Home
Class 12
CHEMISTRY
निम्नलिखित अपचयन प्रक्रमों पर विचार कीजि...

निम्नलिखित अपचयन प्रक्रमों पर विचार कीजिए
`Zn^(2+)+2e^(-) to Zn(s) , E^@`= - 0.76 V
`Ca^(2+) + 2e^(-) to Ca(s) , E^@` = - 2.87 V
`Mg^(2+) + 2e^(-) to Mg(s) , E^@`= - 2.36 V
`Ni^(2+) + 2e^(-) to Ni(s) , E^@`= - 0.25 V
धातुओं के अपचयन सामर्थ्य का बढ़ता हुआ क्रम है:

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
अपचायक सामर्थ्य `prop` ऑक्सीकरण से गुजरने की प्रवृत्ति `prop 1/"अपचयन विभव मान"`
इसलिए, अपचयन सामर्थ्य का बढ़ता हुआ क्रम है:
Ca `gt` Mg `gt` Zn `gt` Ni
Promotional Banner

Similar Questions

Explore conceptually related problems

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 :

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

For Zn^(2+) //Zn, E^@ =- 0. 76 V , for Ag^+ //Af E^@ =0.799 V . The correct statement is .

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 ?