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sergiy2304 [10]
3 years ago
5

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

Chemistry
1 answer:
kari74 [83]3 years ago
8 0

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

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