Calculate the cell potential of each of the following...

Calculate the cell potential of each of the following electrochemical cells at $25^{\circ} \mathrm{C}$. (a) $\mathrm{Pt} \mid \mathrm{H}_{2}(8.00 \mathrm{~atm}), \mathrm{H}^{+}\left(1.00 \times 10^{-3} \mathrm{M}\right) \|$ $$ \mathrm{Ag}^{+}(0.00549 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s}) $$ (b) $\mathrm{Pt} \mid \mathrm{H}_{2}(1.00 \mathrm{~atm}), \mathrm{H}^{+}(\mathrm{pH}=5.97) \|$ $$ \mathrm{H}^{+}(\mathrm{pH}=3.47), \mathrm{H}_{2}(1.00 \mathrm{~atm}) \mid \mathrm{Pt} $$ (c) $\mathrm{Pt} \mid \mathrm{H}_{2}(0.0361 \mathrm{~atm}), \mathrm{H}^{+}(0.0175 \mathrm{M}) \|$ $\mathrm{H}^{+}(0.0175 \mathrm{M}), \mathrm{H}_{2}\left(5.98 \times 10^{-4} \mathrm{~atm}\right) \mid \mathrm{Pt}$

Calculate the cell potential of each of the following electrochemical cells at $25^{\circ} \mathrm{C}$.
(a) $\mathrm{Pt} \mid \mathrm{H}_{2}(8.00 \mathrm{~atm}), \mathrm{H}^{+}\left(1.00 \times 10^{-3} \mathrm{M}\right) \|$
$$
\mathrm{Ag}^{+}(0.00549 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s})
$$
(b) $\mathrm{Pt} \mid \mathrm{H}_{2}(1.00 \mathrm{~atm}), \mathrm{H}^{+}(\mathrm{pH}=5.97) \|$
$$
\mathrm{H}^{+}(\mathrm{pH}=3.47), \mathrm{H}_{2}(1.00 \mathrm{~atm}) \mid \mathrm{Pt}
$$
(c) $\mathrm{Pt} \mid \mathrm{H}_{2}(0.0361 \mathrm{~atm}), \mathrm{H}^{+}(0.0175 \mathrm{M}) \|$
$\mathrm{H}^{+}(0.0175 \mathrm{M}), \mathrm{H}_{2}\left(5.98 \times 10^{-4} \mathrm{~atm}\right) \mid \mathrm{Pt}$

Key Concepts

Electrochemical Cells

These are devices that convert chemical energy into electrical energy through spontaneous redox reactions occurring in two physically separated half-cells. Each half-cell contains an electrode and an electrolyte, and the two are connected by a conductive path and a salt bridge to maintain charge balance.

Cell Potential (EMF)

This is the voltage difference between the cathode and the anode of an electrochemical cell. It represents the driving force behind the electron flow and is determined by the difference in electrode potentials, which can deviate from standard state values when conditions such as concentration or pressure change.

Standard Electrode Potential

These are reference values measured under standard conditions (typically 1 M concentration for solutions, 1 atm for gases, and 25°C) and are used to predict the direction of redox reactions. They serve as a baseline for calculating cell potentials under non-standard conditions.

Hydrogen Electrode

A common reference electrode based on the redox couple of hydrogen. This electrode, often used as the standard, consists of a platinum electrode in contact with H2 gas and a solution containing H+ ions at a defined activity, allowing comparison with other electrode potentials.

Nernst Equation

A fundamental equation in electrochemistry that adjusts the electrode potential from its standard value to account for non-standard conditions like altered concentrations or gas pressures. It quantitatively relates the cell potential to the reaction quotient, thereby predicting the effect of changes in conditions on the voltage output.

Transcript

00:01 In this question we've been asked to apply next equation to calculate the cell potential.

00:05 Remember? next equation.

00:09 This equation enables us to calculate the cell potential when conditions are not standard.

00:16 So remember when we are using standard cell potential or electrode potentials from the handbooks, from handbooks, those were determined under standard conditions.

00:28 But then right now we've been given certain information concerning the concentration of the solutions, and the different pressures so this whatever we've been given there is not standard so we have to use next equation we say is 0 .0 592 over the number of moles of electrons that are transferred say we've got a that is being reduced to form a so the number of moles of electrons is going to be three in this case so this is just the number of moles multiplied by the reaction cautioned.

01:07 So here we are looking at the concentration of the products divided by the concentration of the reactants.

01:16 This is more like kc or kb for solutions and solids.

01:22 So making use of this.

01:26 So first of all we have to calculate.

01:29 We have to calculate the electrode, the standard the standard cell potential and we do this by looking at the half equations and the electrode potentials for each reaction oxidation or reaction oxidation or reduction reaction that is occurring at the electrodes so if we are to look at this if we are to look at this say we've got a say we've got a say we've got a say we've got a those half equations are 2h plus plus two electrons to form h2 and we also have ag plus plus an electron to form ag.

02:25 So at the end of the day, the reactions that we are going to have here, this is going to be in reverse water.

02:34 This is going to be in reverse water.

02:36 This is going to happen instead for this reaction.

02:39 So at the end of the day, the overall reaction is going to be h2 plus 2h.

02:45 To form 2h plus plus 2ag remember for us to cancel out these electrons we have they have to be the same for the first reaction and the other reaction so for us to have two electrons here we have to multiply everything by 2 this is why we are having a 2a g here and 2ag over then so for this reaction the first one the overall cell potential the overall electrical potential here is 0v and the other one is positive 0 .7 994 volts so for this reaction the overall cell potential this is going to be the standard because we are using standard conditions here this is going to be equal to positive 0 .994 0 .7994 and it is very important that we include this sign because it tells us us that this reaction is spontaneous in that direction.

03:50 This is something that can actually okay.

03:53 So now moving on to next, substituting to calculate the cell potential.

03:58 This is going to be 0 .7994 minus 0 .0592 divided by the number of electrons that we are transferring.

04:12 During the process, we are transferring a total of two electrons.

04:16 So this is going to be divided by 2 multiplied by log, the concentration of, the concentration of, remember this is a solid, this is an accius, this is a cacios, and this is going to be gas...

Please give Ace some feedback