Find the amount of silver liberated cathode 0-500 A of current is passed through AgNO-3 electrolyte 1 hour- Atomic weight of silver is 107-9-g-mol-1-

Question

Find the amount of silver liberated cathode 0-500 A  of current is passed through AgNO-3 electrolyte 1 hour- Atomic weight of silver is 107-9-g-mol-1-

Toppr Admin · Accepted Answer

Atomic weight of silvenis - -107-9g-mol-I-0-500A-The ECE of silverEquivalent mass of silver-E-Ag-107-9-160- -160- -160-As Ag is monoatomic-Z-Ag-dfrac - E - f - -dfrac - 107-9 - 96500 - -0-001118-M-Zit-0-0011178-times 0-5-times 3600-2-01g-

Find the amount of silver liberated at cathode if $$0.500 A $$ of current is passed through $$AgNO_3$$ electrolyte for $$1$$ hour. Atomic weight of silver is $$107.9\,g\,mol^{-1}$$

Atomic weight of silvenis = $$107.9g/mol$$
$$I=0.500A$$
The ECE of silver
Equivalent mass of silver,$$E_{Ag}=107.9$$ [As Ag is monoatomic]
$$Z_{Ag}=\dfrac { E }{ f } =\dfrac { 107.9 }{ 96500 } =0.001118$$
$$M=Zit=0.0011178\times 0.5\times 3600=2.01g$$

Find the amount of silver liberated at cathode if $$0.500 A $$ of current is passed through $$AgNO_3$$ electrolyte for $$1$$ hour. Atomic weight of silver is $$107.9\,g\,mol^{-1}$$

Atomic weight of silvenis = $$107.9g/mol$$$$I=0.500A$$The ECE of silverEquivalent mass of silver,$$E_{Ag}=107.9$$ [As Ag is monoatomic]$$Z_{Ag}=\dfrac { E }{ f } =\dfrac { 107.9 }{ 96500 } =0.001118$$$$M=Zit=0.0011178\times 0.5\times 3600=2.01g$$

Atomic weight of silvenis = $$107.9g/mol$$
$$I=0.500A$$
The ECE of silver
Equivalent mass of silver,$$E_{Ag}=107.9$$ [As Ag is monoatomic]
$$Z_{Ag}=\dfrac { E }{ f } =\dfrac { 107.9 }{ 96500 } =0.001118$$
$$M=Zit=0.0011178\times 0.5\times 3600=2.01g$$