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)
do i only need to use G=-nFE to get W max or there is more to do
because it gives 1893.5g, it threw me off. I wasn't sure if it was there to trick me
To calculate the maximum work done by the electrochemical cell, you can indeed use the equation:
Wmax = -nFE
Where:
- Wmax is the maximum work done by the cell,
- n is the number of electrons transferred in the balanced equation of the cell reaction,
- F is the Faraday constant (96,485 C/mol),
- E is the standard cell potential of the electrochemical cell (in volts).
However, before you can use this equation, you need to determine the number of electrons transferred, n, in the reaction. To do this, refer to the balanced equation of the cell reaction:
2Hg2+(aq) + MnO4-(aq) + 4H2O(l) -> 2Hg(l) + MnO42-(aq) + 8H+(aq)
From the equation, you can see that 1 mole of MnO4-(aq) reacts with 2 moles of electrons. Therefore, n = 2.
Next, you need to determine the value of E, the standard cell potential, which is given as -0.35 V in the problem statement.
Now, substitute the values of n and E into the equation:
Wmax = -nFE
Wmax = -(2)(96,485 C/mol)(-0.35 V)
Simplifying the equation gives:
Wmax = (2)(96,485 C/mol)(0.35 V)
Wmax = 67,539.8 C⋅V
Next, convert the units from Coulombs⋅Volts to kilojoules (kJ). To do this, divide by the conversion factor 1 C⋅V = 0.01 kJ:
Wmax = (67,539.8 C⋅V)/(1 C⋅V/0.01 kJ)
Wmax = 6,753.98 kJ
Finally, round the answer to three significant figures:
Wmax ≈ 6,754 kJ
Therefore, the maximum work done by the electrochemical cell is approximately 6,754 kJ.