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Pb^(+2) + 2e^(-) rarr Pb(s), E^(@) = -0....

`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

A

`Pb gt Sn gt Ni gt Cr`

B

`Cr gt Ni gt Sn gt Pb`

C

`Cr gt Sn gt Ni gt Pb`

D

`Sn gt Ni gt Cr gt Pb`

Text Solution

Verified by Experts

The correct Answer is:
B
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