Chemistry, 03.07.2019 01:00 babyboo6745

Write the half reactions for the following equation. identify which half reaction is oxidized and which is reduced. ag + no₃- --> ag+ + no write the half reactions for the following equation. identify which half reaction is oxidized and which is reduced. zn + no₃- --> zn²+ + no₂ write the half reactions for the following equation. identify which half reaction is oxidized and which is reduced. cr₂o₇ ²- (aq) + h₂s(g) --> cr³+(aq) + s(s)

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Answer from: lexhoangg

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 is being oxidized. Oxidation is loss of electrons. 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

$Zn \implies Zn^2^+$

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

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

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

Answer from: Quest

Abbreviated as mg-25.

Answer from: Quest

thats how its supposed to be

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