step 5: Multiply the half reactions so that the number of electrons is balanced.
Since the are 2 electrons involved in the oxidation half reation and there are 8 electrons involved in the reduction half reaction, we need to multiply the oxidation half reaction by 4.
4H2S (aq) → 4S (s) + 8H^1+ (aq) + 8e^1-
step 6: Add the reduction and oxidation half reactions together.
4H2S (aq) + Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- →
4S (s) + 8H^1+ (aq) + 8e^1- + 2Cr^2+ (aq) + 7H2O
step 7: Cancel whatever appears on both sides of the reaction arrow.
step 5: Multiply the half reactions so that the number of electrons is balanced.
Since the are 2 electrons involved in the oxidation half reation and there are 8 electrons involved in the reduction half reaction, we need to multiply the oxidation half reaction by 4.
4H2S (aq) → 4S (s) + 8H^1+ (aq) + 8e^1-
step 6: Add the reduction and oxidation half reactions together.
4H2S (aq) + Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- →
4S (s) + 8H^1+ (aq) + 8e^1- + 2Cr^2+ (aq) + 7H2O
step 7: Cancel whatever appears on both sides of the reaction arrow.
A [math] ext{metathesis}[/math] or [math] ext{partner exchange}[/math] reaction that exploits the INSOLUBILITY of lead iodide in aqueous solution…
[math]Pb(NO_{3})_{2}(aq) + 2KI(aq) longrightarrow PbI_{2}(s)downarrow + 2KNO_{3}(aq)[/math]
…lead iodide precipitates from solution as a yellow solid…
A [math]ext{metathesis}[/math] or [math]ext{partner exchange}[/math] reaction that exploits the INSOLUBILITY of lead iodide in aqueous solution…
[math]Pb(NO_{3})_{2}(aq) + 2KI(aq) longrightarrow PbI_{2}(s)downarrow + 2KNO_{3}(aq)[/math]
…lead iodide precipitates from solution as a yellow solid…
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This is a redox reaction, balance it with the half reaction method. Since we have H2S in the reaction, assume acidic media.
H2S (aq) + Cr2O7^2- (aq) → S (s) + Cr^2+ (aq)
step 1: Identify the species that is oxidized.
The S in H2S has an oxidation number of -2. It is converted to neutral S. S is the oxidized species.
step 2: Balance the oxidation half reaction.
H2S (aq) → S (s) + 2H^1+ (aq) + 2e^1-
step 3: Identify the species that is reduced.
The Cr in Cr2O7^2- has an oxidation number of +6. UIt is converted to Cr^2+ with an oxidation number of +2. Cr is the reduced species.
step 4: Balance the reduction half reaction.
Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- → 2Cr^2+ (aq) + 7H2O
step 5: Multiply the half reactions so that the number of electrons is balanced.
Since the are 2 electrons involved in the oxidation half reation and there are 8 electrons involved in the reduction half reaction, we need to multiply the oxidation half reaction by 4.
4H2S (aq) → 4S (s) + 8H^1+ (aq) + 8e^1-
step 6: Add the reduction and oxidation half reactions together.
4H2S (aq) + Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- →
4S (s) + 8H^1+ (aq) + 8e^1- + 2Cr^2+ (aq) + 7H2O
step 7: Cancel whatever appears on both sides of the reaction arrow.
4H2S (aq) + Cr2O7^2- (aq) + 6H^1+ (aq) → 4S (s) + (aq) + 2Cr^2+ (aq) + 7H2O
step 8: Check that mass (atoms) and charge are balanced.
This is a redox reaction, balance it with the half reaction method. Since we have H2S in the reaction, assume acidic media.
H2S (aq) + Cr2O7^2- (aq) → S (s) + Cr^2+ (aq)
step 1: Identify the species that is oxidized.
The S in H2S has an oxidation number of -2. It is converted to neutral S. S is the oxidized species.
step 2: Balance the oxidation half reaction.
H2S (aq) → S (s) + 2H^1+ (aq) + 2e^1-
step 3: Identify the species that is reduced.
The Cr in Cr2O7^2- has an oxidation number of +6. UIt is converted to Cr^2+ with an oxidation number of +2. Cr is the reduced species.
step 4: Balance the reduction half reaction.
Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- → 2Cr^2+ (aq) + 7H2O
step 5: Multiply the half reactions so that the number of electrons is balanced.
Since the are 2 electrons involved in the oxidation half reation and there are 8 electrons involved in the reduction half reaction, we need to multiply the oxidation half reaction by 4.
4H2S (aq) → 4S (s) + 8H^1+ (aq) + 8e^1-
step 6: Add the reduction and oxidation half reactions together.
4H2S (aq) + Cr2O7^2- (aq) + 14H^1+ (aq) + 8e^1- →
4S (s) + 8H^1+ (aq) + 8e^1- + 2Cr^2+ (aq) + 7H2O
step 7: Cancel whatever appears on both sides of the reaction arrow.
4H2S (aq) + Cr2O7^2- (aq) + 6H^1+ (aq) → 4S (s) + (aq) + 2Cr^2+ (aq) + 7H2O
step 8: Check that mass (atoms) and charge are balanced.
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