Q. What is the potential of platinum wire dipped into Asolution of 0.1 M in ${Sn}^{2 +}$ and 0.01 M in ${Sn}^{4 +}$?
A) $E^{{^\circ}}$
B) $E^{{^\circ}}\$+ 0.059
C) $E^{{^\circ}} + \frac{0.059}{2}$
D) $E^{{^\circ}}$ - 0.059
Explanation
(c)
${Sn}^{2 +}$ → ${Sn}^{4 +}$ + ${2e}^{-}$
$E_{cell}$ = $E^{{^\circ}}$ - $\frac{0.059}{n}\log\frac{\left\lbrack {Sn}^{4 +} \right\rbrack}{\left\lbrack {Sn}^{2 +} \right\rbrack}$
= $E^{{^\circ}}$ - $\frac{0.059}{2}\log\left\lbrack \frac{0.01}{0.1} \right\rbrack$
= $E^{{^\circ}}$ + $\frac{0.059}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.