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Chapter 13: Problem 11

For the following cells, write the half-reactions occurring at each electrode and the complete cell reaction, and calculate the cell potential: (a) \(\mathrm{Pt}, \mathrm{H}_{2}(0.2 \mathrm{~atm})|\mathrm{HCl}(0.5 \mathrm{M})| \mathrm{Cl}_{2}(0.2 \mathrm{~atm}), \mathrm{Pt}\) (b) \(\mathrm{Pt} \mid \mathrm{Fe}^{2+}(0.005 \mathrm{M}), \mathrm{Fe}^{3+}(0.05 \mathrm{M}), \mathrm{HClO}_{4}(0.1 \mathrm{M}) \| \mathrm{HClO}_{4}(0.1 \mathrm{M}), \mathrm{VO}_{2}^{+}\) \((0.001 M), \mathrm{VO}^{2+}(0.002) M \mid \mathrm{Pt}\).

Short Answer

Expert verified
(a) E_cell = 1.36 V. (b) E_cell = 0.23 V.

Step by step solution

01

Identify Reduction and Oxidation Reactions for Cell (a)

For the cell \( \mathrm{Pt}, \mathrm{H}_{2}(0.2 \mathrm{~atm})|\mathrm{HCl}(0.5 \mathrm{M})| \mathrm{Cl}_{2}(0.2 \mathrm{~atm}), \mathrm{Pt} \), we first identify the half-reactions. The reduction half-reaction at the cathode involves chlorine gas:\[ \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^- (aq) \].The oxidation half-reaction at the anode involves hydrogen gas:\[ \mathrm{H}_2(g) \rightarrow 2\mathrm{H}^+(aq) + 2e^- \].
02

Write Complete Cell Reaction for Cell (a)

Combine the half-reactions from Step 1:1. \( \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^- (aq) \)2. \( \mathrm{H}_2(g) \rightarrow 2\mathrm{H}^+(aq) + 2e^- \)The complete cell reaction is:\[ \mathrm{Cl}_2(g) + \mathrm{H}_2(g) \rightarrow 2\mathrm{Cl}^- (aq) + 2\mathrm{H}^+(aq) \].
03

Calculate Cell Potential for Cell (a)

The standard electrode potentials are:- \( \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^- \) has \( E^0 = +1.36 \mathrm{~V} \)- \( \mathrm{H}_2(g) \rightarrow 2\mathrm{H}^+ + 2e^- \) has \( E^0 = 0.00 \mathrm{~V} \)The cell potential is given by the Nernst equation, but under standard conditions, we calculate:\[ E_{cell} = E_{cathode}^0 - E_{anode}^0 = 1.36 \mathrm{~V} - 0.00 \mathrm{~V} = 1.36 \mathrm{~V} \].
04

Identify Half-Reactions for Cell (b)

For the cell \( \mathrm{Pt} \mid \mathrm{Fe}^{2+}(0.005 \mathrm{M}), \mathrm{Fe}^{3+}(0.05 \mathrm{M}), \mathrm{HClO}_{4}(0.1 \mathrm{M}) \| \mathrm{HClO}_{4}(0.1 \mathrm{M}), \mathrm{VO}_{2}^{+}(0.001 M), \mathrm{VO}^{2+}(0.002) M \mid \mathrm{Pt} \), the half-reactions are:- Oxidation half-reaction involving iron: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^- \).- Reduction half-reaction involving vanadium:\( \mathrm{VO}_2^+ + 2\mathrm{H}^+ + e^- \rightarrow \mathrm{VO}^{2+} + \mathrm{H}_2O \).
05

Write Complete Cell Reaction for Cell (b)

Combine the half-reactions from Step 4:1. \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^- \)2. \( \mathrm{VO}_2^+ + 2\mathrm{H}^+ + e^- \rightarrow \mathrm{VO}^{2+} + \mathrm{H}_2O \)The complete cell reaction is:\[ \mathrm{VO}_2^+ + 2\mathrm{H}^+ + \mathrm{Fe}^{2+} \rightarrow \mathrm{VO}^{2+} + \mathrm{H}_2O + \mathrm{Fe}^{3+} \].
06

Calculate Cell Potential for Cell (b)

