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Chapter 11: Q2DQ (page 435)

In making a specific galvanic cell, explain how one determines which electrodes and solutions to use in the cell.

Short Answer

Expert verified

The solution choice and electrode in the cell are used to make the specific galvanic cell has explained.

Step by step solution

01

Definition of galvanic cell and its electrodes

The device in which free energy of chemical and physical process can be converted into electrical energy called galvanic cell. These cells contains two electrodes which is immersed in one or more electrolytes

02

Composition of galvanic cell

The composition of galvanic cell consists salt bridge, solution and two electrodes in which electrodes can be dipped and porous membrane. The solution can be balanced and contain cation and anion electrode in it.

03

Electrodes of galvanic cell

There are two electrodes in galvanic cell are cathode and anode. The positive electrode is cathode in which reduction takes place with high reduction potential. The negative electrode in which oxidation takes place with low reduction potential.

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Most popular questions from this chapter

When magnesium metal is added to a beaker of HCl(aq), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when Mg is added directly to the beaker of HCl? How can you harness this reaction to do useful work?

Consider the following galvanic cell:

A 15.0-mole sample of NH₃ is added to the Ag compartment

(assume 1.00 L of total solution after the addition).

The silver ion reacts with ammonia to form complex

ions as shown:

$A g^{+} + N H_{3} \to A g N H_{3} (K_{1} = 2.1 \times 10^{3}) A g N H_{3}^{+} + N H_{3} \to A g {(N H_{3})}_{2} (K_{2} = 8.2 \times 10^{3})$

Calculate the cell potential after the addition of

15.0 moles of NH₃.

A chemist wishes to determine the concentration of $Cr {O_{4}}^{2 -}$ electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise24) and a silver wire coated with ${Ag}_{2} {CrO}_{4} .$ The value for the following half-reaction is +0.446 V relative to the standard hydrogen electrode:

${Ag}_{2} {CrO}_{4} + 2 e^{-} \to 2 Ag + Cr {O_{4}}^{2 -}$
a. Calculate E_cell and $ΔG$ at 25°C for the cell reaction when $Cr {O_{4}}^{2 -}$ = 1.00 mol/L.
b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant.

c. If the coated silver wire is placed in a solution (at 25°C) in which $Cr {O_{4}}^{2 -}$ = 1.00 x 10^-5 M, what is the expected cell potential?

d. The measured cell potential at 25°C is 0.504 V when the coated wire is dipped into a solution of unknown $Cr {O_{4}}^{2 -}$ . What is the $Cr {O_{4}}^{2 -}$ for this solution?

e. Using data from this problem and from Table 11.1,calculate the solubility product (K_sp) for ${Ag}_{2} {CrO}_{4} .$

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is

$2 CO (g) + O_{2} (g) \to 2 {CO}_{2} (g)$

The two half-cell reactions are

$CO + O^{2 -} \to {CO}_{2} + 2 e^{-} O_{2} + 4 e^{-} \to 2 O^{2 -}$

The two half-reactions are carried out in separate compartments connected with a solid mixture of CeO₂ and Gd₂O₃. Oxide ions can move through this solid at high temperatures (about 800°C). ∆Gfor the overall reaction at 800°C under certain concentration conditions is -380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Question:A galvanic cell is based on the following half-reactions:

${Fe}^{+ 2} + 2 e^{-} \overset{}{\to} Fe (s) E^{\circ} = - 0.440 2 H^{+} + 2 e^{-} \overset{}{\to} H_{2} (g) E^{\circ} = 0$

In this cell, the iron compartment contains an iron electrodeand $[{Fe}^{+ 2}] = 1.00 \times 10^{- 3} atm$ , and the hydrogen compartment contains a platinum electrode,role="math" localid="1663906376319" ${pH}_{2} = 1.00 atm$ and a weak acid HA at an initial concentration of role="math" localid="1663906339580" $1.00 M$ .If the observed cell potential is $0.333 V$ at $25^{\circ} C$ , calculate the Ka value for the weak acid HA atrole="math" localid="1663906523400" $25^{\circ} C$ .

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