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E^(c-) for Cr^(3+)+3e^(-) rarr Cr and Cr...

`E^(c-)` for `Cr^(3+)+3e^(-) rarr Cr `and `Cr^(3+)+e^(-) rarr Cr^(2+)` are `-0.74 V` and `-0.40V`, respectively, `E^(c-)` for the reaction is `Cr^(+2)+2e^(-) rarr Cr`

A

`-0.91V`

B

`+0.91V`

C

`-1.14V`

D

`+0.34V`

Text Solution

Verified by Experts

The correct Answer is:
a

`Cr^(3+)+3e^(-) rarr Cr,`
`Cr^(3+)+e^(-) rarr Cr^(2+), `
`ulbar(Cr^(2+)+2e^(-) rarr Cr,)`
`E^(c-)._(3)=(n_(1)E^(c-)._(1)-nE^(c-)._(2))/(n_(3))`
`=([3xx(-0.74)]-[(-1)xx(-0.40)])/(2)=-0.91V`
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Given that the standard reduction potential (E^(@)) of Cr^(3+) | Cr and Cr^(2+) |Cr are -0.74 and -0.91 V respectively. The E^(@) of Cr^(3+) |Cr^(2+) is:

E^(@) of two reactions are given below : Cr^(3+) + 3e rarr Cr, E^(@) = -0.74 V OCl^(-) + H_(2)O + 2e rarr Cl^(-) + 2OH^(-), E^(@) = 0.94 V What will be the E^(@) for ? 3OCl^(-) + 2Cr + 3H_(2)O rarr 2Cr^(3+) + 3Cl^(-) + 6OH^(-)

Pb^(+2) + 2e^(-) rarr Pb(s), E^(@) = -0.13V Sn^(+2) + 2e^(-) rarr Sn(s), E^(@) = - 0.16V Ni^(+2) + 2e^(-) rarr Ni(s), E^(@) = -0.25V Cr^(+3) + 3e^(-) rarr Cr(s), E^(@) = -0.74V Based on the above data, the reducing power of Pb, Sn, Ni and Cr is in the order

Sn^(4+)(aq) + 2e^(-) rightarrow Sn^(2+)(aq) E^(@) = 0.15V Cr^(3+)(aq) + e^(-) rightarrow Cr^(2+)(aq) E^(@) = 0.41V According to the standard reduction potentials above, what is the value of E^(@) for the reaction below? 2Cr^(3+)(aq) + Sn^(2+)(aq) rightarrow 2Cr^(2+)(aq) + Sn^(4+)(aq)

The standard reduction potential data at 25^(@)C is given below E^(@) (Fe^(3+), Fe^(2+)) = +0.77V , E^(@) (Fe^(2+), Fe) = -0.44V , E^(@) (Cu^(2+),Cu) = +0.34V , E^(@)(Cu^(+),Cu) = +0.52 V , E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V , E^(@) (Cr^(3+), Cr) =- 0.74V , E^(@) (Cr^(2+),Cr) = - 0.91V , Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: {:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):} Codes:

Given E_(Cr^(3+)//Cr)^(@)=-0.72V , and E_(Fe^(2+)//Fe)^(@)=-0.42V The potential for the cell. Cr|Cr^(3+)(0.1M)||Fe^(2+)( 0. 01M)Fe is

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