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.
What is the balanced redox reaction for the following in basic conditions?
Cd (s) + NO3 ^- (aq) → Cd ^+ (aq) +NO (g)
3 answers
Cancel out the 8e^- on both sides:
4Cd(s) + 4H2O(l) + 3NO3^- (aq) → 4Cd^+ (aq) + 3NO(g) + 12OH^-(aq)
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 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)