Consider the following electrochemical cell:
Pt(s) | Hg2^2+ (0.800M) || MnO4^- (0.650M), H^+ (1.00M), Mn^2+ (0.375M) | Pt(s)
Using the half-reaction method, write a balanced chemical equation for the electrochemical cell
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Where's the other half of the Hg^2+
To understand this electrochemical cell, we need to break it down into its components and understand their roles.
The cell notation used to represent this electrochemical cell is:
Pt(s) | Hg2^2+ (0.800M) || MnO4^- (0.650M), H^+ (1.00M), Mn^2+ (0.375M) | Pt(s)
Let's break it down:
1. Electrodes:
- The left electrode is platinum (Pt) in solid state (s).
- The right electrode is also platinum (Pt) in solid state (s). Both electrodes are inert and do not participate in the reaction.
2. Electrolyte Solutions:
- The left electrolyte solution is Hg2^2+ with a concentration of 0.800M. This represents a solution of mercury(II) ions.
- The double vertical line (||) represents a salt bridge or a porous barrier that allows ion flow between the two compartments without mixing the solutions.
- The right electrolyte solution contains manganese(II) ions (Mn^2+) with a concentration of 0.375M, along with MnO4^- ions and hydrogen ions (H^+), both having concentrations of 0.650M and 1.00M, respectively.
Now, let's understand the overall cell reaction and the individual half-reactions happening at each electrode:
Overall Cell Reaction:
The overall cell reaction occurs between Hg2^2+ and MnO4^- to produce H^+ and Mn^2+ ions.
Half-Reactions:
1. At the left electrode:
Hg2^2+ (aq) + 2e^- → 2Hg(l)
Here, mercury(II) ions gain two electrons and are reduced to form liquid mercury metal.
2. At the right electrode:
8H^+ (aq) + MnO4^- (aq) + 5e^- → Mn^2+ (aq) + 4H2O (l)
In this half-reaction, manganese(VII) ions (MnO4^-) are reduced to manganese(II) ions (Mn^2+) in an acidic solution, while hydrogen ions (H^+) from the acidic solution are reduced to form water (H2O).
To determine the cell potential or electromotive force (EMF) of the electrochemical cell, you would need to know the standard reduction potentials for the half-reactions involved. These values are typically given in a standard reduction potential table. By subtracting the reduction potential of the left electrode from the reduction potential of the right electrode, you can determine the cell potential.
Please note that without the standard reduction potentials, it is not possible to calculate the exact cell potential in this case.