Describe electrolytic refining of Copper?
-10 Derive laws of resistance in series so as to explain equivalent resistance of
circuit?
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In electrolytic refining of copper, the electrolyte is a solution of acidified copper sulphate. The anode is impure copper, whereas the cathode is a strip of pure copper on passing electric current, pure copper is deposited on the cathode.
or,
Electrolytic refining is a process of refining a metal (mainly copper) by the process of electrolysis. As far as the mechanism of the process is concerned, during electrolysis, a large chunk or slab of impure metal is used as the anode with a thin strip of pure metal at the cathode.
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The equivalent resistance is the algebraic sum of the resistances (Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance: I=VRS=9V90Ω=0.1A.
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The purification of metals by electrolysis. It is commonly applied to copper. A large piece of impure copper is used as the anode with a thin strip of pure copper as the cathode. ... Gold and silver in the impure copper form a so-called anode sludge at the bottom of the cell, which is recovered.
The equivalent resistance is the algebraic sum of the resistances (Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance: I=VRS=9V90Ω=0.1A.
Explanation:
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