What is the emf of a galvanic cell consisting of a Cd2+/Cd half-cell and a Pt/H+/H2 half-cell if [Cd2+] = 0.28 M, [H+] = 0.21 M, and PH2 = 0.62 atm?
What is V?
Do I use the equation:
E = E - (.0257V / n) lnQ
What do I put in for E initial?
We are looking at these reactions:
Cd2+ + 2e- ---> Cd Eo = - 0.40 V
2H+ + 2e- ---> H2 Eo = 0.00 V
To get a positive voltage in a galvanic cell, reverse the first equation and add to the second:
Cd + 2H+ ---> Cd2+ + H2 ...
Finding the emf of a Cd galvanic cell