A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O...

Question

A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?

Accepted Answer

Identify the half-reactions for the cathode and anode. For the given reaction, the oxidation at the anode is: $$4 	ext{Fe}^{2+} ightarrow 4 	ext{Fe}^{3+} + 4e^-$$ and the reduction at the cathode is: $$	ext{O}_2 + 4 	ext{H}^+ + 4e^- ightarrow 2 	ext{H}_2	ext{O}.. $$Write the Nernst equation for the overall cell reaction. The Nernst equation is given by: $$E = E^\circ - \frac{0.0592}{n} \log Q$$, where $$E^\circ is $$the standard cell potential, $$n is $$the number of moles of electrons transferred, and $$Q is $$the reaction quotient.. Calculate the reaction quotient, $$Q$$, using the given concentrations and pressures. For the reaction, $$Q = \frac{[	ext{Fe}^{3+}]^4 \cdot (	ext{P}_{	ext{O}_2})^{1/2}}{[	ext{Fe}^{2+}]^4 \cdot [	ext{H}^+]^4}. $$Substitute the values $$[	ext{Fe}^{2+}] = 1.3 	ext{ M}$$, $$[	ext{Fe}^{3+}] = 0.010 	ext{ M}$$, $$	ext{P}_{	ext{O}_2} = 0.50 	ext{ atm}$$, and $$[	ext{H}^+] = 10^{-3.50} 	ext{ M}$$ into the equation.. Determine the number of electrons transferred, $$n$$, in the balanced equation. From the half-reactions, you can see that 4 electrons are transferred.. Substitute the values into the Nernst equation to find the emf of the cell. Remember to use the standard emf, $$E^\circ$$, for the reaction which can be found in standard reduction potential tables.

A voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?

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주요 개념

Electrochemical Cells

Nernst Equation

pH and Hydrogen Ion Concentration