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Chapter 20: Problem 120

\(E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) for the reaction, \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\) \(\operatorname{sn}^{4+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) (a) Can a solution be prepared that is \(0.500 \mathrm{M}\) in each of the four ions at \(298 \mathrm{K} ?\) (b) If not, in what direction must a net reaction Occur?

Short Answer

Expert verified
No, a solution cannot be prepared that is 0.500M in each of the four ions at 298K because \(E_{\mathrm{cell}}\) is negative. Therefore, the net reaction must occur in the backward direction.

Step by step solution

01

Nernst Equation Formulation

First, express the Nernst equation, which provides a relation between the cell potential and the concentrations of the species involved in the cell reaction. It follows from the Second Law of thermodynamics. The formula is: \[E_{\mathrm{cell}}= E_{\mathrm{cell}}^{\circ} - \frac {0.0592}{\text{n}} \log\frac{[\text{Products}]^{\text{Stoichiometric coefficient}}}{[\text{Reactants}]^{\text{Stoichiometric coefficient}}}\] Here, \[E_{\mathrm{cell}}^{\circ}\] is the standard cell potential, \[n\] is the number of moles of electrons transferred in the reaction (for this equation, \(n = 2\)), and \[\text{Products}\] and \[\text{Reactants}\] denote the concentration of products and reactants, respectively.
02

Substitute Given Values

Substitute the values related to the exercise into the equation formulated in step 1. For this case, these values are: \(E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\), concentrations are all \(0.500 \mathrm{M}\), and since Cu and Sn transformation involve loss and gain of two electrons respectively, \(n = 2\). The modified Nernst equation for this problem is: \[E_{\mathrm{cell}}= -0.0050 - \frac {0.0592}{2} \log\left(\frac{(0.500)^2}{(0.500)^2}\right)\]
03

Simplify and Solve for E-cell

Simplify the expression and solve for \(E_{\mathrm{cell}}\). Since the logarithm of 1 is zero, the equation simplifies to: \[E_{\mathrm{cell}}= -0.0050 - 0\] Therefore, \(E_{\mathrm{cell}}= -0.0050 \mathrm{V}\]
04

Determine the Direction of the Reaction

Examine the sign of E-cell to determine the direction of the reaction. Here, since \(E_{\mathrm{cell}}\) is negative, it implies that the reaction cannot go in forward direction, but instead has to proceed in backward direction or to the left.

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Most popular questions from this chapter

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

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

Describe a laboratory experiment that you could perform to evaluate the Faraday constant, \(F,\) and then show how you could use this value to determine the Avogadro constant.

Calculate the quantity indicated for each of the following electrolyses. (a) \(\left[\mathrm{Cu}^{2+}\right]\) remaining in \(425 \mathrm{mL}\) of a solution that was originally \(0.366 \mathrm{M} \mathrm{CuSO}_{4},\) after passage of \(2.68 \mathrm{A}\) for 282 s and the deposition of Cu at the cathode (b) the time required to reduce \(\left[\mathrm{Ag}^{+}\right]\) in \(255 \mathrm{mL}\) of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) from 0.196 to \(0.175 \mathrm{M}\) by electrolyzing the solution between \(\mathrm{Pt}\) electrodes with a current of \(1.84 \mathrm{A}\)

Of the following statements concerning electrochemical cells, the correct ones are: (a) The cathode is the negative electrode in both voltaic and electrolytic cells. (b) The function of a salt bridge is to permit the migration of electrons between the half-cell compartments of an electrochemical cell. (c) The anode is the negative electrode in a voltaic cell. (d) Electrons leave the cell from either the cathode or the anode, depending on what electrodes are used. (e) Reduction occurs at the cathode in both voltaic and electrolytic cells. (f) If electric current is drawn from a voltaic cell long enough, the cell becomes an electrolytic cell. (g) The cell reaction is an oxidationreduction reaction.
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