The standard electrode potentials are:- \( \mathrm{VO}_2^+ + 2\mathrm{H}^+ + e^- \rightarrow \mathrm{VO}^{2+} + \mathrm{H}_2O \) has \( E^0 = +1.00 \mathrm{~V} \).- \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) has \( E^0 = +0.77 \mathrm{~V} \).The cell potential is calculated as:\[ E_{cell} = E_{cathode}^0 - E_{anode}^0 = 1.00 \mathrm{~V} - 0.77 \mathrm{~V} = 0.23 \mathrm{~V} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Reactions
In electrochemical cells, half-reactions are key elements that describe what happens at each electrode. Each half-reaction occurs at either the anode or the cathode.
The anode is where oxidation occurs, and the cathode is where reduction takes place. For example, in one of the problems given:
  • The oxidation half-reaction on the anode involves hydrogen gas (\[ \mathrm{H}_2(g) \rightarrow 2\mathrm{H}^+(aq) + 2e^- \]), which releases electrons.
  • Meanwhile, the reduction half-reaction at the cathode involves the gain of electrons by chlorine gas (\[ \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^- (aq) \]).
By writing these half-reactions, you track the transfer of electrons, which helps us understand the complete cell reaction.
Cell Potential
The cell potential (\(E_{cell}\)) is a measure of how much voltage the electrochemical cell can produce. It's essential to calculate this value to understand the effectiveness of the cell.
The potential is determined by comparing the standard electrode potentials of the half-reactions at the cathode and anode.
In short, the equation given by:
  • \[ E_{cell} = E_{cathode}^0 - E_{anode}^0 \].
To calculate the cell potential for one cell in the exercise:
  • The potential for reduction at the cathode (\( \mathrm{Cl}_2(g) \)) was +1.36 V.
  • The potential for oxidation at the anode (\( \mathrm{H}_2(g) \)) was 0.00 V.
Therefore, the cell potential was \(1.36\) volts. This tells us how strongly the cell can "push" electrons through the circuit.
Nernst Equation
The Nernst equation allows you to calculate the cell potential under non-standard conditions. It's particularly useful when concentrations or pressures are not at standard levels, like 1 M concentration or 1 atm pressure.
This equation is:
  • \[ E = E^0 - \frac{RT}{nF} \ln Q \],
where \(E^0\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.
By using the Nernst equation, you can determine the actual cell potential under specific conditions, reflecting how real-world environments might influence a cell's voltage.
Oxidation and Reduction Reactions
Oxidation and reduction reactions are vital components of an electrochemical cell.
These reactions involve the transfer of electrons from one species to another.
  • Oxidation occurs at the anode, where a species loses electrons. For instance, in the reactions given, hydrogen gas underwent oxidation: \[ \mathrm{H}_2(g) \rightarrow 2\mathrm{H}^+(aq) + 2e^- \].
  • Reduction takes place at the cathode, where a species gains electrons. For example, chlorine gas was reduced in the cell reaction: \[ \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^- (aq) \].

In understanding these reactions, you essentially follow the electrons' journey through the cell, which is crucial for grasping how energy is harnessed in technology, such as batteries and fuel cells.

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Most popular questions from this chapter

Calculate the potential of the cell consisting of a hydrogen electrode \(\left(P_{\mathrm{H}_{2}}=\right.\) \(1 \mathrm{~atm}\) ) and a saturated calomel reference electrode (a) in a solution of \(0.00100 M \mathrm{HCl},\) (b) in a solution of \(0.00100 \mathrm{M}\) acetic acid, and (c) in a solution containing equal volumes of \(0.100 M\) acetic acid and \(0.100 M\) sodium acetate. Assume that activities are the same as concentrations.

The nitrate concentration in an industrial effluent is determined using a nitrate ion-selective electrode. Standards and samples are diluted 20 -fold with \(0.1 M\) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) to maintain constant ionic strength. Nitrate standards of 0.0050 and \(0.0100 M\) give potential readings of -108.6 and \(-125.2 \mathrm{mV}\), respectively. The sample gives a reading of \(-119: 6 \mathrm{mV}\). What is the concentration of nitrate in the sample?

A valinomycin-based potassium ion-selective electrode is evaluated for sodium interference using the fixed interference method. A potassium calibration curve is prepared in the presence of \(140 \mathrm{~m} M\) sodium. The straight line obtained from extrapolation of the linear portion deviates from the experimental curve by \(17.4 \mathrm{mV}\) at a potassium concentration corresponding to \(1.5 \times 10^{-5} M .\) If the linear slope is \(57.8 \mathrm{mV}\) per decade, what is \(K_{\mathrm{NaK}}\) for the electrode?

A \(50-\mathrm{mL}\) solution that is \(0.10 \mathrm{M}\) in chloride and iodide ions is titrated with \(0.10 M\) silver nitrate. (a) Calculate the percent iodide remaining unprecipitated when silver chloride begins to precipitate. (b) Calculate the potential of a silver electrode versus the SCE when silver chloride begins to precipitate and compare this with the theoretical potential corresponding to end point for the titration of iodide. (c) Calculate the potential at the end point for chloride. Use concentrations in calculations.

Discuss the mechanism of the glass membrane electrode response for \(\mathrm{pH}\) measurements.
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