The reaction Zn^(2+) + 2e^(–)rarrZn(s) h...

The reaction `Zn^(2+) + 2e^(–)rarrZn(s)` has a standard potential of `– 0.76 V`. This means:

A

Zn can't replace hydrogen from acids

B

Zn is a reducing agent

C

Zn is an oxidising agent

D

`Zn^(2+)` ions are a reducing agent

Verified by Experts

The correct Answer is:

B

Similar Questions

Explore conceptually related problems

The reaction: Zn^(2+)(aq)+2e^(-)toZn (s) has a electrode potential of -0.76 V. This means-

The chemical reaction: Zn^(2+)+2e^(-)rarrZn Is an example of

Standard reduction electrode potential of Zn^(2+)//Zn is -0.76V . This means:

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

Calculate the half cell potential at 298K for the reaction, Zn^(+2) + 2e^(-) rarr Zn " if " [Zn^(+2)] =2M, E_(Zn^(+2)//Zn)^(@) = -0.76V

Use the standard reduction potentials to find the standard cell potential E^(@) , for the reaction: Zn(s)+2Tl^(+)(aq)rarrZn^(2+)(aq)+2Tl(s)