QJEEM08Cha66f: JEE Main 2008 Chemistry - Class 12 JEEM08Cha66f Solution: Given = -0.72 V; = -0.42V, The potential for the cell Cr

Question

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Given

E_{C{r^{3 + }}/Cr}^o

= -0.72 V;

E_{Fe^{2+}/Fe}^o

= -0.42V, The potential for the cell Cr | Cr³⁺ (0.1M) || Fe²⁺ (0.01 M) | Fe is

Accepted Answer

From the given representation of the cell,

{E_{cell}}

can be found as follows.

{E_{cell}} = E_{F{e^{2 + }}/Fe}^ \circ - E_{C{r^{3 + }}/Cr}^ \circ - {{0.059} \over 6}\log {{{{\left[ {C{r^{3 + }}} \right]}^2}} \over {{{\left[ {F{e^{2 + }}} \right]}^3}}}

\left[ {} \right.

Nernst - Equ.

\left. {} \right]

= - 0.42 - \left( { - 0.72} \right) - {{0.059} \over 6}\log {{{{\left( {0.1} \right)}^2}} \over {{{\left( {0.01} \right)}^3}}}

= - 0.42 + 0.72 - {{0.059} \over 6}\log {{0.1 \times 0.1} \over {0.01 \times 0.01 \times 0.01}}

= 0.3 - {{0.059} \over 6}\log {{{{10}^{ - 2}}} \over {{{10}^{ - 6}}}}

= 0.3 - {{0.059} \over 6} \times 4

= 0.30 - 0.0393 = 0.26\,V

Hence option (d) is correct answer.

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