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A cell is set up between ‘Zn’ and ‘Cu’ e...

A cell is set up between ‘Zn’ and ‘Cu’ electrode. If the two half cells work under standard condition, calculate the cell potential. Given `E^@(Zn^(2+)//Zn)=-0.76V` and `E^@(Cu^(2+)//Cu)=+0.34V`.

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The standard cell potential for the cell is , Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu [E^@ for Zn^(2+)//Zn=-0.76V, E^@ for Cu^(2+)//Cu=+0.34V]

Write the cell reaction and cell potential of the cell Zn//Zn^(2+)(0.01 M)abs"Ag^+(0.1M)//Ag. Given E^0(Zn^(2+)//Zn)=-0.76V E^0(Ag^+//Ag)=+0.80V .

Write the cell reaction and calculate the emf of the Cell Zn|Co^2+(aq)|| Zn^2+(aq)| Co (Given E_(Zn|Zn^2+)^@=+0.76V and E_(Co|Co^2+)^@=+0.28V .

Find the cell potential at 25^@C for the cell- Mg//Mg^(2+)(0.001)//Cu^(2+)(0.0001 M)//Cu Given E^@(Mg^(2+)//Mg) = - 2.37V , E^@(Cu^(2+)//Cu) =+0.34V

Calculate the half cell potential at 25^@C for the cell reaction Cu^(2+)(aq)+2erarrCu , When [Cu^(2+)]=4M and E^@(Cu^(2+)//Cu)=0.34V .

Predict whether we can store CuSO_(4) solution in a zinc vessel from the following data. Show your calculation. E_(Zn^(2+)//Zn)^(@) = 0.76V E_(Cu^(2+)//Cu)^(@) = 0.34V

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