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Chapter 19: Problem 29

What is the potential of a cell made up of \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cells at \(25^{\circ} \mathrm{C}\) if \(\left[\mathrm{Zn}^{2+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{Cu}^{2+}\right]=0.15 \mathrm{M} ?\)

Short Answer

Expert verified
The potential of the cell is 1.087V.

Step by step solution

01

Identify the half-cell reactions

The half-cell reactions can be represented as follows: \(Zn → Zn^{2+} + 2e^-\) and \(Cu^{2+} + 2e^- → Cu\). The former is the anode (Zn) and the latter is the cathode (Cu). The anode experiences oxidation while the cathode experiences reduction.
02

Construct the Cell Reaction

We combine the two half-cell reactions to give the overall cell reaction: \(Zn + Cu^{2+} → Zn^{2+} + Cu\).
03

Identify Standard Reduction Potentials

Checking a Reduction Potential Table, we have: E°(Cu2+/Cu) = +0.34V and E°(Zn2+/Zn) = -0.76V.
04

Calculate the Standard Cell Potential

We compute the standard cell potential, E°cell, using the formula: E°cell = E°(cathode) - E°(anode). Substituting, we have: E°cell = +0.34V - (-0.76V) = +1.1V.
05

Calculate the Cell Potential at 25°C

By using the Nernst equation: \(Ecell = E°cell - \frac{(RT/nF) lnQ}\), where R=8.31 (J/mol×K); T=298K (temperature in Kelvin); n=2 (number of electrons transferred); F=96485 C/mol (Faraday’s Constant), and Q (reaction quotient) = [Zn2+]/[Cu2+]. Substituting the given values: \(Ecell = 1.1V - \frac{(8.31 × 298)/(2 × 96485) ln(0.25/0.15)}\). This simplifies to \(Ecell=1.1V - 0.013V = 1.087V\).

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Most popular questions from this chapter

How many faradays of electricity are required to produce (a) \(0.84 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at exactly \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) from aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution; (b) \(1.50 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) at \(750 \mathrm{mmHg}\) and \(20^{\circ} \mathrm{C}\) from molten \(\mathrm{NaCl}\); (c) \(6.0 \mathrm{~g}\) of Sn from molten \(\mathrm{SnCl}_{2} ?\)

The cathode reaction in the Leclanché cell is given by $$ 2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s) $$ If a Leclanché cell produces a current of \(0.0050 \mathrm{~A}\) calculate how many hours this current supply will last if there are initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \left|E^{\circ}\right|=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \left|E^{\circ}\right|=0.25 \mathrm{~V} \end{array} $$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to SHE. (a) Are the \(E^{\circ}\) values of the half- reactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge?

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to \(\mathrm{Fe}(\mathrm{II})\) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.
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