
During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’. These are
(A) Fe and Ni
(B) Ag and Au
(C) Pb and Zn
(D) Se and Ag
Answer
576.3k+ views
Hint: Only the ions which are less basic than copper and are present as the impurity in copper ore can be found in the anodic mud. Higher the reduction potential of the ion, more basic it will be.
Complete step by step solution:
- In electrolytic refining of the metal, impure metal rod is used as anode and pure metal is used as cathode. So, copper first gets oxidized and gets reduced at cathode and this process we can separate pure copper. The solution we use is an acidified solution of the metal we are refining. So, we use an acidified solution of copper salt.
- Now, we generally refine blister copper by electrolytic method to gain pure copper. Blister copper has usually around 98.5% of purity but in other parts, impurities may be present. These impurities include Selenium, Tellurium, Gold, Silver, Platinum, Antimony and Selenium.
- So, the more basic metals than copper remain present in the solution in the form of ions and the metals that are less basic than copper will get deposited as anode mud.
- More basic metals than copper will have lower oxidation potentials than copper and hence they will not get oxidized in the process and will get collected at the bottom in the form of anodic mud.
- Let’s find which of the given metals have lower oxidation potential than copper.
\[Fe \to F{e^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.44V
\[Ni \to N{i^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.25V
\[Pb \to P{b^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.13V
\[Zn \to Z{n^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.76V
\[Ag \to A{g^ + } + {e^ - }\] \[{E_{oxi}}\]= -0.80V
\[Au \to A{u^{3 + }} + 3{e^ - }\] \[{E_{oxi}}\]= -1.40V
\[Cu \to C{u^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]= -0.34V
We can clearly see that Ag and Au only have lower oxidation potential than copper and hence they will not get oxidized at all and so they will get deposited as anodic mud.
Note: Do not get confused between oxidation potential and reduction potential of metals. You do not need to remember the exact oxidation potential values of all metals, if you can remember the oxidation potential of metals comparatively.
Complete step by step solution:
- In electrolytic refining of the metal, impure metal rod is used as anode and pure metal is used as cathode. So, copper first gets oxidized and gets reduced at cathode and this process we can separate pure copper. The solution we use is an acidified solution of the metal we are refining. So, we use an acidified solution of copper salt.
- Now, we generally refine blister copper by electrolytic method to gain pure copper. Blister copper has usually around 98.5% of purity but in other parts, impurities may be present. These impurities include Selenium, Tellurium, Gold, Silver, Platinum, Antimony and Selenium.
- So, the more basic metals than copper remain present in the solution in the form of ions and the metals that are less basic than copper will get deposited as anode mud.
- More basic metals than copper will have lower oxidation potentials than copper and hence they will not get oxidized in the process and will get collected at the bottom in the form of anodic mud.
- Let’s find which of the given metals have lower oxidation potential than copper.
\[Fe \to F{e^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.44V
\[Ni \to N{i^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.25V
\[Pb \to P{b^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.13V
\[Zn \to Z{n^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]=0.76V
\[Ag \to A{g^ + } + {e^ - }\] \[{E_{oxi}}\]= -0.80V
\[Au \to A{u^{3 + }} + 3{e^ - }\] \[{E_{oxi}}\]= -1.40V
\[Cu \to C{u^{2 + }} + 2{e^ - }\] \[{E_{oxi}}\]= -0.34V
We can clearly see that Ag and Au only have lower oxidation potential than copper and hence they will not get oxidized at all and so they will get deposited as anodic mud.
Note: Do not get confused between oxidation potential and reduction potential of metals. You do not need to remember the exact oxidation potential values of all metals, if you can remember the oxidation potential of metals comparatively.
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