The cell reaction is
Zn + 2H^+ ==> Zn^2+ + H2
Calculate Ecell = 0.76v + 0 = 0.76v.
Then Ecell = Eocell + (0.05916/2)*log(Q) whee Q = (pH2)*(Zn^2+)/(H^+)^2(Zn)(s)
I assume this is done at STP so pH2 = 1.00 atm and (Zn)(s) = 1.00 by definition. Plug in H^+ and Zn^2+ and solve for Ecell.
Consider the voltaic cell:
Zn(s) / Zn2+(aq) // H+(aq) / H2(g) / Pt(s)
Calculate the value of Ecell, in V, when [Zn2+] =1.00 M and [H+] = 0.00100 M
DATA:
Zn2+(aq) +2 e¯ ---> Zn(s);Eo = - 0.76 V
2 H+(aq) + 2 e¯ ---> H2(g);Eo = 0.00 V
1 answer