Number of coulombs required to deposit 50g of silver at cathode from silver nitrate solution (atomic mass of Ag = 108)

44675.9 C

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96500 C

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208,440 C

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193,000 C

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Solution

The correct option is A 44675.9 C
$A g^{+} + e^{-} \to A g 96.500 C - 108 g o f A g ? - 50 g = \frac{50 \times 96500}{108} = 44675.9 C$

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Similar questions

Q. Calculate the number of coulombs required to deposit

50 g

of silver at cathode from silver nitrate solution.
(Atomic mass of silver

= 108

When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.3, approximately 108) the amount of silver deposited is:

When 9.65 coulombs of electricity is passed through a solution of silver nitrate

(atomic weight of taking as 108) the amount of silver deposited is

[EAMCET 1992; KCET 2000]

Q. How many kgs of silver would be deposited at cathode if

0.0342 d m^{3}

of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolyzed between platinum electrodes?(Given : Atomic masses of Ag = 108, O =16)

Q. How many grams of silver would be deposited at cathode if

0.05 d m^{3}

of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolysed between platinum electrodes?
(Given: Atomic mass of

A g = 108, O = 16

)

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Number of coulombs required to deposit 50g of silver at cathode from silver nitrate solution (atomic mass of Ag = 108)

The correct option is A 44675.9 CAg++e−→Ag96.500 C−108 g of Ag ?−50 g=50×96500108=44675.9 C

The correct option is A 44675.9 C
$A g^{+} + e^{-} \to A g 96.500 C - 108 g o f A g ? - 50 g = \frac{50 \times 96500}{108} = 44675.9 C$