For the redox reaction :
`Zn(s)+Cu^(2+)(aq) hArr Zn^(2+)(aq)+Cu(s)`
Reaction quotient (Q)`=([Zn^(2+)(aq)])/([Cu^(2+)(aq)])=0.01`. What will be the value of `E_(cell)` ?
Given that OA=1.10 V.

Explanation

OA represents `E_(cell)^(@)`. The value of `E_(cell)` can be calculated with the help of Nernst equation :
`E_(cell) = E_(cell)^(@) - (0.0591)/(n) log ([Zn^(2+)(aq)])/([Cu^(2+)(aq)])`
`=(1.10V)-(0.0591)/(2)log(0.01)=1.10 V=((0.0591))/(2)xx(-2)=1.159 V`