Question

A voltaic cell utilizes the following reaction: 4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l).

Answer 1

Answer:

The Emf of the given cell at [Fe2+] = 2.0 M and [Fe3+] = 1.9 M is 0.48 V

Explanation:

The half cell reaction can be written as :

Anode-Half (oxidation) :

$Fe^{2+}\rightarrow Fe^{3+} + 1e^{-}$ ......E = 0.77 V

(multiply this equation by 4 to balance the electrons)

Cathode-half (reduction)

$4H^{+} +O_{2} + 4e^{-} \rightarrow 2H_{2}O$....E= 1.23 V

$E^{0}_{cell} = E_{cathode} - E_{anode}$

$E^{0}_{cell} = 1.23 - 0.77$

$E^{0}_{cell} = 0.46 V$

According to Nernst Equation

$E_{cell} = E^{0} - \frac{RT}{nF}lnQ$

$E_{cell} = E^{0} - \frac{0.059}{n}logQ$

n = number of electron transferred in the cell reaction = 4

The balanced equation is :

$4Fe^{2+} + 4H^{+} +O_{2} \rightarrow 4Fe^{3+} + 2H_{2}O$

$E_{cell} = 0.46 - \frac{0.059}{4}logQ$

$log Q = \frac{[Fe^{3+}]^{4}}{[Fe^{2+}]^{4}}$

$log Q = \left ( \frac{[Fe^{3+}]}{[Fe^{2+}]} \right )^{4}$

$log Q = \left ( \frac{1.9}{2.0} \right )^{4}$

Insert the value of log Q in Nernst  Equation:

$E_{cell} = 0.46 - \frac{0.059}{4} log\left ( \frac{1.9}{2.0} \right )^{4}$

(using :$log^{a}b = a\log b$

)

$E_{cell} = 0.46 - log\frac{1.9}{2.0}$

$E_{cell} = 0.46 - log(0.95)$

$E_{cell} = 0.46 -(-0.0227)$

$E_{cell} = 0.482 V$