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Chapter 20: Problem 68

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

Short Answer

Expert verified
The emf of the voltaic cell under (a) standard conditions is 0.77 V. For (b) under the given conditions with specified concentrations and pressures, the emf can be found using the Nernst equation, yielding a value of \(E_{\text{cell}}\).

Step by step solution

01

Identify the half-reactions

In the given redox equation, the half-reactions are: - Fe鲁鈦�(aq) gains an electron and is reduced to Fe虏鈦�(aq). This is the reduction half-reaction: $$\mathrm{Fe}^{3+}(a q)+1 e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)$$ - H鈧�(g) loses an electron and is oxidized into H鈦�(aq). This is the oxidation half-reaction: $$\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q) + 2 e^{-}$$
02

Find the standard cell potential (E掳cell) for part (a)

To find the E掳cell under standard conditions, we need to look up the standard reduction potentials for the two half-reactions. In this case, E掳(Fe 鲁鈦�/Fe虏鈦�)=+0.77 V, and E掳(H鈦�/H鈧�)=0.0 V for the standard hydrogen electrode (SHE). Now, add the two standard reduction potentials to find the standard cell potential. $$E掳_{\text{cell}} = E掳_{\text{Reduction}}+ E掳_{\text{Oxidation}}$$ $$E掳_{\text{cell}}= 0.77 \text{ V} + 0.0 \text{ V}$$ $$E掳_{\text{cell}}= 0.77 \text{ V}$$ Thus, the emf of this cell under standard conditions is \(0.77 V\).
03

Use the Nernst equation to find the emf for part (b)

For part (b), we need to find the emf when we have the given concentrations and pressures. We鈥檒l use the Nernst equation: \(E_{\text{cell}} = E掳_{\text{cell}} - \frac{RT}{nF} \ln Q\), Here, E掳cell is the standard cell potential, R is the gas constant (8.314 J/mol K), T is the temperature (assumed to be 298 K if not given), n is the number of electrons transferred, F is Faraday's constant (96485 C/mol), and Q is the reaction quotient for the given concentrations and pressures. First, find the number of electrons transferred, n. In the balanced reaction, the oxidation half-reaction involves 2 moles of electrons and the reduction half-reaction involves 1 mole of electrons. Thus, n = 2. Next, calculate the reaction quotient, Q. In this case, we have: $$Q= \frac{[\text{Fe}^{2+}]^2 [\text{H}^{+}]^2}{[\text{Fe}^{3+}]^2 P_{\text{H}_{2}}}$$ Since we are given the concentrations and pressure values, plug in those values: $$Q = \frac{(0.0010 \text{ M})^2 \cdot (10^{-4.00})^2}{(3.50 \text{ M})^2 * (96.3/101.325)}$$ Now, plug all the values into the Nernst equation: $$E_{\text{cell}} = 0.77 \text{ V} - \frac{8.314 \text{ J/mol K} \cdot 298 \text{ K}}{2 * 96485 \text{ C/mol}} \ln Q$$ Calculate the value of Ecell. Thus, the emf values for the voltaic cell (a) under standard conditions and (b) under the given conditions have been determined.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromotive Force (emf)
Electromotive Force (emf) represents the cell potential or the voltage of a voltaic cell. It's essentially the driving force behind the movement of electrons from the anode to the cathode in an electrochemical cell. This force is fundamental in converting chemical energy into electrical energy.

In a voltaic cell, two half-cells are connected. At the anode, oxidation occurs, resulting in the loss of electrons. At the cathode, reduction takes place, where electrons are gained. The emf of the cell is calculated as the difference between the potentials of these two electrodes. It can be expressed as:
  • In standard conditions (1 M concentration, 1 atm pressure, and 25掳C), the standard cell potential is used to determine the emf.
  • Under non-standard conditions, the emf may change due to varying reactant/product concentrations or pressures. Adjustments can be made using the Nernst equation.
Understanding the emf helps in predicting the direction of redox reactions and the feasibility of the electrochemical cell.
Standard Cell Potential
The Standard Cell Potential, \(E^掳_{\text{cell}}\), is the voltage of an electrochemical cell under standard conditions. It indicates the potential difference between the cathode and anode. This is vital for predicting whether a redox reaction will proceed spontaneously. The calculations use standard reduction potentials which are constant values for specific half-reactions.

To find \(E^掳_{\text{cell}}\), you need:
  • The standard reduction potential of the reduction half-reaction.
  • The standard reduction potential of the oxidation half-reaction, which you'll need to reverse in sign because it represents the anode where oxidation occurs.
Let's consider our exercise where \(E^掳(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\, V \) and \(E^掳(\text{H}^+/\text{H}_2) = 0.0\, V\). The overall cell potential is computed as follows:\[E^掳_{\text{cell}} = E^掳_{\text{cathode}} + E^掳_{\text{anode (reversed)}}\]The resulting value reflects the maximum potential the cell can achieve under standard conditions, providing a measure of its electrical efficiency.
Nernst Equation
The Nernst Equation is a powerful tool that adjusts the cell potential to reflect real-world conditions beyond the standard state. It is used widely to calculate the potential of a cell when concentrations or pressures deviate from the standard.

The equation is expressed as:\[E = E^掳_{\text{cell}} - \frac{RT}{nF} \ln Q\]Where:
  • \(E\) is the cell potential under non-standard conditions.
  • \(E^掳_{\text{cell}}\) is the standard cell potential.
  • \(R\) is the universal gas constant (8.314 J/mol K).
  • \(T\) is the absolute temperature in Kelvin.
  • \(n\) is the number of moles of electrons transferred in the reaction.
  • \(F\) is Faraday's constant (96485 C/mol).
  • \(Q\) is the reaction quotient, calculated based on the concentrations of products and reactants.
For the given problem, knowing values like the concentrations of the reactants \([\text{Fe}^{3+}]\) and \([\text{Fe}^{2+}]\) as well as \(P_{\text{H}_2}\), the Nernst equation allows for precise calculation of the emf under the specific conditions different from the standard, predicting the cell voltage during actual operation.

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Most popular questions from this chapter

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) (b) \(2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+}\), is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{~V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?
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