Consider the following voltaic cell. What is the half reaction that takes place at the cathode? What is "E" for the cell? How many electrons are exchanged (n)?

#Fe(s) + 2H^(+)(aq) -> Fe^(2+)(aq) + H_2(g)#

1 Answer
May 17, 2017

Answer:

Cathodic Reduction => #2H^+(aq) + 2e^- => H_2(g)#
Anodic Oxidation => #Fe^o(s) => Fe^2+ + 2 e-#
#E_(cell) = 0.41v#

Explanation:

Cathodic Reduction => #2H^+(aq) + 2e^- => H_2(g)#
Anodic Oxidation => #Fe^o(s) => Fe^2+ + 2 e-#
#E_(cell)^o = E_(redn)^o - E_("oxidn")^o# = #E_((H^+|H_2)# - #E_(Fe(s)|Fe^"2+")#
From Standard Reduction Potential table
#E_H = 0.00v# and #E_(Fe) = -0.41v#
#E_(cell) = [0.00v - (- 0.41v)] = 0.41v#