Question

Calculate \[E^\circ \] and \[E\] for the cell \[Sn|S{n^{2 + }}(1M)||P{b^{2 + }}({10^{ - 3}}M)|Pb\] \[E^\circ (S{n^{2 + }}/Sn) = - 0.14V,E^\circ (P{b^{2 + }}/Pb) = - 0.13V\]. Is cell representation correct?

Answer 1

Hint: We can solve this question if we have a good understanding of representation of the galvanic cell and Nernst equation. By finding the standard electrode potential for the full cell we can calculate emf of the cell. Oxidation and reduction concepts help us in representation of the cell.

Complete Step by step answer:
Let us first discuss the representation of the cell and Nernst equation.
Cell is represented as:
- A double vertical line (||) is used to separate the anode half-cell from the cathode reaction. This is known as salt bridge.
- Anode is always on the left hand side of the salt bridge. Cathode is always on the right hand side.
A single vertical line (|) is used to separate different states of matter on the same side, and a comma is used to separate like states of matter on the same side.

Nernst Equation: Nernst showed that for a electrode reaction:
 $ {M^{n + }}(aq) + n{e^ - } \to M(s) $ the electrode potential at any concentration is measured with respect to standard hydrogen electrode can be represented by:
 $ {E_{({M^{n + }}/M)}} = E{^\circ _{({M^{n + }}/M)}} - \dfrac{{RT}}{{nF}}\ln \dfrac{{[M]}}{{{M^{n + }}}} $, R is gas constant(8.314),F is faraday constant(96487 C),T is temperature in kelvin and $ [{M^{n + }}] $ is the concentration of the species.
Now hope you all understand the concept behind this question.
\[Sn|S{n^{2 + }}(1M)||P{b^{2 + }}({10^{ - 3}}M)|Pb\]
In this cell we can see Sn is oxidizing to $ S{n^{2 + }} $ at anode and it is at left side so it correct and $ P{b^{2 + }} $ is reducing at cathode in right side so it is also correct and salt bridge is also present.
So the representation of the cell is correct.
Given:\[E^\circ (S{n^{2 + }}/Sn) = - 0.14V,E^\circ (P{b^{2 + }}/Pb) = - 0.13V\]
 $ {E_{cell}} = E^\circ (cathode) - E^\circ (anode) $
 $
 E{^\circ _{cell}} = E^\circ (cathode) - E^\circ (anode) \\
 \Rightarrow E{^\circ _{cell}} = - 0.13 - (0.14) = 0.01V \\
 $
Using Nernst equation:
 $ {E_{({M^{n + }}/M)}} = E{^\circ _{({M^{n + }}/M)}} - \dfrac{{RT}}{{nF}}\ln \dfrac{{[M]}}{{{M^{n + }}}} $
 $ \Rightarrow {E_{cell}} = 0.01 - \dfrac{{0.059}}{2}\log \left[ {\dfrac{{(S{n^{2 + }})}}{{(P{b^{2 + }})}}} \right] $, n=2 because 2e are taking place in electrode reaction.
 $ \Rightarrow {E_{cell}} = 0.01 - \dfrac{{0.059}}{2}\log \left[ {\dfrac{1}{{{{10}^{ - 3}}}}} \right] $
 $
 \Rightarrow {E_{cell}} = 0.01 - 0.0885 \\
 \Rightarrow {E_{cell}} = - 0.0785V \\
 $
Note: Always write the electrode reaction for cathode and anode to calculate the number of electrons transfer. You can use the LOAN trick word where L stands for left, O-oxidation, N-negative. Writing cathode and anode reactions are important to solve this question.