Question

Given : $E_{Cr^{3 +}/Cr}^{0} = - 0.72,E_{Fe^{2 +}/Fe}^{0} = - 0.42V$The

potential for the cell $\text{Cr }|\text{ C}\text{r}^{\text{3+}}(\text{0.1 M})\ ||\text{ F}\text{e}^{\text{2+}}(\text{0.01 M})\ |$ Fe at 298 K is : $(Take\frac{2.303R(298)}{F} = 0.06)$

• A. 0.339 V
• B. – 0.339 V
• C. – 0.26 V
• D. 0.26 V

Answer 1

Answer: (D)

0.26 V

Answer 2

$E_{cell} = E^{0}cell - \frac{0.059}{6}\log\frac{\left\lbrack Cr^{+ 3} \right\rbrack^{2}}{\left\lbrack Fe^{+ 2} \right\rbrack^{3}}$

$\text{= 0.3 - }\frac{\text{0.056}}{6}\text{log}\frac{\left( \text{0.1} \right)^{2}}{(0.01)^{3}}\text{= 0.3 - 0.04}$

$\text{= 0.26 V}$