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Using the data find out the strongest re...

Using the data find out the strongest reducing agent -
`E_(Cr_(2)O_(7)^(2-)|Cr^(+))^(@)=+1.33V, E_(Cl_(2)|Cl^(-))^(@)=+1.36V`
`E_(MnO_(4)^(-)|Mn^(2+))^(@)=+1.51V, E_(Cr^(2+)|Cr)^(@)=-0.74V`

A

`Cl^(-)`

B

`Cr`

C

`Cr^(3+)`

D

`Mn^(+2)`

Text Solution

Verified by Experts

The correct Answer is:
B
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Use the given data - E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=+1.33V, E_(Cl_(2)|Cl^(-))^(@)=+1.36V E_(MnO_(4)^(-)|Mn^(2+))^(@)=+1.51V, E_(Cr^(3+)|Cr)^(@)=-0.74V and find out the most stable ion in its reduced form -

Use the given data - E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=+1.33V, E_(Cl_(2)|Cl^(-))^(@)=+1.36V E_(MnO_(4)^(-)|Mn^(2+))^(@)=+1.51V, E_(Cr^(3+)|Cr)^(@)=-0.74V and find out the most stable oxidised species -

Use the given data E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=+1.33V, E_(Cl_(2)|Cl^(-))^(@)=+1.36V E_(MnO_(4)^(-)|Mn^(2+))^(@)=+1.51V, E_(Cr^(3+)|Cr)^(@)=-0.74V find out in which option the order of reducing power is correct -

Find the stronger oxidising agent in each pair, stating reason. (d) Cr^(3+), Fe^(3+) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Find the stronger oxidising agent in each pair, stating reason. (a) Cl_(2), Br_(2) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Fe has a much lower tendency to get oxidised than Mn and Cr - explain. [Given : E_(Cr^(2+)|Cr)^(0)=-0.9V, E_(Cr^(3+)|Cr^(2+))^(0)=-0.4V E_(Mn^(2+)|Mn)^(0)=-1.2V, E_(Mn^(3+)|Mn^(2+))^(0)=+1.5V E_(Fe^(2+)|Fe)^(0)=-0.4V, E_(Fe^(3+)|Fe^(2+))^(0)=+0.8V ]

Find the stronger oxidising agent in each pair, stating reason. (c) Ag^(+), Cu^(2+) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Stability of Fe^(3+) ion is greater than Mn^(3+) ion but less than Cr^(3+) - explain. [Given : E_(Cr^(2+)|Cr)^(0)=-0.9V, E_(Cr^(3+)|Cr^(2+))^(0)=-0.4V E_(Mn^(2+)|Mn)^(0)=-1.2V, E_(Mn^(3+)|Mn^(2+))^(0)=+1.5V E_(Fe^(2+)|Fe)^(0)=-0.4V, E_(Fe^(3+)|Fe^(2+))^(0)=+0.8V ]

Identify X^(-) In the following spontaneous reactions Cl_(2)+2X^(-)+X_(2) , where X^(-) is a halide ion . [ Given E_((F_(2))|2F^(-))^(@)= +2.87 V , E_(Cl_(2)|2Cl^(-))^(@) = +1.36 V, E_(Br_(2)|2Br^(-))^(@)=+1.06 V and E_(I_(2)|2I^(-))^(@)=+0.54 ]

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