically occurs with this stoichiometry.