Question

For the cell: Ag(s) | AgNO3(aq) || HCl(aq) | AgCl(s), Ag(s) a) Write down the half-cell and overall cell reactions for the above cell and define the symbols in the notation. b) Using the Nernst equation, show how the EMF of each half-cell may be written in terms of the concentration of each species. c) At equilibrium, the cell will have 0 V. From the following data, calculate the equilibrium constant for the reaction at 298 K: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) at 298 K, EoAgCl, Cl-, Ag = 0.2223 V and EoAg+, Ag = 0.8 V RT/F = 0.0257 V In terms of the reaction at the cathode, please consider reacting chloride with silver to form silver chloride, balancing the charges with electrons. Also note that the cell can be written either way round; it just depends on whether the reaction is spontaneous or not.

          For the cell:
Ag(s) | AgNO3(aq) || HCl(aq) | AgCl(s), Ag(s)

a) Write down the half-cell and overall cell reactions for the above cell and define the symbols in the notation.
b) Using the Nernst equation, show how the EMF of each half-cell may be written in terms of the concentration of each species.
c) At equilibrium, the cell will have 0 V. From the following data, calculate the equilibrium constant for the reaction at 298 K:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
at 298 K, EoAgCl, Cl-, Ag = 0.2223 V and EoAg+, Ag = 0.8 V
RT/F = 0.0257 V

In terms of the reaction at the cathode, please consider reacting chloride with silver to form silver chloride, balancing the charges with electrons. Also note that the cell can be written either way round; it just depends on whether the reaction is spontaneous or not.

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Added by Elizabeth G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Ali Azmy

05/14/2023

Step 1

Half-cell reactions: Cathode: AgCl(s) + e- → Ag(s) + Cl-(aq) Anode: Ag(s) → Ag+(aq) + e- Overall cell reaction: AgCl(s) + Ag(s) → Ag(s) + Ag+(aq) + Cl-(aq) Cell notation: Ag(s) | AgNO3(aq) || HCl(aq) | AgCl(s), Ag(s) b) Now, we will use the Nernst equation to Show more…

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For the cell: Ag(s) | AgNO3(aq) || HCl(aq) | AgCl(s), Ag(s) a) Write down the half-cell and overall cell reactions for the above cell and define the symbols in the notation. b) Using the Nernst equation, show how the EMF of each half-cell may be written in terms of the concentration of each species. c) At equilibrium, the cell will have 0 V. From the following data, calculate the equilibrium constant for the reaction at 298 K: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) at 298 K, EoAgCl, Cl-, Ag = 0.2223 V and EoAg+, Ag = 0.8 V RT/F = 0.0257 V In terms of the reaction at the cathode, please consider reacting chloride with silver to form silver chloride, balancing the charges with electrons. Also note that the cell can be written either way round; it just depends on whether the reaction is spontaneous or not.

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Transcript

00:01 So we have for this galvanic cell, or electrochemical cell, we have silver solid, silver nitrate, and then we have hcl.

00:20 Like this, we have agcl solid and ag solid.

00:29 Basically, before we go further to define the symbols, single line means a phase transition, and double line means salt bridge.

00:41 So this is what happens here is oxidation.

00:44 So this is the anode.

00:47 We have oxidation, and this is the cathode.

00:53 We have reduction.

00:57 So we have reduction happening in the cathode and oxidation happening in the anode, and the symbols are single line is phase transition.

01:05 Double line is a salt bridge.

01:09 So to get the half cell equation, we have ag solid is going to give us ag plus an electron.

01:22 And then we have silver chloride solid plus this electron is going to give us silver solid plus chloride minus aqueous.

01:35 So basically, the half cell equation is going to be agcl solid, ag plus aqueous plus cl minus aqueous, and this is the first question.

01:53 The second one says, using nernst equation, show how emf for each half cell is going to be written down.

02:03 So emf of the anode, e anode, basically from nernst, equals a standard of the anode equation, ag solid, ag plus aqueous minus rt over nf, but n here is 1.

02:34 The number of electrons is only 1.

02:37 So we have this as 1.

02:40 Then the concentration of silver ion, and this is for anode...

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