Discussion Forum

Standard cell voltage for the cell Pb | Pb²⁺ ll Sn²⁺ | Sn is - 0.01 V. If the cell is to exhibit E_Cell  = 0, the value of [Sn²⁺]/ [Pb²⁺] should be antilog of -

Answer:

Explanation:

Apply Nernst equation to the reaction

$ Pb +Sn^{2+}\rightarrow Pb^{2+}+Sn$

Or $ E^{\circ}+\frac{0.059}{2}\log_{}{}\frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} =E_{Cell}$

Or $log_{}{}\frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} = \frac{0.01\times2}{0.059}=0.3(\because E_{Cell} =0)$

Or $\frac{\left[Sn^{2+}\right]}{\left[Pb^{2+}\right]} = antilog(0.3)$