Question

Blood alcohol $\left(C_2{H}_{5}OH\right)$ levels can be determined by the following titration:
$16H^{+}+2C{r}_2{O}_7^{2-}+C_2{H}_{5}OH\to 4C{r}^{3+}+2C{O}_2+11H_{2}O$
The endpoint is reached when the orange (dichromate ion) solution turns green (chromium ion).
How many electrons are transferred when a mole of alcohol is titrated to its endpoint?

• A. $1.8\times {10}^{24}$
• B. $3.6\times {10}^{24}$
• C. $7.2\times {10}^{24}$
• D. $9.6\times {10}^{24}$

Answer 1

The correct option is C $7.2\times {10}^{24}$

Before balancing the primary chemical reaction for this alcoholic test is

$C{r}_2{O}_7^{-2}+C_2{H}_{5}OH⟶C{r}^{+3}+C{O}_2$

Now, on balancing this equation with ion-electron method we see results:-

Oxidation half reaction:-

$12e^{-}+28H^{+}+2Cr{O}_7^{2-}⟶4C{r}^{+3}+14H_{2}O$

Reduction half reaction:- $n=12e^{-}$

$3H_{2}O+C_2{H}_{5}OH⟶2C{O}_2+12H^{+}+12e^{-}$

On combining we get given above equation

For $1$ mole number of electrons transferred is $\underset{\Downarrow }{12\times 6.023\times {10}^{23}}=7.2\times {10}^{24}$

(Avogadro number)