Question

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)`

Answer 1

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`