Question

For the galvanic (voltaic) cell fe(s) + mn²⁺(aq) ⟶ fe²⁺(aq) + mn(s) (e° = 0.77 v at 25°c), what is [fe²⁺] if [mn²⁺] = 0.050 m and e = 0.78 v? assume t is 298 k

Answer 1

The concentration of Iron in the galvanic (voltaic) cell Fe(s) + Mn²⁺(aq) ⟶ Fe²⁺(aq) + Mn(s) is 0.02297 M.

What is the Nernst Equation?

The Nernst equation enables us to identify the cell potential(voltage) in presence of non-standard conditions in a galvanic cell. It can be expressed by using the formula:

$\mathbf{E_{cell} = E_o - \dfrac{0.059}{n} \times log \dfrac{[Fe^+]}{[Mn^{2+}]}}$

where;

• n = Number of electrons = 2
• $\mathbf{E_o}$ = Initial voltage = 0.77 V
• $\mathbf{E_{cell}}$ = Cell voltage = 0.78 V
• $\mathbf{[Mn^{2+}]}$ = Manganese concentration = 0.050 M

Replacing the values into the above equation, we have:

$\mathbf{0.78 = 0.77 - \dfrac{0.059}{2} \times log \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{0.78 -0.77= -0.0296\times log \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{log^{-1} (-0.3378) = \dfrac{[Fe^{2+}]}{[0.050]}}$

$\mathbf{Fe^{2+} = (0.4594\times 0.050 )\ M}$

$\mathbf{Fe^{2+} =0.02297 \ M }$

Learn more about using the Nernst equation here:

brainly.com/question/24258023