JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 13)
Consider the following reduction processes :
Zn2+ + 2e– $$ \to $$ Zn(s) ; Eo = – 0.76 V
Ca2+ + 2e– $$ \to $$ Ca(s); Eo = –2.87 V
Mg2+ + 2e– $$ \to $$ Mg(s) ; Eo = – 2.36 V
Ni2 + 2e– $$ \to $$ Ni(s) ; Eo = – 0.25
The reducing power of the metals increases in the order :
Zn2+ + 2e– $$ \to $$ Zn(s) ; Eo = – 0.76 V
Ca2+ + 2e– $$ \to $$ Ca(s); Eo = –2.87 V
Mg2+ + 2e– $$ \to $$ Mg(s) ; Eo = – 2.36 V
Ni2 + 2e– $$ \to $$ Ni(s) ; Eo = – 0.25
The reducing power of the metals increases in the order :
Ca < Mg < Zn < Ni
Ni < Zn < Mg < Ca
Zn < Mg < Ni < Ca
Ca < Zn < Mg < Ni
Explanation
Higher the oxidation potential better will be reducing power.
Comments (0)
