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An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V?
Verified Answer
An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit c...
At the anode, either Ag can get oxidized to Ag+ or Sn2+ to Sn4+ or CI− to CI2 or SO2−4 to S2O2−8.
Their respective oxidation potential values are 0.799V,−0.13V,−13.6V, and −2V.
From these values, it is evident that Sn2+ would be oxidized first, followed by Ag at the anode.
At the cathode, either Na+ can get reduced to Na or Sn+ to Sn or H+ to H2.
The reduction potential value of Na+ is highly negative while for Sn2+∣Sn is −0.14V and for H+ +e− →12 H2:
(EH+/H2=−0.059log.110−7)−0.413V.
Thus, Sn2+ will get reduced at cathode followed by H+.
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An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit c...
Changes at the Anode:
1. Oxidation of Cl- ions: The Cl- ions present in the solution are more easily oxidized than water molecules. Therefore, the chloride ions will be oxidized to form Cl2 gas at the anode according to the half-reaction: 2Cl- -> Cl2 + 2e-.
2. Oxidation of S2O8 2- ions (persulfate ions): The persulfate ions are also more easily oxidized than water molecules. Therefore, they will be oxidized to form SO4 2- ions at the anode according to the half-reaction: S2O8 2- -> 2SO4 2- + 2e-.
Changes at the Cathode:
1. Reduction of Ag+ ions: The Ag+ ions present in the solution are more easily reduced than water molecules. Therefore, the silver ions will be reduced to form solid Ag at the cathode according to the half-reaction: Ag+ + e- -> Ag.
2. Reduction of Sn2+ ions: The Sn2+ ions present in the solution are less easily reduced than water molecules. Therefore, they will not be reduced at the cathode.
Summary of Electrode Reactions:
- Anode reactions:
- 2Cl- -> Cl2 + 2e-
- S2O8 2- -> 2SO4 2- + 2e-
- Cathode reactions:
- Ag+ + e- -> Ag
Explanation:
- The anode is the positive electrode where oxidation occurs, while the cathode is the negative electrode where reduction occurs.
- At the anode, both Cl- ions and S2O8 2- ions are oxidized. This means that these ions lose electrons and are converted into Cl2 gas and SO4 2- ions, respectively.
- At the cathode, only the Ag+ ions are reduced. This means that these ions gain electrons and are converted into solid Ag.
- The reduction of Sn2+ ions does not occur at the cathode because their reduction potential (-0.14V) is higher than the reduction potential of water (-0.83V). Therefore, water molecules are preferentially reduced at the cathode instead of Sn2+ ions.
- The overall cell reaction can be represented as:
- 2Ag+ + 2Cl- -> 2Ag + Cl2 (at the anode)
- Sn2+ and SO4 2- ions do not participate in the overall cell reaction.
- The standard electrode potentials provided can be used to determine the likelihood of the electrode reactions occurring. The reactions with more positive standard electrode potentials are more likely to occur.
1. Oxidation of Cl- ions: The Cl- ions present in the solution are more easily oxidized than water molecules. Therefore, the chloride ions will be oxidized to form Cl2 gas at the anode according to the half-reaction: 2Cl- -> Cl2 + 2e-.
2. Oxidation of S2O8 2- ions (persulfate ions): The persulfate ions are also more easily oxidized than water molecules. Therefore, they will be oxidized to form SO4 2- ions at the anode according to the half-reaction: S2O8 2- -> 2SO4 2- + 2e-.
Changes at the Cathode:
1. Reduction of Ag+ ions: The Ag+ ions present in the solution are more easily reduced than water molecules. Therefore, the silver ions will be reduced to form solid Ag at the cathode according to the half-reaction: Ag+ + e- -> Ag.
2. Reduction of Sn2+ ions: The Sn2+ ions present in the solution are less easily reduced than water molecules. Therefore, they will not be reduced at the cathode.
Summary of Electrode Reactions:
- Anode reactions:
- 2Cl- -> Cl2 + 2e-
- S2O8 2- -> 2SO4 2- + 2e-
- Cathode reactions:
- Ag+ + e- -> Ag
Explanation:
- The anode is the positive electrode where oxidation occurs, while the cathode is the negative electrode where reduction occurs.
- At the anode, both Cl- ions and S2O8 2- ions are oxidized. This means that these ions lose electrons and are converted into Cl2 gas and SO4 2- ions, respectively.
- At the cathode, only the Ag+ ions are reduced. This means that these ions gain electrons and are converted into solid Ag.
- The reduction of Sn2+ ions does not occur at the cathode because their reduction potential (-0.14V) is higher than the reduction potential of water (-0.83V). Therefore, water molecules are preferentially reduced at the cathode instead of Sn2+ ions.
- The overall cell reaction can be represented as:
- 2Ag+ + 2Cl- -> 2Ag + Cl2 (at the anode)
- Sn2+ and SO4 2- ions do not participate in the overall cell reaction.
- The standard electrode potentials provided can be used to determine the likelihood of the electrode reactions occurring. The reactions with more positive standard electrode potentials are more likely to occur.
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An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V?
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An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V?.
An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution containing Na ,Sn2 ,Cl-,SO4 2- ions, all at unit concentration is electrolyzed between a silver anode and a platinum cathode. What changes occur at the electrode when current is passed through a cell? Given: E Ag /Ag = 0.799V E Sn2 /Sn= -0.14V E Cl2/Cl-= 1.36V E S2O8 2-/SO4 2- =2V E Sn4 /Sn2 = 0.13V?.
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