Question

Write the half reactions for the following equation. Please identify which half reaction is oxidized and which is reduced.

Ag + NO₃- --> Ag+ + NO

Write the half reactions for the following equation. Please identify which half reaction is oxidized and which is reduced.



Zn + NO₃- --> Zn²+ + NO₂

Write the half reactions for the following equation. Please identify which half reaction is oxidized and which is reduced.



Cr₂O₇ ²- (aq) + H₂S(g) --> Cr³+(aq) + S(s)

Answer 1

Question 1

We start with the half reaction for silver (Ag) is

`Ag \implies Ag^+`

There is an imbalance of charge with left hand side having a charge of 0 and right hand side having a charge of +1. So we add an electron on the right hand side to balance the charges.

`Ag \implies Ag^+ + e`

This would be the oxidation half reaction since
`Ag` is being oxidized. Oxidation is loss of electrons.
`Ag` loses electrons to form the
`Ag^+`

The next half reaction is
`NO_3^-`

`NO_3^- \implies NO`

There are 3 oxygen atoms on left hand side and 1 on right hand side. So we add 2 water molecules on right hand side to balance oxygen.

`NO_3^- \implies NO + 2H_2O`

Then we add
`H^+` ions on left hand side to balance hyrdogen.

`NO_3^- + 4H^+ \implies NO + 2H_2O\\`

We have imbalance of charge with left hand side having an overall +3 (-1 + 4) while right hand side has 0. So we add 3 electrons on left hand side to get the charge to equal on each side.

`NO_3^- + 4H^+ + 3e \implies NO + 2H_2O`

The
`NO_3^-` ion is getting reduced because it is gaining electrons. Reduction is gain of electrons.

Question 2

We start with the half reaction of
`Zn`

`Zn \implies Zn^2^+`

There is an imbalance of charge so we add 2 electrons on right hand side to balance the charge.

`Zn \implies Zn^2^+ + 2e`

Zn is being oxidized in the reaction since it is losing electrons to obtain a higher oxidation state.

The half reaction for
`NO_3^-` is similar as the one obtained above

`NO_3^- + 4H^+ + 3e \implies NO + 2H_2O`

`N` is being reduced since it is moving from a +5 oxidation state to a +2 oxidation state by gaining electrons.

Question 3

We start with the dichromate ion
`Cr_2O_7^2^-`

`Cr_2O_7^2^- \implies Cr^3^+`

First we balance the chromium ions by adding a coefficient of 2 on right hand side.

`Cr_2O_7^2^- \implies 2Cr^3^+`

Then we add 7 water molecules on right hand side to balance the oxygen

`Cr_2O_7^2^- \implies 2Cr^3^+ + 7H_2O`

Now we add 14H^+ ions on left hand side to balance hydrogen

`Cr_2O_7^2^- + 14H^+ \implies 2Cr^3^+ + 7H_2O`

The we add 6 electrons on left hand side to give a +6 charge on each side

`Cr_2O_7^2^- + 14H^+ + 6e \implies 2Cr^3^+ + 7H_2O`

The chromium ion is being reduced since it moves from oxidation state of +6 to a +3 oxidation state.

The second half reaction:

`H_2S \implies S`

We balance the hydrogen by adding H^+ ions on right hand side

`H_2S \implies S + 2H^+`

Now we balance the charge by adding 2 electrons on right hand side

`H_2S \implies S + 2H^+ + 2e`

Hydrogen is being oxidized since it moves from oxidation state of 0 to an oxidation state of +2