Home
Class 12
CHEMISTRY
Predict whether we can store CuSO(4) sol...

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`

Text Solution

Verified by Experts

`CuSO_(4)` solution can not be stored in zinc vessel, because Zn displaces Cu from `CuSO_(4)` solution giving `ZnSO_(4)` and copper metal. The reaction is as follows
`Zn + CuSO_(4)(aq) rarr ZnSO_(4)(aq) + Cu `
Here, Zn is anode and Cu is cathode.
Therefore, `E_("cell")^(@) = E_(Cu^(2+) //Cu) ^(@)- E_(Zn ^(2+)//Zn)^(@)`
`= 0.34 - ( 0.76) = 1.10 V`
Hence, `E_("cell")^(@)` is positive , so `Delta G` will be negative this reaction is feasible.
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE ( PART III ELECTROLCHEMICAL CELLS AND SERIES , NERNST EQUATIONS, BATTERIES AND CORROSION ) ( SHORT ANSWER TYPE Ii QUESTIONS) |8 Videos

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE ( PART III ELECTROLCHEMICAL CELLS AND SERIES , NERNST EQUATIONS, BATTERIES AND CORROSION ) ( LONG ANSWER TYPE QUESTIONS) |3 Videos

  • ELECTROCHEMISTRY

    ARIHANT PUBLICATION|Exercise QUESTIONS FOR PRACTICE ( PART III ELECTROLCHEMICAL CELLS AND SERIES , NERNST EQUATIONS, BATTERIES AND CORROSION ) ( VERY SHORT ANSWER TYPE QUESTIONS) |17 Videos

  • D-BLOCK ELEMENTS

    ARIHANT PUBLICATION|Exercise Chapter practice (Long answer type questions)|2 Videos

  • ELEMENTS : NITROGEN FAMILY

    ARIHANT PUBLICATION|Exercise CHAPTER PRACTICE ( Long Answer Type Questions ) |11 Videos

Similar Questions

Explore conceptually related problems

What does the negative sign in the expression E_(Zn^(2+) //Zn)^(@) = -0.76V mean ?

The emf of a cell corresponding to the reaction, Zn(s) + 2H^(+) ( aq) rarr Zn^(2+) ( 0.1 M ) + H_(2) ( g , 1 atm) is 0.28 V at 25^(@)C . Write the half-cell reaction and calculate the pH of the solution at the hydrogen electrode. E_(Zn^(2+)//Zn)^(@) = - 0.76V, E_(H^(+) //H_(2))^(@) = 0

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 .

Calculate the standard emf of the cell having the cell reaction. Zn(s) + Co^(2+) ( aq) rarr Zn^(2+) ( aq) + Co(s) E_(Zn//Zn^(2+)) ^(@) = 0.76V, E_(Co//Co^(2+))^(@) = 0.25V

E^@ Fe //Fe^(2+) = 0.44 whereas E^@ Cu // Cu^2+ = -0.32 v Then

Calculate the cell emf and Delta_(r ) G^(@) for the cell reation at 25^(@)C . Zn(s) | Zn^(@+) ( 0.1 M )|| Cd^(2+) ( 0.01M) | Cd(s) Given, E_(Zn^(2+) //Zn)^(@) = - 0.763V , E_(Cd^(2+) //Cd) ^(@) = - 0.403 V 1F = 96500 C "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) ], Find E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) then Delta _(r ) G^(@) by using formula, Delta _(r ) G^(@) = - n FE_("cell")^(@)