Question

For the reaction, shown below, determine which statements are True and which are False.

3Fe(s) + Cr2O72-(aq) +14H+(aq) → 3Fe2+(aq) + 2Cr3+(aq) + 7H2O(l) ξo=1.77V

The oxidizing agent is dichromate (Cr2O72-).

The iron half-reaction takes place in a neutral solution.

The oxidation state of chromium in dichromate is +3.

The reducing agent is Fe2+ (aq)

The lowest oxidation state for hydrogen in this reaction is 0

The oxidation state of oxygen changes from 0 to -2.

Answer 1

Separating the oxidation -half reaction and reduction-half reaction:

Oxidation -half reaction:

3Fe(s) ---->3Fe²⁺(aq) + 6e^-

^{reduction -half reaction:}

Cr2O7^2-(aq) +14H⁺(aq) +6e---->2Cr³⁺(aq)+7H2O(l)

a) True,

Dichromate (Cr2O7^2-) has oxidation number of +6 and it is getting reduced to Cr3+ with oxidation number +3, so, it is an oxidizing agent .It oxidizes Fe(s) but itself gets reduced

b) True

3Fe(s) ---->3Fe²⁺(aq) + 6e^{- (neutral medium)}

^c)False

Let the oxidation state of Cr in Cr2O7^2- be x, 2x-7(-2)-2,or, x+6   [Note O has -2 oxidation state]

d)False

Reducing agent is the one that reduces other species and itself gets oxidized in the process.

Fe(s) is getting oxidized to Fe2+, it is the reducing agent.

e)False

In H2O , oxidation state of hydrogen is +1

In H+ also it is +1

f)false

In Cr2O7^2-, the oxidation state of oxygen is -2, and so it is in H2O