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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 \(\left[\mathrm{Fe}^{3+}\right]=3.50 M, P_{\mathrm{H}_{2}}=\) \(0.95 \mathrm{atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 M,\) and the \(\mathrm{pH}\) in both half-cells is 4.00\(?\)

Short Answer

Expert verified
(a) The standard emf of the cell under standard conditions is 0.77 V. (b) The emf of the cell under the given conditions (\([Fe^{3+}]=3.50 M, P_{H_2}=0.95 atm, [Fe^{2+}]=0.0010 M\), and pH in both half-cells is 4.00) is approximately 0.785 V.

Step by step solution

01

Write the Nernst equation

The Nernst equation is given by: \[E_{cell} = E_{cell}^o - \frac{RT}{nF} lnQ\] Where: - \(E_{cell}\) is the emf of the cell - \(E_{cell}^o\) is the standard emf of the cell - R is the gas constant (8.314 J/mol K) - T is the temperature in Kelvin (assumed as 298 K) - n is the number of moles of electrons transferred in the reaction - F is the Faraday's constant (96485 C/mol) - Q is the reaction quotient
02

Identify the half-reactions and their standard reduction potentials

We can split the given reaction into two half-reactions: (1) \(Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)\) (2) \(2H^+(aq) + 2e^- \rightarrow H_2(g)\) Look up the standard reduction potentials (E°) for these half-reactions: \(E^o_1\) = +0.77 V (for half-reaction 1) \(E^o_2\) = 0 V (for half-reaction 2, as H+ is the reference for standard reduction potentials)
03

Calculate the standard emf of the cell (part a)

Now we can calculate the standard emf of the cell by subtracting the standard reduction potential of half-reaction 2 from half-reaction 1: \(E_{cell}^o = E^o_1 - E^o_2 = 0.77V - 0V = 0.77V\) So, the standard emf of the cell is 0.77 V.
04

Calculate the reaction quotient (Q) under given conditions (part b)

The reaction quotient Q is calculated as follows: \[Q= \frac{[Fe^{2+}]^2[H^+]^2}{[Fe^{3+}]^2[P(H_2)]}\] Using the given values: - \([Fe^{3+}] = 3.50 M\) - \(P_{H_2} = 0.95 atm\) - \([Fe^{2+}] = 0.0010 M\) The half-cells have a pH of 4.00, so the concentration of H+ ions can be calculated using the relation: \([H^+] = 10^{-pH} = 10^{-4.00} = 1.00 \times 10^{-4} M\) Now, we can calculate Q: \[Q = \frac{(0.0010)^2(1.00 \times 10^{-4})^2}{(3.50)^2(0.95)} = 2.04 \times 10^{-14}\]
05

Calculate the non-standard emf of the cell (part b)

Plug the values into the Nernst equation to find the emf of the cell under the given conditions: \[E_{cell} = E_{cell}^o - \frac{RT}{nF} lnQ\] n = 2 (as there are 2 moles of electrons transferred in the reaction) \[E_{cell} = 0.77 - \frac{8.314 \times 298 K}{2 \times 96485 C/mol} ln(2.04 \times 10^{-14})\] \[E_{cell} = 0.77 + 0.0145 \approx 0.785 V\] So, the emf of the cell under the given conditions is approximately 0.785 V.

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Most popular questions from this chapter

In the Bronsted-Lowry concept of acids and bases, acid-base reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases? [Sections 20.1 and 20.2\(]\)

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?

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r}{\operatorname{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{V}\) . Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

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

The electrodes in a silver oxide battery are silver oxide \(\left(\mathrm{Ag}_{2} \mathrm{O}\right)\) and zinc. (a) Which electrode acts as the anode? (b) Which battery do you think has an energy density most similar to the silver oxide battery: a Li-ion battery, a nickel-cadmium battery, or a lead-acid battery? [ Section 20.7]
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