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

Short Answer

Expert verified
For the given reactions, equilibrium constants are: (a) \(K = 1.64 \times 10^{13}\), (b) \(K = 1.16 \times 10^{14}\), and (c) \(K = 2.45 \times 10^{61}\).

Step by step solution

01

Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: Fe(s) 鈫 Fe虏鈦(aq) + 2e鈦 Reduction half-reaction: Ni虏鈦(aq) + 2e鈦 鈫 Ni(s)
02

Determine the standard cell potential

Now, we will calculate the standard cell potential (E掳) using the reduction potentials given in Appendix E. For Fe虏鈦(aq)/Fe(s): E掳 = -0.44V For Ni虏鈦(aq)/Ni(s): E掳 = -0.23V Since we are subtracting E掳 values, E掳 (cell) = E掳 (cathode) - E掳 (anode) = -0.23 - (-0.44) = 0.21V
03

Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (螖G掳) using the equation: 螖G掳 = -nFE掳 Where n is the number of moles of electrons transferred and F is the Faraday's constant (96485 C/mol). 螖G掳 = -2 * 96485 * 0.21 = -40525.7 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G掳}}{RT}} = e^{-\frac{-40525.7}{8.314 \times 298}} = 1.64 脳 10鹿鲁 (b) Co(s) + 2 H鈦(aq) 鈫 Co虏鈦(aq) + H鈧(g)
04

Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: Co(s) 鈫 Co虏鈦(aq) + 2e鈦 Reduction half-reaction: 2 H鈦(aq) + 2e鈦 鈫 H鈧(g)
05

Determine the standard cell potential

Now, we will calculate the standard cell potential (E掳) using the reduction potentials given in Appendix E. For Co虏鈦(aq)/Co(s): E掳 = -0.28V For H鈧(g)/H鈦(aq): E掳 = 0.00V E掳 (cell) = E掳 (cathode) - E掳 (anode) = 0.00 - (-0.28) = 0.28V
06

Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (螖G掳) using the equation: 螖G掳 = -nFE掳 螖G掳 = -2 * 96485 * 0.28 = -54021.2 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G掳}}{RT}} = e^{-\frac{-54021.2}{8.314 \times 298}} = 1.16 脳 10鹿鈦 (c) 10 Br鈦(aq) + 2 MnO鈧勨伝(aq) + 16 H鈦(aq) 鈫 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 Br鈧(l)
07

Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: 10 Br鈦(aq) 鈫 5 Br鈧(l) + 10e鈦 Reduction half-reaction: 2 MnO鈧勨伝(aq) + 10e鈦 + 16 H鈦(aq) 鈫 2 Mn虏鈦(aq) + 8 H鈧侽(l)
08

Determine the standard cell potential

Now, we will calculate the standard cell potential (E掳) using the reduction potentials given in Appendix E. For 5 Br鈧(l)/10 Br鈦(aq): E掳 = +1.07V For 2 MnO鈧勨伝(aq)/2 Mn虏鈦(aq): E掳 = +1.51V E掳 (cell) = E掳 (cathode) - E掳 (anode) = 1.51 - 1.07 = 0.44V
09

Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (螖G掳) using the equation: 螖G掳 = -nFE掳 螖G掳 = -10 * 96485 * 0.44 = -424171 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G掳}}{RT}} = e^{-\frac{-424171}{8.314 \times 298}} = 2.45 脳 10鈦堵 Equilibrium constants for reactions (a), (b), and (c) are 1.64 脳 10鹿鲁, 1.16 脳 10鹿鈦, and 2.45 脳 10鈦堵, respectively.

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Key Concepts

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

Equilibrium Constant
In an electrochemical reaction, the equilibrium constant \((K)\) is a crucial factor that links the reaction quotient at equilibrium to the Gibbs Free Energy change. An equilibrium constant provides insights into where the equilibrium of a redox reaction lies, suggesting whether the products or reactants are favored at equilibrium.

The formula to connect the equilibrium constant and Gibbs Free Energy change is:
  • \( K = e^{- rac{\Delta G^\circ}{RT}} \)
Here:
  • \( \Delta G^\circ \) is the standard Gibbs Free Energy change of the reaction.
  • \( R \) is the universal gas constant (8.314 J/(mol路K)).
  • \( T \) is the temperature in Kelvin.
If \( K \) is significantly greater than 1, it implies the reaction favors the formation of products at equilibrium. Conversely, if \( K \) is much less than 1, the reactants are favored. A moderate value of \( K \) suggests a more balanced state at equilibrium.

Through calculating \( K \) for reactions, it becomes easier to predict the extent to which a redox reaction will proceed.
Reduction Potentials
Reduction potentials, often denoted as \( E^\circ \), play a vital role in electrochemistry, acting as a measure of the tendency of a chemical species to gain electrons and undergo reduction. Each half-reaction is associated with a specific reduction potential. It is vital to note:
  • Positive \( E^\circ \) values suggest a substance readily gains electrons (strong oxidizing agent).
  • Negative \( E^\circ \) values indicate a substance less willing to receive electrons.
To calculate the cell potential \( E^\circ_{cell} \) for a complete redox reaction, we utilize the reduction potentials of both the oxidation and reduction reactions. This is done using:
  • \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)
By determining \( E^\circ_{cell} \), you can infer the spontaneity of a reaction. A positive \( E^\circ_{cell} \) indicates that the reaction is spontaneous under standard conditions.
Gibbs Free Energy Change
Gibbs Free Energy change, represented as \( \Delta G \), forms a cornerstone concept in understanding reaction spontaneity. This thermodynamic quantity helps determine whether a reaction can proceed without needing external energy.
  • A negative \( \Delta G^\circ \) signifies a spontaneous process.
  • A positive \( \Delta G^\circ \) means the process is not spontaneous.
  • When \( \Delta G^\circ \) is zero, the system is at equilibrium.
For electrochemical reactions, work or energy changes can be calculated:
  • \( \Delta G^\circ = -nFE^\circ_{cell} \)
Where:
  • \( n \) is the number of moles of electrons exchanged.
  • \( F \) is Faraday's constant (96485 C/mol).
  • \( E^\circ_{cell} \) is the standard cell potential.
This relationship helps us connect Gibbs Free Energy to electrochemical cell potential, providing a deeper insight into the feasibility and direction of a reaction.

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Most popular questions from this chapter

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for \(2.00 \mathrm{~min}, 12.3 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(25.5^{\circ} \mathrm{C}\), and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) Oxidizing agents can convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

(a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metalhydride batteries over the nickel-cadmium batteries?

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of \(4.55\) A through molten \(\mathrm{MgCl}_{2}\), for \(3.50\) days? (b) How many minutes are needed to plate out \(10.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\), using \(3.50 \mathrm{~A}\) of current?

If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(8.7 \times 10^{4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(\bar{E}_{\text {cell }}^{\circ}\).

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