Respuesta :
Question 1
We start with the half reaction for silver (Ag) is
[tex]Ag \implies Ag^+[/tex]
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.
[tex]Ag \implies Ag^+ + e[/tex]
This would be the oxidation half reaction since [tex]Ag[/tex] is being oxidized. Oxidation is loss of electrons. [tex]Ag[/tex] loses electrons to form the [tex]Ag^+[/tex]
The next half reaction is [tex]NO_3^-[/tex]
[tex]NO_3^- \implies NO[/tex]
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.
[tex]NO_3^- \implies NO + 2H_2O[/tex]
Then we add [tex]H^+[/tex] ions on left hand side to balance hyrdogen.
[tex]NO_3^- + 4H^+ \implies NO + 2H_2O\\[/tex]
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.
[tex]NO_3^- + 4H^+ + 3e \implies NO + 2H_2O[/tex]
The [tex]NO_3^-[/tex] ion is getting reduced because it is gaining electrons. Reduction is gain of electrons.
Question 2
We start with the half reaction of [tex]Zn[/tex]
[tex]Zn \implies Zn^2^+[/tex]
There is an imbalance of charge so we add 2 electrons on right hand side to balance the charge.
[tex]Zn \implies Zn^2^+ + 2e[/tex]
Zn is being oxidized in the reaction since it is losing electrons to obtain a higher oxidation state.
The half reaction for [tex]NO_3^-[/tex] is similar as the one obtained above
[tex]NO_3^- + 4H^+ + 3e \implies NO + 2H_2O[/tex]
[tex]N[/tex] 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 [tex]Cr_2O_7^2^-[/tex]
[tex]Cr_2O_7^2^- \implies Cr^3^+[/tex]
First we balance the chromium ions by adding a coefficient of 2 on right hand side.
[tex]Cr_2O_7^2^- \implies 2Cr^3^+[/tex]
Then we add 7 water molecules on right hand side to balance the oxygen
[tex]Cr_2O_7^2^- \implies 2Cr^3^+ + 7H_2O[/tex]
Now we add 14H^+ ions on left hand side to balance hydrogen
[tex]Cr_2O_7^2^- + 14H^+ \implies 2Cr^3^+ + 7H_2O[/tex]
The we add 6 electrons on left hand side to give a +6 charge on each side
[tex]Cr_2O_7^2^- + 14H^+ + 6e \implies 2Cr^3^+ + 7H_2O[/tex]
The chromium ion is being reduced since it moves from oxidation state of +6 to a +3 oxidation state.
The second half reaction:
[tex]H_2S \implies S[/tex]
We balance the hydrogen by adding H^+ ions on right hand side
[tex]H_2S \implies S + 2H^+[/tex]
Now we balance the charge by adding 2 electrons on right hand side
[tex]H_2S \implies S + 2H^+ + 2e[/tex]
Hydrogen is being oxidized since it moves from oxidation state of 0 to an oxidation state of +2
