Question

For ${Cr}_2{O}_7^{-2}+14H^{+}+6e^{-}\to 2{Cr}^{+3}+7H_{2}O;E^o=1.33V$.

At $\left[{Cr}_2{O}_7^{-2}\right]=4.5$ millimole, $\left[{Cr}^{+3}\right]=15$ millimole, $E$ is $1.067V$. The pH of the solution is nearly equal to :

• A. $3$
• B. $4$
• C. $2$
• D. $5$

Answer 1

Answer: (C)

$2$

Using the Nernst equation,

$E=E^o-\frac{0.059}{nF}log\frac{[C{r}^{+3}{]}^2}{\left[C{r}_2{O}_7^{2-}\right][H^{+}{]}^{14}}$

$1.067=1.33-\frac{0.06}{6\times 96500}log\frac{(15\times {10}^{-3}{)}^2}{(4.5\times {10}^{-3})[H^{+}{]}^{14}}$

$\left[H^{+}\right]\simeq {10}^{-2}$

$pH=-log\left[H^{+}\right]=2$

Hence, option C is correct.