Consider the cell Cd|CdCl2.2H2O|AgCl(s)| HgThe emf of the cell at 15°C is 0.67531 V and temperature coefficient of emf is 0.00065 V deg–1. Calculate the value of heat flow (in Joules/mol) if the process is carried reversibly: [rounded up to first decimal places]Correct answer is between '36127.0,36133.0'. Can you explain this answer?

Question

Consider the cell Cd|CdCl2.2H2O|AgCl(s)| HgThe emf of the cell at 15&deg;C is 0.67531 V and temperature coefficient of emf is 0.00065 V deg&ndash;1. Calculate the value of heat flow (in Joules/mol) if the process is carried reversibly: [rounded up to first decimal places]Correct answer is between '36127.0,36133.0'. Can you explain this answer?

Answer

°C can be determined using the Nernst equation:

E = E° - (0.0592/n) * log(Q)

Where:
E is the cell potential
E° is the standard cell potential
n is the number of electrons involved in the reaction
Q is the reaction quotient

The given cell can be represented by the following half-reactions:

Cd + 2e- → Cd(s) (oxidation half-reaction)
AgCl(s) + e- → Ag(s) + Cl- (reduction half-reaction)

The standard cell potential (E°) can be determined using standard reduction potentials:

E°cell = E°reduction - E°oxidation

From standard reduction potentials:

E°Ag+ + e- → Ag(s): +0.799 V
E°Cd2+ + 2e- → Cd(s): -0.403 V

E°cell = 0.799 V - (-0.403 V) = 1.202 V

Now, let's calculate the reaction quotient (Q) using the concentrations of the species involved:

Q = [Cd][Cl-]^2 / [Ag+]

Assuming the concentrations are 1 M for Cd, 1 M for Cl-, and 1 M for Ag+, we have:

Q = (1)(1)^2 / (1) = 1

Finally, we can substitute the values into the Nernst equation to calculate the cell potential (E):

E = 1.202 V - (0.0592/2) * log(1)
E = 1.202 V

Therefore, the emf of the cell at 15°C is 1.202 V.

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