Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E⊖) of two half cell reactions decided which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half cell reactions (acidic medium) along with their E⊖ (V with respect to normal hydrogen electrode) values. Using this data, obtain correct explanations for the following questionI2+2e−→2I⊖; E⊖=0.54Cl2+2e−→2Cl⊖; E⊖=1.36Mn3++e−→Mn2+; E⊖=1.50Fe3++e−→Fe2+; E⊖=0.77O2+4H⨁+4e−→2H2O; E⊖=1.23While Fe3+ is stable, Mn3+ is not stable in acid solution because

Redox reactions play a pivotal role in chemistry and biology. The valu

1.	Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E⊖) of two half cell reactions decided which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half cell reactions (acidic medium) along with their E⊖ (V with respect to normal hydrogen electrode) values. Using this data, obtain correct explanations for the following questionI2+2e−→2I⊖; E⊖=0.54Cl2+2e−→2Cl⊖; E⊖=1.36Mn3++e−→Mn2+; E⊖=1.50Fe3++e−→Fe2+; E⊖=0.77O2+4H⨁+4e−→2H2O; E⊖=1.23While Fe3+ is stable, Mn3+ is not stable in acid solution because
Answer» Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential $(E_{⊖})$ of two half cell reactions decided which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half cell reactions (acidic medium) along with their $E_{⊖}$ (V with respect to normal hydrogen electrode) values. Using this data, obtain correct explanations for the following question $I_{2} + 2 e^{-} \to 2 I^{⊖}; E^{⊖} = 0.54$ $C l_{2} + 2 e^{-} \to 2 C l^{⊖}; E^{⊖} = 1.36$ $M n^{3 +} + e^{-} \to M n^{2 +}; E^{⊖} = 1.50$ $F e^{3 +} + e^{-} \to F e^{2 +}; E^{⊖} = 0.77$ $O_{2} + 4 H^{⨁} + 4 e^{-} \to 2 H_{2} O; E^{⊖} = 1.23$ While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stable in acid solution because

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