Question

The emf of a particular voltaic cell with the cell reaction \[ \text{Hg}_2^{2+} + \text{H}_2 \rightleftharpoons 2 \text{Hg} + 2 \text{H}^+ \] is 0.65 V. The maximum electrical work of this cell when 0.5 g of \( \text{H}_2 \) is consumed is

• A. \( -3.12 \times 10^4 \, \text{J} \)
• B. \( -1.25 \times 10^6 \, \text{J} \)
• C. \( 25.0 \times 10^6 \, \text{J} \)
• D. None

Hint:

For electrochemical reactions, the maximum work is calculated using the number of moles of electrons, Faraday's constant, and the cell potential.

Answer 1

The Correct Option is B

Step 1: Calculate the number of moles of hydrogen.
The molar mass of hydrogen is 1 g/mol, so 0.5 g corresponds to 0.5 moles of \( H_2 \).
Step 2: Calculate the maximum electrical work using the formula.
The work done by the cell is given by: \[ W = n F E \] where \( n \) is the number of moles of electrons, \( F \) is the Faraday constant, and \( E \) is the cell potential. The maximum work comes out to be \( -1.25 \times 10^6 \, \text{J} \).
Final Answer: \[ \boxed{-1.25 \times 10^6 \, \text{J}} \]