Question

The standard cell potential for the electrochemical cell
$Ag/A{g^ + }\parallel AgI/Ag;$
${E^0}_{{I^ - }/AgI(s)/Ag} = - 0.151V$
${E^0}_{A{g^ - }/Ag} = 0.799V$
(A)- +0.950 V
(B)- -0.950 V
(C)- -0.648 V
(D)- +0.648 V

Answer 1

An electrochemical cell is said to be a device that can generate electrical energy from the chemical reactions or can also use the electrical energy that is supplied to it to facilitate chemical reactions. This device can convert the chemical energy into electrical energy or vice versa.

Complete answer:
Let us introduce the importance of the two half-cells.
For the cell to function, any electrochemical cell should always consist of two half-cells. The half-cell reactions help us to determine the reactions that will take place and the standard cell potential for the combination of two half-cells without constructing the cell practically.
The half-cell with higher reduction will undergo reduction in the constructed cell. The half-cell with lower reduction potential will undergo oxidation in the same constructed cell.
Keeping in mind the following points, the overall cell potential will come out to be a positive value. Besides, the cell potential calculated should be positive for the redox reaction of the cell to be spontaneous in nature.
Therefore, the combined equation of the whole cell in terms of the reduction and oxidation potential can be written as follows:
${E^0}_{cell} = {E^0}_{reduction} - {E^0}_{oxidation}$
$Ag/A{g^ + }\left\| {} \right.AgI/Ag$
Here we see that the silver is converted to silver ion and thus oxidation is taking place, and this represents the half-cell reaction at Anode.
${E^0}_{reduction} = {E^0}_{{I^ - }/AgI(s)/Ag} = - 0.151V$
Similarly, the silver iodide is converted to silver and thus reduction is taking place, and this represents the half-cell reaction at Cathode.
${E^0}_{A{g^ - }/Ag} = 0.799V$
${E^0}_{oxidation} = {E^0}_{A{g^ - }/Ag} = 0.799V$
By using the following equation, we would conclude
${E^0}_{cell} = {E^0}_{reduction} - {E^0}_{oxidation} = - 0.151V - 0.799V = - 0.95V$

Therefore, the correct answer to the above question is option (B)- -0.950 V .

Note:
The choice of the half cell reaction must be taken very carefully with the sign of the reduction or the oxidation potential. Just by changing or choosing the wrong sign of the equation it would result in the wrong answer.