The emf (in V) of a Daniell cell containing 0.1 M `ZnSO_(4)` and 0.01 M `CuSO_(4)` solutions at their respective electrodes is `(E_((Cu^(2+))/(Cu))^@ = +0.34 V, E_((Zn^(2+))/(Zn))^@ = -0.76 V)`

Options

A. 1.1
B. 1.16
C. 1.13
D. 1.07

Answer

D

Explanation

`E_(cell)^@=(E_((Cu^(2+))/(Cu))- E_((Zn^(2+))/(Zn)))`But, `CuSO_(4) + Zn to ZnSO_(4) + Cu`
but
`E_(cell)= E_(cell)^@- 0.059/n "log"([Zn^(2+)])/([Cu^(2+)])`
( `because Zn^(2+)` is the product and `Cu^(2+)` is the reactant)
` E_(cell)=1.1-0.059/2 "log"0.1/0.01`1.1-0.0295
`E_(cell)=1.07V`