The cell potential for Zn|Zn²⁺(aq)||Snˣ⁺|Sn is 0.801 V at 298 K. The reaction quotient for the above reaction is 10⁻². The number of electrons involved in the given electrochemical cell reaction is......
(Given E°_Zn²⁺|Zn = -0.763 V, E°_Snˣ⁺|Sn = +0.008 V and (2.303RT)/F = 0.06 V)
(Given E°_Zn²⁺|Zn = -0.763 V, E°_Snˣ⁺|Sn = +0.008 V and (2.303RT)/F = 0.06 V)
Answer: 4
Solution
Using the Nernst equation:
E_cell = E°_cell - (2.303RT)/(nF) log Q
Given:
Substituting:
0.801 = 0.771 - (0.06/n) log(10⁻²)
0.801 = 0.771 - (0.06/n)(-2)
0.801 = 0.771 + 0.12/n
0.030 = 0.12/n
n = 0.12/0.030 = 4
Therefore, 4 electrons are involved in the cell reaction.
E_cell = E°_cell - (2.303RT)/(nF) log Q
Given:
- E_cell = 0.801 V
- E°_cell = E°_cathode - E°_anode = 0.008 - (-0.763) = 0.771 V
- Q = 10⁻²
- (2.303RT)/F = 0.06 V
Substituting:
0.801 = 0.771 - (0.06/n) log(10⁻²)
0.801 = 0.771 - (0.06/n)(-2)
0.801 = 0.771 + 0.12/n
0.030 = 0.12/n
n = 0.12/0.030 = 4
Therefore, 4 electrons are involved in the cell reaction.
JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer