The standard reduction potentials at 298 K for the following half cells are given below:
NO3−+4H++3e−→NO(g)+2H2OEθ=0.97 V
V2+(aq)+2e−→VEθ=−1.19 V
Fe3+(aq)+3e−→FeEθ=−0.04 V
Ag+(aq)+e−→Ag(s)Eθ=0.80 V
Au3+(aq)+3e−→Au(s)Eθ=1.40 V
The number of metal(s) which will be oxidized by NO3− in aqueous solution is __________.
Step-by-Step Verified Solution
To determine the metals that can be oxidized by NO3− , we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since NO3− has an SRP of 0.97 V).
From the given data, we can find the metals that satisfy this condition:
- V2+(aq)+2e−→V , Eθ=−1.19V (less than 0.97 V, hence can be oxidized)
- Fe3+(aq)+3e−→Fe , Eθ=−0.04V (less than 0.97 V, hence can be oxidized)
- Ag+(aq)+e−→Ag(s) , Eθ=0.80V (less than 0.97 V, hence can be oxidized)
- Au3+(aq)+3e−→Au(s) , Eθ=1.40V (greater than 0.97 V, hence cannot be oxidized)
Therefore, V , Fe , and Ag are the metals that can be oxidized by NO3− .
The total number of such metals is 3.