JEE MAIN - Chemistry (2007 - No. 27)
The cell, Zn | Zn2+ (1M) || Cu2+ (1M) | Cu($$E_{cell}^o$$ = 1.10V) was allowed to be completely discharged at 298 K. The relative concentration of Zn2+ to Cu2+ $$\left[ {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right]$$ is
antilog (24.08)
37.3
1037.3
9.65 $$\times$$ 104
Explanation
$${E_{cell}} = 0;\,\,$$ when cell is completely discharged.
$${E_{cell}} = {E^ \circ }_{cell} - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)$$
or$$\,\,\,\,0 - 1.1 - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)$$
$$\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {{2 \times 1.1} \over {0.059}} = 37.3$$
$$\therefore$$ $$\,\,\,\,\left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {10^{37.3}}$$
$${E_{cell}} = {E^ \circ }_{cell} - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)$$
or$$\,\,\,\,0 - 1.1 - {{0.059} \over 2}\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right)$$
$$\log \left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {{2 \times 1.1} \over {0.059}} = 37.3$$
$$\therefore$$ $$\,\,\,\,\left( {{{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}} \right) = {10^{37.3}}$$
Comments (0)
