Home
Class 12
CHEMISTRY
Standard reduction potential values for ...

Standard reduction potential values for the electrodes are given below.
`Mg^(2+)2e^(-) rarrMg " is" -2.37V`
`Zn^(2+) +2e^(-) rarrZn " is"-0.76V`
`Fe^(2+)+2e^(-)rarrFe " is"-0.44V`
Which of the following statement is correct

A

Zince will reduce `Fe^(2+)`

B

Zinc will reduce `Mg^(2+)`

C

Mg oxidises Fe

D

Zinc oxidises Fe

Text Solution

Verified by Experts

The correct Answer is:
A

Low SRP will reduces high SRP ion.
Promotional Banner

Similar Questions

Explore conceptually related problems

E^(@) values of Mg^(2+)//Mg is -2.37V, of Zn^(2+)//Zn is -0.76V and Fe^(2+)// Fe is - 0.44V Which of the statement is correct ?

For a reaction Zn^(+2) + 2e^(-) rarr Zn, E^(@) = -0.76V which of the following statement is true? (E = Electrode potential)

The standard potentials at 25^@C for the following half reactions are given against them Zn^(2+) +2e^(-) rarr Zn, E^0 =-0 . 762 V Mg^(2+) +2e^(-) rarr Mg, =- 2.37 V . When zinc dust is added to a solution of magnesium chloride .

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:

The voltage of a cell whose half-cells are given below is Mg^(2+)+2e^(-)rarrMg(s),E^(@)=-2.37V Cu^(2+)+2e^(-)rarrCu(s),E^(@)=+0.34V standard EMF of the cell is

E^(c-) of Mg^(2+)|Mg,Zn^(2+)|Zn , and Fe^(2+)|Fe are -2.37V,-0.76V , and -0.44V , respectively. Which of the following is correct ?

The standard electrode potential of the half cells are given below. ZnrarrZn^(2+)+2e^(-):E^(@)=0.76V FerarrFe^(2+)+2e^(-):E^(@)=0.44V The emf of the cell Fe^(2+)+ZnrarrZn^(2+)+Fe is

The standard reduction for the following reactions are : Fe^(3+) + 3e^(-) rarr Fe with E^(@) = - 0.036 V Fe^(2+) + 2e^(-) rarr Fe with E^(@) = - 0.44 V What would be the standard electrode potential for the reaction Fe^(3+) + e^(-) rarr Fe^(2+) ?