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Chapter 20: Problem 57

A cell has a standard cell potential of \(+0.177 \mathrm{V}\) at 298 \(\mathrm{K}\) . What is the value of the equilibrium constant for the reaction ( a ) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Short Answer

Expert verified
For the given standard cell potential of +0.177 V at 298 K, the equilibrium constant (K) can be calculated using the Nernst Equation for different values of n. When n=1, K ≈ 1.96×10^(-9); when n=2, K ≈ 3.84×10^(-18); and when n=3, K ≈ 7.61×10^(-27).

Step by step solution

01

Write the Nernst Equation and relationship between E and K

Use the Nernst equation to relate the cell potential, number of electrons transferred, and the reaction quotient to the equilibrium constant: E° = - (\(R * T\)) / (\(n * F\)) * ln K Where E° is the standard cell potential in volts (V), R is the gas constant equal to \(8.314\frac{J}{K\cdot mol}\), T is the absolute temperature in Kelvin (K), n is the number of moles of electrons transferred, F is the Faraday's constant equal to \(9.6485 * 10^4\frac{C}{mol}\), and ln K is the natural logarithm of the equilibrium constant.
02

Solve for ln K

Rearrange the equation to solve for ln K: ln K = - (\(n * F\)) / (\(R * T\)) * E°
03

Calculate K for n=1

Plug in the given values of E°=0.177 V, T=298 K, and n=1 into the equation and calculate K: ln K = - (\(1 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -20.099 K = e^(-20.099) K ≈ 1.96×10^(-9) Thus, when n=1, the equilibrium constant (K) is approximately 1.96×10^(-9).
04

Calculate K for n=2

Plug in the given values of E°=0.177 V, T=298 K, and n=2 into the equation and calculate K: ln K = - (\(2 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -40.198 K = e^(-40.198) K ≈ 3.84×10^(-18) Thus, when n=2, the equilibrium constant (K) is approximately 3.84×10^(-18).
05

Calculate K for n=3

Plug in the given values of E°=0.177 V, T=298 K, and n=3 into the equation and calculate K: ln K = - (\(3 * (9.6485 * 10^4\frac{C}{mol})\)) / (\(8.314\frac{J}{K\cdot mol} * 298K\)) * (0.177 V) ln K = -60.297 K = e^(-60.297) K ≈ 7.61×10^(-27) Thus, when n=3, the equilibrium constant (K) is approximately 7.61×10^(-27).

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Most popular questions from this chapter

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{znO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text { red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{V}\) . The overall cell potential is \(+1.35 \mathrm{V}\) . Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid?

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

(a) What conditions must be met for a reduction potential to be a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode? (c) Why is it impossible to measure the standard reduction potential of a single half reaction?

If the equilibrium constant for a two-electron redox reaction at 298 \(\mathrm{K}\) is \(1.5 \times 10^{-4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)
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