# 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)?

## $F e \left(s\right) + 2 {H}^{+} \left(a q\right) \to F {e}^{2 +} \left(a q\right) + {H}_{2} \left(g\right)$

May 17, 2017

Cathodic Reduction => $2 {H}^{+} \left(a q\right) + 2 {e}^{-} \implies {H}_{2} \left(g\right)$
Anodic Oxidation => $F {e}^{o} \left(s\right) \implies F {e}^{2} + + 2 e -$
${E}_{c e l l} = 0.41 v$

#### Explanation:

Cathodic Reduction => $2 {H}^{+} \left(a q\right) + 2 {e}^{-} \implies {H}_{2} \left(g\right)$
Anodic Oxidation => $F {e}^{o} \left(s\right) \implies F {e}^{2} + + 2 e -$
${E}_{c e l l}^{o} = {E}_{red n}^{o} - {E}_{\text{oxidn}}^{o}$ = E_((H^+|H_2) - ${E}_{F e \left(s\right) | F {e}^{\text{2+}}}$
From Standard Reduction Potential table
${E}_{H} = 0.00 v$ and ${E}_{F e} = - 0.41 v$
${E}_{c e l l} = \left[0.00 v - \left(- 0.41 v\right)\right] = 0.41 v$