The given reaction is:
Cd(S)+2H+⟶Cd2++H2(g)The Nernst equation is used to calculate the EMF of the cell:
E=Eo−‌‌ln‌QWhere:
Where:
E is the cell potential under non-standard conditions
E∘ is the standard cell potential
R is the ideal gas constant
(8.314J∕mol⋅K)T is the temperature in Kelvin (assumed to be 298 K )
n is the number of moles of electrons transferred in the balanced reaction (2 in this case)
F is Faraday's constant
(96,485C∕mol )
Q is the reaction quotient
The standard cell potential
(E∘) can be calculated as follows:
Eo=Eo(‌ cathode ‌)−Eo(‌ anode ‌) In this case, the cathode is the hydrogen electrode, and the anode is the cadmium electrode. The standard reduction potential of the hydrogen electrode is 0 V , and the standard reduction potential of the cadmium electrode is -0.4 V . Therefore, the standard cell potential is:
Eo=0−(−0.4)=0.4VThe reaction quotient
(Q) is calculated as follows:
Q=‌Plugging in the given values:
Q=‌=20Now, we can plug all the values into the Nernst equation:
E=0.4−‌‌ln(20)Solving for E :
E=0.3616VTherefore, the EMF of the cell is 0.3616 V . So the correct answer is Option C.