JIPMER 2016
- Exam
- JIPMER
- Year
- 2016
- Subject
- Chemistry
- Chapter
- Electrochemistry
JIPMER 2016 · Chemistry · Electrochemistry
For the cell reaction Pb+Sn^ 2+ Pb^ 2+ +Sn Given that, Pb Pb^ 2+ ,...
For the cell reaction
Given that,
What would be the ratio of cation concentration for which ?
Options
- A1/4
- B1/2
- C1/3
- D1/1
Correct answer
B. 1/2
Step-by-step solution
For the given cell reaction,
E_{\text{cell }^{\circ}=-0.14 \mathrm{ V}+0.13 \mathrm{ V}=-0.01 \mathrm{ V}
According to \begin{aligned} E_{\text{cell }} & =E_{\text{cell }^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Pb}^{2+}\right]}{\left[\mathrm{Sn}^{2+}\right]} \\ \Rightarrow \quad 0 & =-0.01=\frac{0.0591}{2} \log \frac{\left[\mathrm{Pb}^{2+}\right]}{\left[\mathrm{Sn}^{2+}\right]}\end{aligned}$$