For the cell ` Zn(s) | Zn^(2+) || Cu^(2+) | Cu(s)` , the standard cell voltage ` E_("cell")^(0)` is 1.10 V . When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.97 V . One possible explanaiton of the observed voltage is

Options

A. there are 2.00 mol of `Cu^(2+) ` bu only 1.00 mol of `Zn^(2+) `
B. the Zn electrode had twice the surface of the Cu electrode
C. the `[Zn^(2+)]` was larger than the `[Cu^(2+)]`
D. the volume of the `Zn^(2+)` solution was larger than the volume of the `Cu^(2+)` solution

Answer

C

Explanation

`Zn + Cu^(+2) to Cu + Zn^(+2) , E=E^(0) + (0.0591)/(2) log""((Cu^(+2))/(Zn^(+2)))`
`0.98 = 1.1 + (0.0591)/(2) log""([Cu^(+2)])/([Zn^(+2)]) ,` So , ` [Zn^(+2)] gt [Cu^(+2)]^(2)`