Calculate ΔG for the cell reaction at 298 KZn(s)/Zn (0.0004 M) || Cd (0.2 M)/Cd(s)Given= + 0.76 V= –0.403 V(1) –42960 J (2) –84500 J(3) +84920 J (4) +42960 J+2 +2E oZn/Zn+2E oC /Cd

To calculate ΔG for the cell reaction at 298 K, we can use the equation:

ΔG = -nFΔE

where ΔG is the change in Gibbs free energy, n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant (96485 C/mol), and ΔE is the cell potential.

In this case, the balanced equation for the cell reaction is:

Zn(s) + Cd2+ → Zn2+ + Cd(s)

From the given information, we have:

EoZn/Zn2+ = +0.76 V EoCd/Cd2+ = -0.403 V

To calculate ΔE, we can use the Nernst equation:

ΔE = Eo - (RT/nF) * ln(Q)

where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.

Given that the concentrations of Zn2+ and Cd2+ are 0.0004 M and 0.2 M, respectively, we can calculate Q as follows:

Q = [Zn2+]/[Cd2+] = (0.0004)/(0.2) = 0.002

Substituting the values into the Nernst equation, we get:

ΔE = 0.76 - (8.314 * 298 / (2 * 96485)) * ln(0.002)

Simplifying the equation, we find:

ΔE = 0.76 - (0.0257) * ln(0.002)

Calculating the natural logarithm and multiplying by -0.0257, we get:

ΔE = 0.76 - (-0.0627) = 0.8227 V

Now, we can calculate ΔG using the equation:

ΔG = -nFΔE

Since the balanced equation involves the transfer of 2 moles of electrons, we have:

ΔG = -2 * 96485 * 0.8227 = -159,920 J

Therefore, the correct answer is option (2) -84500 J.