JEE MAIN - Chemistry (2003 - No. 35)
For the redox reaction Zn(s) + Cu2+(0.1 M) $$\to$$ Zn2+(1M) + Cu(s) taking place in a cell, $$E_{cell}^o$$ is 1.10 volt. Ecell for the cell will be ($$2.303{{RT} \over F}$$ = 0.0591)
1.80 volt
1.07 volt
0.82 volt
2.14 volt
Explanation
$${E_{cell}} = {E^ \circ }_{cell} + {{0.059} \over n}\log {{\left[ {C{u^{ + 2}}} \right]} \over {\left[ {Z{n^{ + 2}}} \right]}}$$
$$ = 1.10 + {{0.059} \over 2}\log \left[ {0.1} \right]$$
$$ = 1.10 - 0.0295$$
$$ = 1.07V$$
$$ = 1.10 + {{0.059} \over 2}\log \left[ {0.1} \right]$$
$$ = 1.10 - 0.0295$$
$$ = 1.07V$$
Comments (0)
