JEE MAIN - Chemistry (2008 - No. 23)
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
0.26 V
0.399 V
−0.339 V
−0.26 V
Explanation
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.
$${E_{cell}} = E_{F{e^{2 + }}/Fe}^ \circ - E_{C{r^{3 + }}/Cr}^ \circ - {{0.059} \over 6}\log {{{{\left[ {C{r^{3 + }}} \right]}^2}} \over {{{\left[ {F{e^{2 + }}} \right]}^3}}}$$ $$\left[ {} \right.$$ 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.
Comments (0)
