Question

For $C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
$E^0=+1.33V$ at $\left[C{r}_2{O}_7^{2-}\right]=4.5$ millimole per ltr $\left[C{r}^{3+}\right]=15$ millimole per ltr , $E$ is $1.06V$. The $pH$ of the solution is nearly equal to:

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

Answer: (A)

$E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.059}{n}$ \log $\frac{{\left[C{r}^{3+}\right]}^2{\left[H_{2}O\right]}^7}{\left[C{r}_2{O}_7^{2-}\right]{\left[H^{+}\right]}^{14}}$

$1.06=1.33-\frac{0.059}{6}$

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

$0.27=\frac{0.059}{6}\log \left[\frac{225\times 10^{-6}}{\left(4.5\times 10^{-3}\right){\left[H^{+}\right]}^{14}}\right]$

$27.45=\log \left[\frac{50\times 10^{-3}}{{\left[H^{+}\right]}^{14}}\right]$

$27.45=\log \left[50\times 10^{-3}\right]-\log {\left[H^{+}\right]}^{14}$

$27.45=-1.3010-14\log \left[H^{+}\right]$

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

$28.751=14\times -\log \left[H^{+}\right]$

$28.041=14pH$

$pH=2$