2 electrode cell containing silver nitrate solution and copper sulphate solution in series a current of 2.5 ampere pass through till0.78 gram of silver was deposited how llong will the current flow and how long will be copper deposited

Open in App

Solution

1 mole of silver contains 47 g
Therefore, 0.78 g of silver contains 0.016 moles of silver: 0.016 moles of silver electrons
Therefore quantity of charge Q = 0.016 * 96500
= 1544 C
Q = I * t
t = Q / I
= 1544 / 2
= 772 seconds

Suggest Corrections

Similar questions

Q. Two electrolytic cells containing silver nitrate & copper sulphate solution are connected in a series. The steady current of

2.5

A was passed thought them till 1.018g of silver were deposited. How long did the current flow & what is the mass of Cu deposited?

Q. An electric current is passed through silver nitrate solution using silver electrodes.

15.28

g of silver was found to be deposited on cathode. What will be the weight of copper deposited on cathode if same amount of electricity is passed through copper sulphate solution using copper electrodes?

Three electrolytic cells A,B,C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Three elctrolytic cells A,B,C containing solutions of $Z n S O_{4}, A g N O_{3}$ and $C u S O_{4}$ respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver deposited at the cathode3 of cell B. How long did the current flow ? What mass of copper and zinc were deposited ? [Atomic masses; Cu = 63,5, Zn = 65.3; Ag = 108]

When an electric current is passed through a solution containing copper sulphate, copper gets deposited on which electrode?

Join BYJU'S Learning Program

2 electrode cell containing silver nitrate solution and copper sulphate solution in series a current of 2.5 ampere pass through till0.78 gram of silver was deposited how llong will the current flow and how long will be copper deposited

1 mole of silver contains 47 gTherefore, 0.78 g of silver contains 0.016 moles of silver: 0.016 moles of silver electronsTherefore quantity of charge Q = 0.016 * 96500= 1544 CQ = I * tt = Q / I= 1544 / 2= 772 seconds

1 mole of silver contains 47 g
Therefore, 0.78 g of silver contains 0.016 moles of silver: 0.016 moles of silver electrons
Therefore quantity of charge Q = 0.016 * 96500
= 1544 C
Q = I * t
t = Q / I
= 1544 / 2
= 772 seconds