1.

The emf in V of Danneil cell containing 0.1 M `ZnSO_(4)` and 0.01 M `CuSO_(4)` solutions their respective electrodes is `E_(Cu^(2+) | Cu)^(@) = 0.34`V and `E_(Zn^(2+) | Zn)^(@) = -0.76 ` VA. 1.10 VB. 1.16 VC. 1.13 VD. 1.07 V

Answer» Correct Answer - D
The Daniel cell is represented as
`Zn | Zn ^(2+) ||Cu^(2+) | Cu` and cell reactions
`Zn + Cu^(2+) to Zn^(2+) + Cu`
`E_(cell)^(@) = E_(RHS)^(@) - E_(LHS)^(@) = 0.34 - (-0.76) = 1.1 V , n = 2 `
A/C to Nernst equation
`E_(cell) = E_(cell)^(@) - (0.0592)/(n) log ([Zn^(2+)])/([Cu^(2+)]) because [Zn (s) ] = [ Cu(s) ] =1 `
`= 1.1 V - (0.0592)/(2) "log" (0.1)/(0.01)`
`= 1.1 V - (0.0592)/(2) "log" 10^(-1)`
=`1.1 V - (0.0592)/(2) = 1.07` V


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