A voltaic cell is constructed that uses the following...

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at $298 \mathrm{~K}$ with $\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}$ and $\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}$. (a) Determine $E$ for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With $\left[\mathrm{Ag}^{+}\right]$ equal to $0.15 \mathrm{M}$, at what concentration of $\mathrm{I}^{-}$ would the cell have zero potential?

A voltaic cell is constructed that uses the following half-cell reactions:
$$
\begin{array}{l}
\mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\
\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q)
\end{array}
$$
The cell is operated at $298 \mathrm{~K}$ with $\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}$ and $\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}$. (a) