first reactlon, H2(g) ----->2H+ +2e- , Eo=0 V (1)
Cd+2+2e- -------->Cd, Eo=-0.403V (2)
Adding Eq.1 and Eq.2 gives
H2+Cd+2 -------->2H+ + Cd, Eo= -0.403V
E= Eo-(0.0591/2)*logQ
-0.3657= -0.403 -(0.0591/2)* logQ, Q= [H+]2/ [H2][Cd+2]
0.0373= -0.02955* logQ
logQ= -1.26
Q= 0.055 = [H+]2/ (0.801*1)
[H+]2= 0.055*0.801 , [H+] =0.21 M
Given the measured cell potential, E_cell, is -0.3657 V at 25 degree C in the following...
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Given the measured cell potential, Ecell, is-0.3583 V at 25 °C in the following cell, calculate the Ht concentration Pt (s)|H2lg, 0.795 atm)lH (aq, ? M)l|Cd2 (aq, 1.00 M)|Cd (s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, E°, are as follows. 2H+(aq) + 2e- H2(g) E0.00 V E-0.403 V ? 2 + Number H0.21