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E^(c-) of some elements are given as : ...

`E^(c-)` of some elements are given as `:`
`{:(I_(2)+2e^(-)rarr 2I^(c-),,,,E^(c-)=0.54V),(MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(c-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(c-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(c-)=0.1V):}`
Select the strongest oxidant and weakest oxidant among these elements.

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To solve the problem of identifying the strongest and weakest oxidants from the given standard reduction potentials (E° values), we can follow these steps: ### Step 1: Understand the Concept of Oxidants An oxidant is a substance that gains electrons in a chemical reaction, causing another substance to be oxidized. The strength of an oxidant is determined by its standard reduction potential (E°). The higher (more positive) the E° value, the stronger the oxidant. ### Step 2: List the Given Reduction Potentials From the question, we have the following standard reduction potentials: 1. \( \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \), \( E° = 0.54 \, \text{V} \) ...
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E^(-) of some elements are given as : {:(I_(2)+2e^(-)rarr 2I^(-),,,,E^(-)=0.54V),(MnO_(4)^(-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(-)=0.1V):} a. Select the stronges reductant and weakes oxidant among these elements. b. Select the weakest reductant and strongest oxidant among these elements.

E^(o) of some elements are given as : {:(I_(2)+2e^(-)rarr 2I^(-),,,,E^(o)=0.54V),(MnO_(4)^(-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(o)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(o)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(o)=0.1V):} a. Select the strongest reductant and weakest oxidant among these elements. b. Select the weakest reductant and strongest oxidant among these elements.

If {:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):} then which of these is true?

Given below are a set of half-cell reactions (acidic medium) along with their E_(@) with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. {:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} Among the following, identify the correct statement:

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Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential (E^(c-)) of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions ( acidic medium ) along with their E^(c-)(V with respect to normal hydrogen electrode ) values. Using this data, obtain correct explanations for Question. I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54 Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36 Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50 Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77 O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23 Among the following, identify the correct statement. (a)Chloride ion is oxidized by O_(2) . (b) Fe^(2+) is oxidized by iodine. (c)Iodide ion is oxidized by chlorine (d) Mn^(2+) is oxidized by chlorine.

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