Question

Aluminum displaces hydrogen from dilute HCl whereas silver does not. The emf of a cell prepared by combining Al/$A{l^{3 + }}$ and Ag/$A{g^ + }$ is 2.46V. The reduction potential of silver electrodes is +0.80 V. The reduction potential of aluminum electrode is:
A. +1.66 V
B. -3.26 V
C. 3.26 V
D. -1.66 V

Answer 1

To calculate the reduction potential of aluminum electrode use the formula for calculating the cell potential where the reduction potential of silver is subtracted by the reduction potential of aluminum.

Complete step by step answer: Given,
The emf of a cell prepared by combining Al/$A{l^{3 + }}$ and Ag/$A{g^ + }$ is 2.46 V.
The reduction potential of silver electrodes is +0.80 V.
An electrochemical cell is a device used to produce electricity by the chemical reaction. The electrochemical cell converts chemical energy into electrical energy.
When a metal electrode is dipped in the electrolyte of its own ions , some potential difference is seen across the interface. This potential difference generated is known as electrode potential.
The electrochemical cell is represented with the help of special notation.
The tendency of an electrode to gain electrons is called its reduction potential.
The electrochemical cell is shown below.
$Al|A{l^{3 + }}||A{g^ + }|Ag$
The cell potential is calculated as shown below.
$E_{cell}^0 = E_{A{g^ + }}^0{/_{Ag}} - E_{A{l^{3 + }}}^0{/_{Al}}$
To calculate the reduction potential of silver electrode, substitute the values in the above equation.
$2.46 = 0.80 - E_{A{l^{3 + }}}^0{/_{Al}}$
$E_{A{l^{3 + }}}^0{/_{Al}} = - 1.66$
Therefore, the correct option is D.

Note:
The given cell potential is of the daniell cell. For calculating the cell potential of a galvanic cell the potential of the half-cell on the right side (cathode) is subtracted by the Potential of the half-cell on the left (anode).