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Consider the following reduction process...

Consider the following reduction processes :
`Zn^(2+)+2e^(-)toZn(s), E^(o)=-0.76 V`
`Ca^(2+)+2e^(-)toCa(s),E^(o)=-2.87 V`
`Mg^(2+)+2e^(-)toMg(s),E^(o)=-2.36 V`
`Ni^(2+)+2e^(-)toNi(s), E^(o)=-0.25 V
The reducing power of the metals increases in the order :

A

`Ca lt Zn lt Mg lt Ni`

B

`Ni lt Zn lt Mg lt Ca`

C

`Zn lt Mg lt Ni lt Ca`

D

`Ca lt Mg lt Zn lt Ni`

Text Solution

Verified by Experts

The correct Answer is:
B
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The standard reduction potential at 298 K for the following half reactions are given below: Fe^(3+)(1M)+e^(-)rarrFe^(2+)(aq),E^(o)=0.770 V, Zn^(2+)(1M)+2e^(-)rarrZn(s),E^(o)=-0.762 V Cd^(2+)(1M)+2e^(-)rarrCd(s),E^(o)=-0.402 V, 2H^(+)(1M)+2e^(-)rarrH_(2)(g),E^(o) = 0.00 V. The strpongest reducing agent is:

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

The standard reduction potentials at 298K for the following half cells are given : ZN^(2+)(aq)+ 2e^(-)Zn(s) , E^(@) = - 0.762V Cr^(3+)(aq)+ 3e^(-) Cr(s) , E^(@) = - 0.740V 2H^(+) (aq)+2e^(-) H_(2)(g) , E^(@)= 0.000V Fe^(3+)(aq)+ e^(-) Fe^(2+)(aq) , E^(@) = 0.770V which is the strongest reducing agent ?

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