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Balance the following redox reaction which takes place in a basic solution using the oxidation number method: [5 Marks]

Cr2O7^2-(aq) + Ag(s) -> Ag^+(aq) + Cr^3+(aq) (basic)

User Rajahsekar
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To balance the redox reaction in a basic solution using the oxidation number method, follow these steps:

Step 1: Assign oxidation numbers to all elements.
Cr2O7^2-: The oxidation number of oxygen is -2, so the total oxidation number for (Cr2O7)^2- is -2 × 7 + 2 = -12. Since the overall charge of the ion is -2, the oxidation number of chromium (Cr) is +6.
Ag: The oxidation number of silver (Ag) is 0 since it is in its elemental form.
Ag+: The oxidation number of silver (Ag) is +1 in the Ag+ ion.
Cr^3+: The oxidation number of chromium (Cr) is +3 in the Cr^3+ ion.

Step 2: Identify the atoms undergoing oxidation and reduction.
In the reaction, Cr is reduced from +6 to +3 (gaining electrons), while Ag is oxidized from 0 to +1 (losing electrons).

Step 3: Write two half-reactions for oxidation and reduction.
Oxidation half-reaction: Ag(s) -> Ag^+(aq) + 1 e^-
Reduction half-reaction: Cr2O7^2-(aq) + 14 H2O(l) -> 2 Cr^3+(aq) + 7 OH^-(aq)

Step 4: Balance the elements except for H and O in each half-reaction.
Oxidation half-reaction: Ag(s) -> Ag^+(aq) + 1 e^-
Reduction half-reaction: Cr2O7^2-(aq) + 14 H2O(l) -> 2 Cr^3+(aq) + 7 OH^-(aq) + 6 e^-

Step 5: Balance the oxygen atoms by adding H2O to the half-reactions.
Oxidation half-reaction: Ag(s) -> Ag^+(aq) + 1 e^-
Reduction half-reaction: Cr2O7^2-(aq) + 14 H2O(l) -> 2 Cr^3+(aq) + 7 OH^-(aq) + 6 e^-

Step 6: Balance the hydrogen atoms by adding H+ to the half-reactions.
Oxidation half-reaction: Ag(s) -> Ag^+(aq) + 1 e^-
Reduction half-reaction: Cr2O7^2-(aq) + 14 H2O(l) -> 2 Cr^3+(aq) + 7 OH^-(aq) + 6 e^- + 14 H+(aq)

Step 7: Balance the charges by adding electrons to one of the half-reactions.
Oxidation half-reaction: Ag(s) -> Ag^+(aq) + 1 e^-
Reduction half-reaction: Cr2O7^2-(aq) + 14 H2O(l) -> 2 Cr^3+(aq) + 7 OH^-(aq) + 6 e^- + 14 H+(aq)

Step 8: Balance the number of electrons in both half-reactions by multiplying them as needed.
Oxidation half-reaction: 6 Ag(s) -> 6 Ag^+(aq) + 6 e^-
Reduction half-reaction: 3 Cr2O7^2-(aq) + 42 H2O(l) -> 6 Cr^3+(aq) + 21 OH^-(aq) + 6 e^- + 42 H+(aq)

Step 9: Combine the half-reactions and cancel out any common terms.
6 Ag(s) + 3 Cr2O7^2-(aq) + 42 H2O(l) -> 6 Ag^+(aq) + 6 Cr^3+(aq) + 21 OH^-(aq) + 42 H+(aq)

Step 10: Simplify the equation and ensure it is balanced.
2 Ag(s) + 3 Cr2O7^2-(aq) + 21 H2O(l) -> 2 Ag^+(aq) + 3 Cr^3+(aq) + 7 OH^-(aq) + 42 H+(aq)

The balanced redox reaction in a basic solution is:
2 Ag(s) + 3 Cr2O7^2-(aq) + 21 H2O(l) -> 2 Ag^+(aq) + 3 Cr^3+(aq) + 7 OH^-(aq) + 42 H+(aq).
User DGuntoju
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