Write the half reactions and the balanced equation for the galvanic cell
Hg(ℓ)|Hg2+2(aq)||MnO4(aq),Mn2+(aq),H+(aq)|Pt(s). What is the smallest possible integer coefficient of H2O(ℓ) in the combined balanced equation?
It would help if you included the potentials from the problem (or from your text/notes) so we use the same set of numbers.
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To determine the half reactions and write the balanced equation, we need to identify the oxidation and reduction half reactions. Let's break down the given galvanic cell and its components.
The anode half-reaction occurs at the left side of the cell:
Hg(ℓ) → Hg2+2(aq) + 2e-
The cathode half-reaction occurs at the right side of the cell:
MnO4(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(ℓ)
To balance the number of electrons transferred in both half-reactions, we will multiply the first reaction by 5 and the second reaction by 2:
5Hg(ℓ) → 5Hg2+2(aq) + 10e-
2MnO4(aq) + 16H+(aq) + 10e- → 2Mn2+(aq) + 8H2O(ℓ)
Now, we add the two half-reactions together:
5Hg(ℓ) + 2MnO4(aq) + 16H+(aq) → 5Hg2+2(aq) + 2Mn2+(aq) + 8H2O(ℓ)
Therefore, the balanced equation for the galvanic cell is:
5Hg(ℓ) + 2MnO4(aq) + 16H+(aq) → 5Hg2+2(aq) + 2Mn2+(aq) + 8H2O(ℓ)
The smallest possible integer coefficient for H2O(ℓ) in the combined balanced equation is 8.
To determine the half reactions and the balanced equation for the given galvanic cell, we need to follow a few steps:
Step 1: Identify the reduction and oxidation half reactions. Reduction involves gaining electrons, while oxidation involves losing electrons.
In the given cell, we have:
Hg(ℓ)|Hg2+2(aq) || MnO4(aq), Mn2+(aq), H+(aq)| Pt(s)
The half reaction occurring at the left electrode (anode) is the oxidation half reaction, while the one occurring at the right electrode (cathode) is the reduction half reaction.
Step 2: Representative the oxidation and reduction half reactions.
Anode (Oxidation): Hg(ℓ) -> Hg2+2(aq) + 2e-
Cathode (Reduction): MnO4(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(ℓ)
Step 3: Balance the number of atoms and charges in each half reaction.
Anode (Oxidation): Hg(ℓ) -> Hg2+2(aq) + 2e-
Since the number of atoms is already balanced, we only need to balance the charges by multiplying Hg2+(aq) by 2.
Hg(ℓ) -> 2Hg2+2(aq) + 2e-
Cathode (Reduction): MnO4(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(ℓ)
Here, the number of atoms and charges are not balanced. We can start by balancing oxygen atoms by adding H2O to the right-hand side.
MnO4(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(ℓ)
Now, we can balance hydrogen atoms by adding H+ to the left-hand side.
MnO4(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(ℓ)
Finally, we balance the charges by multiplying Mn2+(aq) by 5.
MnO4(aq) + 8H+(aq) + 5e- -> 5Mn2+(aq) + 4H2O(ℓ)
Step 4: Combine the two half reactions to form the balanced equation.
To combine the two half reactions, we need to ensure that both half reactions have the same number of electrons. In this case, multiplying the oxidation half reaction by 5 achieves the desired result.
5Hg(ℓ) -> 10Hg2+2(aq) + 10e-
MnO4(aq) + 8H+(aq) + 5e- -> 5Mn2+(aq) + 4H2O(ℓ)
Now, we can add the two half reactions together:
5Hg(ℓ) + MnO4(aq) + 8H+(aq) + 5e- -> 10Hg2+(aq) + 5Mn2+(aq) + 4H2O(ℓ)
Step 5: Determine the smallest possible integer coefficient of H2O(ℓ) in the combined balanced equation.
Looking at the balanced equation, we can observe that the coefficient for H2O(ℓ) is already at its smallest possible integer value, which is 4.
Therefore, the smallest possible integer coefficient of H2O(ℓ) in the combined balanced equation is 4.