Pt, (H2) 1atm | CH3COOH 0.1 M, CH3COONa 0.15M || NaBr 0.1M, AgBr(S) | Ag
write the cathodic, anodic and final reactions.
Well, let me break it down for you, equation by equation, with a touch of humor!
First, the cathodic reaction (where reduction occurs):
Ag+ + e- → Ag (silver gets its groove back!)
Next, the anodic reaction (where oxidation happens):
2 Br- → Br2 + 2 e- (bromine gets a little carried away and forms a duo)
Now, let's put them together for the final reaction:
2 Ag+ + 2 Br- → 2 Ag + Br2 (silver and bromine team up for an electrifying performance!)
Remember, chemistry is all about creating reactions, both chemical and emotional!
To determine the cathodic, anodic, and final reactions in this electrochemical cell, we need to consider the redox reactions happening at the electrodes.
Given:
Pt, (H2) 1atm | CH3COOH 0.1 M, CH3COONa 0.15 M || NaBr 0.1 M, AgBr(S) | Ag
Let's break down the cell components and reactions step by step:
1. On the left side of the cell (anode): Pt, (H2) 1atm | CH3COOH 0.1 M, CH3COONa 0.15 M
This implies that the anode is made of platinum (Pt) and a hydrogen (H2) gas electrode, surrounded by a solution containing acetic acid (CH3COOH) and sodium acetate (CH3COONa).
2. On the right side of the cell (cathode): NaBr 0.1 M, AgBr(S) | Ag
This indicates that the cathode consists of a silver (Ag) electrode in contact with a solution containing sodium bromide (NaBr) and solid silver bromide (AgBr).
Now, let's determine the half-reactions at each electrode:
Anode (oxidation half-reaction):
In the anode compartment, acetic acid can act as a reducing agent, which means it gets oxidized. Acetic acid (CH3COOH) can lose electrons to form products. The half-reaction can be written as follows:
CH3COOH --> 2CO2 + 4H+ + 4e-
Cathode (reduction half-reaction):
In the cathode compartment, the sodium bromide (NaBr) solution can act as an oxidizing agent, which means it gets reduced. The reduction half-reaction involving bromide ions can be written as follows:
2Br- --> Br2 + 2e-
Note: The solid silver bromide (AgBr) on the right side of the cathode is not involved in the half-reaction as it does not participate directly in the redox process. It serves as a source for bromide ions.
Finally, the overall balanced reaction can be obtained by multiplying the half-reactions appropriately to balance the electrons transferred:
2(CH3COOH) + 2NaBr --> 2CO2 + 2H2O + Br2 + 2CH3COONa
In this reaction, acetic acid is oxidized to carbon dioxide and water, while bromide ions are reduced to form bromine.
To summarize:
Anode (oxidation): 2CH3COOH --> 2CO2 + 4H+ + 4e-
Cathode (reduction): 2Br- --> Br2 + 2e-
Overall reaction: 2(CH3COOH) + 2NaBr --> 2CO2 + 2H2O + Br2 + 2CH3COONa