Question

In concentration cell Fe/Fe2+(C) || Fe2+(C2)/Fe, for spontaneous cell reaction ​

Answer 1

Answer:

a.c2>c1 is the correct answer

Answer 2

Answer:

The correct option is (1)

Explanation:

$\[Fe/F{e^{2 + }}({C_1})||F{e^{2 + }}/Fe({C_2})\]$

First, write the balanced chemical reaction and find n-factor

Oxidation half:

$\[Fe \to F{e^{2 + }} + 2{e^ - }\]$

Reduction half:

$\[F{e^{2 + }} + 2{e^ - } \to Fe\]$

Final reaction:

$\[F{e^{2 + }}({C_2}) \to F{e^{2 + }}({C_1})\]$

Since the total number of electron changes is 2, therefore, n-factor is 2.

Using the Nernst equation:

${E_{cell}} = E_{cell}^0 - \frac{{0.0591}}{n}\log \left( {\frac{{{\text{Concentration of product}}}}{{{\text{Concentration of reactant}}}}} \right) \hfill$

${E_{cell}} = E_{cell}^0 - \frac{{0.0591}}{n}\log \left( {\frac{{{\text{Concentration of F}}{{\text{e}}^{2 + }}(P)}}{{{\text{Concentration of F}}{{\text{e}}^{2 + }}(R)}}} \right) \hfill$

${E_{cell}} = E_{cell}^0 - \frac{{0.0591}}{2}\log \left( {\frac{{{C_1}}}{{{C_2}}}} \right) \hfill$

For spontaneous cells, the reaction $E_{cell}$ must be positive.

So, $log \left( {\frac{{{C_1}}}{{{C_2}}}} \right) \hfill$ must be negative to make $- \frac{{0.0591}}{2}\log \left( {\frac{{{C_1}}}{{{C_2}}}} \right) \hfill$ positive and the result  $E_{cell}$ will be positive and cell reaction will be spontaneous.

Therefore, $C_2 > C_1$