QJEEAdv03Chcf01: JEE Advanced 2003 Chemistry - Class 12 JEEAdv03Chcf01 Solution: Two students use the same stock solution of ZnSO 4 and solution …

The two given cells are represented as :

$Zn(s)|Z{n^{2 + }}({C_1})||C{u^{2 + }}(aq)(C = ?)|Cu(s)$ , ${E_{cell}} = {E_1}$

$Zn(s)|Z{n^{2 + }}({C_2})||C{u^{2 + }}(aq)(C = 0.5\,M)|Cu(s)$ , ${E_{cell}} = {E_2}$

Given, E₂ > E₁ , E₂ $-$ E₁ = 0.03 and C₁ = C₂ [concentration of Zn²⁺ is the same in both the solutions]

$\therefore$ Cell reaction : $Zn(s) + C{u^{2 + }}(aq)$ $\rightleftharpoons$ $Z{n^{2 + }}(aq) + Cu(s)$

$\therefore$ ${E_{cell}} = E_{cell}^0 - {{2.303RT} \over {2F}}\log {{[Z{n^{2 + }}]} \over {[C{u^{2 + }}]}}$

For cell 1, ${E_1} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}$ [Given : ${{2.303RT} \over F} = 0.06$ ]

For cell 2, ${E_2} = E_{cell}^0 - {{2.303RT} \over {nF}}\log {{{C_2}} \over {0.05}}$

or, ${E_2} = E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}$

$\therefore$ ${E_2} - {E_1} = \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_2}} \over {0.05}}} \right) - \left( {E_{cell}^0 - {{0.06} \over 2}\log {{{C_1}} \over C}} \right)$

or, $0.03 = {{0.06} \over 2}\left( {\log {{{C_2}} \over C} \times {{0.5} \over {{C_1}}}} \right) = {{0.06} \over 2}\log {{0.5} \over C}$ [ $\because$ C₁ = C₂]

or, $\log {{0.5} \over C} = {{0.03 \times 2} \over {0.06}}$ or, ${{0.5} \over C} = 10$

$\therefore$ C = 0.05 M