What is the balanced redox reaction for the following in basic conditions?
Cd (s) + NO3 ^- (aq) → Cd ^+ (aq) +NO (g)
To balance the redox reaction in basic conditions, follow these steps:
1. Write down the unbalanced equation:
Cd (s) + NO3^- (aq) → Cd^+ (aq) + NO (g)
2. Split the reaction into two half-reactions: oxidation and reduction.
Oxidation half-reaction:
Cd(s) → Cd^+ (aq)
Reduction half-reaction:
NO3^- (aq) → NO(g)
3. Balance the atoms in the half-reactions:
Oxidation half-reaction:
Cd(s) → Cd^+ (aq) + 2e^-
Reduction half-reaction:
4H2O(l) + 3NO3^- (aq) + 8e^- → 3NO(g) + 12OH^-(aq)
4. Balance the charge in the half-reactions by adding electrons:
Oxidation half-reaction:
Cd(s) → Cd^+ (aq) + 2e^-
Reduction half-reaction:
4H2O(l) + 3NO3^- (aq) + 8e^- → 3NO(g) + 12OH^-(aq) + 8e^-
5. Multiply the half-reactions by appropriate numbers to make the number of electrons equal:
Multiplying the oxidation half-reaction by 4 to balance the electrons:
4Cd(s) → 4Cd^+ (aq) + 8e^-
Multiplying the reduction half-reaction by 1 to balance the electrons:
4H2O(l) + 3NO3^- (aq) + 8e^- → 3NO(g) + 12OH^-(aq) + 8e^-
6. Combine the half-reactions:
4Cd(s) + 4H2O(l) + 3NO3^- (aq) → 4Cd^+ (aq) + 3NO(g) + 12OH^-(aq)
7. Simplify by removing spectator ions or molecules:
Cancel out the 8e^- on both sides:
4Cd(s) + 4H2O(l) + 3NO3^- (aq) → 4Cd^+ (aq) + 3NO(g) + 12OH^-(aq)
8. Check the overall charge and mass on both sides to ensure the equation is balanced.
Cancel out the 8e^- on both sides:
4Cd(s) + 4H2O(l) + 3NO3^- (aq) → 4Cd^+ (aq) + 3NO(g) + 12OH^-(aq)
Apologies for the mistake. Let's correct the balanced equation:
To cancel out the 8 electrons, we can multiply the oxidation half-reaction by 8 and the reduction half-reaction by 3:
Oxidation half-reaction:
8Cd(s) → 8Cd^+ (aq) + 16e^-
Reduction half-reaction:
12H2O(l) + 9NO3^- (aq) + 24e^- → 9NO(g) + 36OH^-(aq)
Now, let's combine the half-reactions:
8Cd(s) + 12H2O(l) + 9NO3^- (aq) → 8Cd^+ (aq) + 9NO(g) + 36OH^-(aq)
The balanced redox reaction in basic conditions is:
8Cd(s) + 12H2O(l) + 9NO3^- (aq) → 8Cd^+ (aq) + 9NO(g) + 36OH^-(aq)
To balance a redox reaction in basic conditions, follow these steps:
Step 1: Write the unbalanced equation.
Cd (s) + NO3^- (aq) → Cd^+ (aq) + NO (g)
Step 2: Separate the reaction into two half-reactions: the oxidation half-reaction (loss of electrons) and the reduction half-reaction (gain of electrons).
Oxidation half-reaction: Cd (s) → Cd^+ (aq)
Reduction half-reaction: NO3^- (aq) → NO (g)
Step 3: Balance the atoms in each half-reaction, excluding oxygen and hydrogen.
Oxidation half-reaction: Cd (s) → Cd^2+ (aq)
Reduction half-reaction: 2(NO3^-) (aq) → 2(NO) (g)
Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the side that is deficient in oxygen.
Oxidation half-reaction: Cd (s) → Cd^2+ (aq)
Reduction half-reaction: 2(NO3^-) (aq) + 4(H2O) (l) → 2(NO) (g) + 6(OH^-) (aq)
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that is deficient in hydrogen.
Oxidation half-reaction: Cd (s) → Cd^2+ (aq)
Reduction half-reaction: 2(NO3^-) (aq) + 4(H2O) (l) → 2(NO) (g) + 6(OH^-) (aq) + 4(H+) (aq)
Step 6: Balance the charge by adding electrons (e-) to the appropriate side of each half-reaction.
Oxidation half-reaction: Cd (s) → Cd^2+ (aq) + 2(e-)
Reduction half-reaction: 2(NO3^-) (aq) + 4(H2O) (l) + 4(e-) → 2(NO) (g) + 6(OH^-) (aq) + 4(H+) (aq)
Step 7: Multiply the half-reactions by appropriate coefficients to ensure that the number of electrons lost equals the number of electrons gained. In this case, since the Cd loses 2 electrons and the NO3^- gains 4 electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
2(Cd) (s) → 2(Cd^2+) (aq) + 4(e-)
2(NO3^-) (aq) + 4(H2O) (l) + 4(e-) → 2(NO) (g) + 6(OH^-) (aq) + 4(H+) (aq)
Step 8: Combine the two half-reactions, canceling out electrons.
2(Cd) (s) + 2(NO3^-) (aq) + 4(H2O) (l) → 2(Cd^2+) (aq) + 2(NO) (g) + 6(OH^-) (aq) + 4(H+) (aq)
Step 9: Simplify the equation by canceling out any common species on both sides of the reaction.
Cd (s) + NO3^- (aq) + 2(H2O) (l) → Cd^2+ (aq) + NO (g) + 3(OH^-) (aq) + 2(H+) (aq)
Thus, the balanced redox reaction in basic conditions is:
Cd (s) + NO3^- (aq) + 2(H2O) (l) → Cd^2+ (aq) + NO (g) + 3(OH^-) (aq) + 2(H+) (aq)