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Chapter 20: Problem 14

Consider the reaction \(\operatorname{Co}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{Ni}(\mathrm{s}), \quad\) with \(\quad E_{\mathrm{cell}}^{\circ}=0.02 \mathrm{V} . \quad\) If \(\quad \mathrm{Co}(\mathrm{s}) \quad\) is added to a solution with \(\left[\mathrm{Ni}^{2+}\right]=1 \mathrm{M},\) should the reaction go to completion? Explain.

Short Answer

Expert verified
Yes. Since the calculated cell potential is positive, the reaction is spontaneous and will go to completion.

Step by step solution

01

Determining the Nernst Equation

The Nernst equation is used to calculate cell potentials under non-standard conditions. It is given by \(E = E° - (RT/nF) * lnQ\) where E is the cell potential, E° is the standard cell potential, R is the gas constant (8.314 J/mol.K), T is the temperature in Kelvin (assuming it to be 298K as it is not given), n is the number of electron transfers in the reaction (here it is 2 as observed from the balanced reaction), F is Faraday's constant (96485 C/mol) and Q is the reaction quotient.
02

Determination of the Reaction Quotient (Q)

Q is the ratio of the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. In this case, as the reactant Co(solid) is in solid form and the product Ni(solid) is also in solid form their concentrations are omitted from the reaction quotient. Also, Co²⁺ is produced in the reaction. However, since its concentration is not given, we assume Q to be 1 by default.
03

Substituting values into the Nernst Equation

Substituting the values into the Nernst equation, we obtain \( E = 0.02 - (8.314*298/2*96485)*ln(1)\).
04

Calculation and interpretation of results

Since ln(1) = 0, the equation simplifies to \( E=0.02V \). The cell potential is positive, indicating that the reaction is spontaneous and would go to completion.

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Most popular questions from this chapter

Describe how you might construct batteries with each of the following voltages: (a) \(0.10 \mathrm{V} ;\) (b) \(2.5 \mathrm{V} ;\) (c) \(10.0 \mathrm{V}\). Be as specific as you can about the electrodes and solution concentrations you would use, and indicate whether the battery would consist of a single cell or two or more cells connected in series.

A solution containing both \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) ions is subjected to electrolysis. (a) Which metal should plate out first? (b) Plating out is finished after a current of \(0.75 \mathrm{A}\) is passed through the solution for 2.50 hours. If the total mass of metal is \(3.50 \mathrm{g},\) what is the mass percent of silver in the product?

Predict whether, to any significant extent, (a) \(\mathrm{Fe}(\mathrm{s})\) will displace \(\mathrm{Zn}^{2+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution; (c) \(\mathrm{Ag}(\mathrm{s})\) will react with \(1 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (d) \(\mathrm{O}_{2}(\mathrm{g})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution.

Calculate the quantity indicated for each of the following electrolyses. (a) the mass of \(\mathrm{Zn}\) deposited at the cathode in 42.5 min when 1.87 A of current is passed through an aqueous solution of \(\mathrm{Zn}^{2+}\) (b) the time required to produce \(2.79 \mathrm{g} \mathrm{I}_{2}\) at the anode if a current of \(1.75 \mathrm{A}\) is passed through \(\mathrm{KI}(\mathrm{aq})\)

$$E_{\text {cathode }}^{\circ}=(2.71-2.310) V=+0.40 V$$VVV
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