Respuesta :
The equation is as follow,
Cr₂O₇²⁻ → Cr³⁺
Balance the Cr atoms,
Cr₂O₇²⁻ → 2 Cr³⁺
Identify Oxidation Number of Elements on Both sides,
Left Hand Side;
Cr = +6
O = -2
Right Hand Side;
Cr = +3
O = -2
Means each Cr is gaining 3 electrons to change into +3 state, Hence, 6 electrons are gained by two Cr atoms, So,
Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺
Balance Oxygen atoms, as it is acidic medium, add H⁺ ion on the side containing large number of Oxygen atoms and H₂O on side containing less number of Oxygen atoms, So,
H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺ + H₂O
Now Balance H and O,
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺ + 7 H₂O
6 electrons are transferred in this reaction.
Cr₂O₇²⁻ → Cr³⁺
Balance the Cr atoms,
Cr₂O₇²⁻ → 2 Cr³⁺
Identify Oxidation Number of Elements on Both sides,
Left Hand Side;
Cr = +6
O = -2
Right Hand Side;
Cr = +3
O = -2
Means each Cr is gaining 3 electrons to change into +3 state, Hence, 6 electrons are gained by two Cr atoms, So,
Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺
Balance Oxygen atoms, as it is acidic medium, add H⁺ ion on the side containing large number of Oxygen atoms and H₂O on side containing less number of Oxygen atoms, So,
H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺ + H₂O
Now Balance H and O,
14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2 Cr³⁺ + 7 H₂O
6 electrons are transferred in this reaction.
[tex]\rm \bold{ 14H^+ + Cr_2O_7^2^- \rightarrow 2Cr^3^+ + 7H_2O}[/tex] is the full balanced reaction. In the balanced half-reaction 3 electron will transfer.
The reaction in acidic condition,
[tex]\rm \bold{ H^+ + Cr_2O_7^2^- \rightarrow Cr^3^+ + H_2O}[/tex]
Balance the equation,
[tex]\rm \bold{ 14H^+ + Cr_2O_7^2^- \rightarrow 2Cr^3^+ + 7H_2O}[/tex]
Means each Cr is gaining 3 electrons to change into +3 state, Hence, 6 electrons are gained by two Cr atoms,
[tex]\rm \bold{ 14H^+ + Cr_2O_7^2^- + 6e^- \rightarrow 2Cr^3^+ + 7H_2O}[/tex]
If 6 electrons are transferring at balanced equation,
Therefore 3 electron will transfer in the balanced half-reaction.
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