Question

For $C{r}_2{O}_7^{-2}+14H^{+}+6e^{-}\to 2C{r}^{+3}+7H_{2}O;E$${}^{\circ }$= $1.33VAt[C{r}_2{O}_7^{-2}]=4.5$ milli mole $[C{r}^{+3}]=15$ milli mole, $E$ is $1.067V$.The pH of the solution is nearly equal to

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (C)

$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}⟶2C{r}^{3+}+7H_{2}O$

$E=E^{\circ }-\frac{0.059}{n}\log \frac{{\left[C{r}^{3+}\right]}^2{\left[H_{2}O\right]}^7}{\left[C{r}_2{O}_7^{2-}\right]{\left[H^{+}\right]}^{14}}$

$1.067=1.33-\frac{0.059}{6}$

$\log \frac{{\left(15\times 10^{-3}\right)}^2(1{)}^7}{\left(4.5\times 10^{-3}\right){\left[H^{+}\right]}^{14}}$

$\frac{0.059}{6}\log \left[\frac{225\times 10^{-6}}{\left(4.5\times 10^{-3}\right)\times {\left[H^{+}\right]}^{14}}\right]=0.263$

$\log \left[\frac{225\times 10^{-6}}{\left(4.5\times 10^{-3}\right)\times {\left[H^{+}\right]}^{14}}\right]$

$=\frac{0.263\times 6}{0.059}=26.74$

$\log \left[\frac{50\times 10^{-3}}{{\left[H^{+}\right]}^{14}}\right]=26.74$

$\log \left(50\times 10^{-3}\right)-\log {\left(H^{+}\right)}^{14}=26.74$

$-1.3010-14\log \left[H^{+}\right]=26.74$

$-14\log \left[H^{+}\right]=26.74+1.3010$

or $+14pH=28.041$

$pH=\frac{28.041}{14}$

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

$=2$