a) [tex]2Na(s) +2H_2O(I)[/tex]→ [tex]2Na^+(aq) + OH^-(aq)+ H_2(g)[/tex]
b) [tex]2Ag (s) +2H^(aq)[/tex] → [tex]2 Ag^+ (aq) +H_2(g)[/tex]
The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.
a)
[tex]2Na(s) +2H_2O(I)[/tex]→ [tex]2Na^+(aq) + OH^-(aq)+ H_2(g)[/tex]
[tex]E^0 cell = E^0 (reduction) - E^0 (oxidation)[/tex]
= [tex]E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)[/tex]
= -0.83 - (-2.71) =1.88V
b)
[tex]2Ag (s) +2H^(aq)[/tex] → [tex]2 Ag^+ (aq) +H_2(g)[/tex]
[tex]E^0cell[/tex]= [tex]E^0(H^+/H_2) -E^0(Ag^+/Ag)[/tex]
[tex]E^0cell[/tex]=-0. - (0.8) =-0.8V
Learn more about the half-reactions here:
https://brainly.in/question/18053421
#SPJ1
