Answer
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Hint: The metals that has higher oxidation potential than the metal that is getting refined electrolytically, is collected in their ionic form in the solution and the metal that has lower oxidation potential then the metal getting purified, gets deposited at the bottom as anodic mud.
Complete answer:
In the electrolytic refining of any metal, the impure metal is taken as anode and the pure metal we need to extract is taken as cathode. In refining metal, we take pure metal as cathode because reduction occurs at cathode and so the metal we want in pure form will also get collected at cathode.
- The electrolytic refining of metal removes the impurities from the impure metal by either disposing them as anode mud or in solution as ionic form of them.
- Now, all the metal atoms that have higher oxidation potential than the metal that we are refining, will get oxidised but not get reduced at the cathode. So, they will remain in their ionic form in the solution. The metal that has lower oxidation potential than the metal getting purified, does not oxidise at all and gets deposited at the bottom of the container as anodic mud. So, let’s compare the oxidation potentials of the metals given and decide which will be there in the anodic mud.
\[{E^ \circ }_{Cu/C{u^{2 + }}} = - 0.34V\]
\[{E^ \circ }_{Au/A{u^{3 + }}} = - 1.40V\]
\[{E^ \circ }_{Ag/A{g^ + }} = - 0.80V\]
\[{E^ \circ }_{Zn/Z{n^{2 + }}} = + 0.76V\]
We can say that higher the oxidation potential of the metal, higher its tendency is to get oxidised. So, as we earlier studied, the metals that have higher oxidation potential than the metal getting purified means Silver remains as ions in the solution.
- That means Cu and Zn if present as an impurity in the metal, then they will be there in the solution only. The metal that has lower oxidation potential than the metal getting purified, then it will not get oxidised at get collected at the bottom of the container which is called anode mud. So, only Au will be there in anodic mud.
So, the correct answer is “Option c”.
Note: Remember that here the tendency of metals to get oxidised and not getting oxidised is explained with reference to its oxidation potential, do not get confused by reduction potential of the metal as this will inverse the property. It is okay if you do not remember the exact values of the electrode potential values of the metals, but you should be able to say whose electrode potential will be higher or lower with respect to other methods.
Complete answer:
In the electrolytic refining of any metal, the impure metal is taken as anode and the pure metal we need to extract is taken as cathode. In refining metal, we take pure metal as cathode because reduction occurs at cathode and so the metal we want in pure form will also get collected at cathode.
- The electrolytic refining of metal removes the impurities from the impure metal by either disposing them as anode mud or in solution as ionic form of them.
- Now, all the metal atoms that have higher oxidation potential than the metal that we are refining, will get oxidised but not get reduced at the cathode. So, they will remain in their ionic form in the solution. The metal that has lower oxidation potential than the metal getting purified, does not oxidise at all and gets deposited at the bottom of the container as anodic mud. So, let’s compare the oxidation potentials of the metals given and decide which will be there in the anodic mud.
\[{E^ \circ }_{Cu/C{u^{2 + }}} = - 0.34V\]
\[{E^ \circ }_{Au/A{u^{3 + }}} = - 1.40V\]
\[{E^ \circ }_{Ag/A{g^ + }} = - 0.80V\]
\[{E^ \circ }_{Zn/Z{n^{2 + }}} = + 0.76V\]
We can say that higher the oxidation potential of the metal, higher its tendency is to get oxidised. So, as we earlier studied, the metals that have higher oxidation potential than the metal getting purified means Silver remains as ions in the solution.
- That means Cu and Zn if present as an impurity in the metal, then they will be there in the solution only. The metal that has lower oxidation potential than the metal getting purified, then it will not get oxidised at get collected at the bottom of the container which is called anode mud. So, only Au will be there in anodic mud.
So, the correct answer is “Option c”.
Note: Remember that here the tendency of metals to get oxidised and not getting oxidised is explained with reference to its oxidation potential, do not get confused by reduction potential of the metal as this will inverse the property. It is okay if you do not remember the exact values of the electrode potential values of the metals, but you should be able to say whose electrode potential will be higher or lower with respect to other methods.
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