`Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq).`
Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])` . Variation of `E_(cell)` with log `Q` is of the type with `OA=1.10` `V.E_(cell) ` will be `1.1591V` when

Options

A. `0.1`
B. `0.01`
C. `100`
D. `1000`

Answer

C

Explanation

`E = E^(@) + (0.059)/(2)"log"([Cu^(2+)])/([Zn^(2+)])`
`= E^(@) + (0.059)/(2)logtheta`
`y = c + mx`
intercept `c = E^(@) = 1.10 V`
if `E = 1.1591 V` then,
`1.1591 = 1.10 + (0.059)/(2)log theta`
`log theta = (0.059 xx 2)/(0.059) = 2`
`theta = 100^(@)`