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Chapter 20: Problem 51

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 \(E_{\text {cell }}^{\circ}\).

Short Answer

Expert verified
Using the given equilibrium constant (K=1.5 x 10鈦烩伌) and temperature (298 K), we calculate the corresponding Gibbs free energy change (螖G掳) using the formula 螖G掳 = -RT ln(K), which yields 螖G掳 = 32,839.57 J/mol. We then use the formula 螖G掳 = -nFE掳鈧坈ell to find the standard cell potential (E掳鈧坈ell), resulting in E掳鈧坈ell = -0.17 V.

Step by step solution

01

Calculate 螖G掳 using the equilibrium constant

We can use the given equilibrium constant (K=1.5 x 10鈦烩伌) and the formula 螖G掳 = -RT ln(K) to find the Gibbs free energy change. First, plug in the values for R (8.314 J/mol路K) and T (298 K) into the formula: 螖G掳 = - (8.314 J/mol路K) x (298 K) x ln(1.5 x 10鈦烩伌) Now, calculate the natural logarithm of K: ln(1.5 x 10鈦烩伌) = -13.41 So, we can plug this value back into the formula for 螖G掳 : 螖G掳 = - (8.314 J/mol路K) x (298 K) x (-13.41) Finally, calculate the value of 螖G掳 : 螖G掳 = 32839.57 J/mol
02

Calculate E掳鈧坈ell using the 螖G掳 value

Now that we have the value for 螖G掳, we can use the formula 螖G掳 = -nFE掳鈧坈ell to find the standard cell potential. First, plug in the values for n (2) and F (96485 C/mol) into the formula: 32839.57 J/mol = - (2) x (96485 C/mol) x E掳鈧坈ell Now, isolate E掳鈧坈ell by dividing both sides by -2F: E掳鈧坈ell = (32839.57 J/mol) / (-2 x 96485 C/mol) Finally, calculate the value of E掳鈧坈ell : E掳鈧坈ell = -0.17 V So, the corresponding Gibbs free energy change (螖G掳) is 32,839.57 J/mol, and the standard cell potential (E掳鈧坈ell) is -0.17 V.

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Most popular questions from this chapter

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the total energy requirement for this electrolysis if the applied emf is \(+5.00 \mathrm{~V} ?\)

(a) What is the maximum amount of work that a 6 -V lead-acid battery of a golf cart can accomplish if it is rated at \(300 \mathrm{~A}-\mathrm{h} ?(\mathrm{~b})\) List some of the reasons why this amount of work is never realized.

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l)--\rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (b) \(\mathrm{NaI}(a q)+3 \mathrm{HOCl}(a q)-\cdots \rightarrow \mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)--\rightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g) $$ (d) \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s)--\rightarrow\) \(\mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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?

(a) How many coulombs are required to plate a layer of chromium metal \(0.25 \mathrm{~mm}\) thick on an auto bumper with a total area of \(0.32 \mathrm{~m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in \(10.0 \mathrm{~s} ?(\mathrm{c})\) If the external source has an emf of \(+6.0 \mathrm{~V}\) and the electrolytic cell is \(65 \%\) efficient, how much electrical power is expended to electroplate the bumper?
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