The electrochemical cell described by the cell notation has a standard cell potential of -0.35 V. Calculate the Wmax (kJ) the cell has done if 1893.5 g of MnO42-(aq) (Molar Mass - 118.94 g/mol) forms. Round your answer to 3 significant figures.

Pt(s) | Hg22+(aq), Hg2+(aq)||MnO4-(aq), MnO42-(aq) | Pt(s)

since there are 2 transferred eletrons
then G=-nFE=(2)(.35)(96485)

Other than omitting a negative sign on the -nEF I don't see anything wrong but you haven't answered the question.

To calculate the maximum work done by the electrochemical cell, we need to use the formula:

Wmax = -nFEcell

First, we need to determine the value of n (the number of electrons transferred). Looking at the given cell notation:

Pt(s) | Hg22+(aq), Hg2+(aq) || MnO4-(aq), MnO42-(aq) | Pt(s)

We can see that two electrons are transferred, because MnO4- is reduced to MnO42-. Therefore, n = 2.

Next, we need to determine the value of Ecell (the standard cell potential) which is given as -0.35 V.

Now, we can calculate Wmax:

Wmax = (-n) x F x Ecell

Where F is the Faraday constant, which is approximately 96485 C/mol.

Plugging in the values:

Wmax = (-2) x 96485 C/mol x (-0.35 V)

Simplifying:

Wmax = 67339 C/mol

Now, we need to convert from C to kJ.

To do this, we use the conversion factor that 1 C is equal to 1 J/C, and 1 kJ is equal to 1000 J:

Wmax = (67339 C/mol) x (1 J/1 C) x (1 kJ/1000 J)

Simplifying:

Wmax = 67.339 kJ

Rounding to three significant figures, the maximum work done by the cell is 67.339 kJ.