Question

Calculate the potential of the following cell :

$E_{C{o}^{2+}|C{O}^{3+}}^{\ominus }=-2V,E_{C{r}_2{O}_7^{2-}|C{r}^{3+}}^{\ominus }=+1.0V$

$Pt\left|\begin{array}{c}C{o}^{2+},C{o}^{3+}\\ \left(2.0M\right)\left(1.0M\right)\end{array}\right|\begin{array}{c}C{r}^{3+},C{r}_2{O}_7^{2-},H^{\oplus }\\ \left(1.0M\right)\left(4.0M\right)\left(1.0M\right)\end{array}|Pt$

• A. 1.024 V
• B. -1.024 V
• C. 0.976 V
• D. -0.976 V

Answer 1

The correct option is D -0.976 V

$C{o}^{2+}\to C{o}^{3+}+e^{-}]\times 6$

$14H^{\oplus }+6e^{-}+C{r}_{2}O{7}^{2-}\to 2C{r}^{3+}+7H_{2}O$

$E_{cell}^{\ominus }=E_c^{\ominus }-E_a^{\ominus }(E_{red}^{\ominus }C{o}^{3+}|C{o}^{2+}=2V)$
$=1-2=-1V$
$E=E^{\ominus }-\frac{0.059}{6}log\frac{\left[C{o}^{3+}{]}^6\right[C{r}^{3+}{]}^2}{\left[C{o}^{2+}{]}^6\right[H^{\oplus }{]}^{14}\left[C{r}_2{O}_7^{2-}\right]}$
$=-1-0.01log\frac{(1{)}^6\times (1{)}^2}{(2{)}^6\times (1{)}^{14}\times \left(4\right)}$
$=-1-0.01log(2{)}^{-8}$
$=-1+0.01\times 8\times 0.3$
$=-0.976V$