'Aluminium displaces hydrogen from dilute HCl whereas silver does not. The E.M.F. of a cell prepared by combining Al /Al 3 andAg/Ag is 2.46 V. The reduction potential of silver electrode is 0.80 V. The reduction potential of aluminium electrode is :
To find the reduction potential of the aluminum electrode, we need to use the given information about the electrochemical cell.
First, let's understand the given information:
1. Aluminum displaces hydrogen from dilute HCl: This means that aluminum has a higher tendency to accept electrons and get reduced compared to hydrogen, so it displaces hydrogen from the acid.
2. Silver does not displace hydrogen from dilute HCl: This means that silver has a lower tendency to accept electrons and get reduced compared to hydrogen, so it does not displace hydrogen from the acid.
3. The electromotive force (EMF) of the cell is 2.46 V: EMF is a measure of the potential difference between the two electrodes in the cell when no current is flowing.
Now, let's use this information to find the reduction potential of the aluminum electrode:
1. We know that the standard reduction potential of the silver electrode is 0.80 V. This means that when Ag+ ions from a solution are reduced at the silver electrode, the potential difference between the silver electrode and the standard hydrogen electrode (SHE) is 0.80 V.
2. The EMF of the cell is 2.46 V. This means that the potential difference between the aluminum electrode and the SHE is 2.46 V when no current is flowing.
3. Since the silver electrode does not displace hydrogen from dilute HCl, the potential difference between the silver electrode and the aluminum electrode is 0.80 V, which is equal to the reduction potential of the silver electrode.
4. Therefore, the potential difference between the aluminum electrode and the SHE is 2.46 V - 0.80 V = 1.66 V.
5. Therefore, the reduction potential of the aluminum electrode is 1.66 V.
So, the reduction potential of the aluminum electrode is 1.66 V.
Al ==> Al^+ + 3e Eo ox = ?
Ag^+ + e ==> Ag Eo red = 0.8
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Al + 3Ag^+ ==> 3 Ag + Al^3+ E=2.46
? + 0.8 = 2.46
? in this set up is the oxidation potential. Change the sign to make it the reduction potential.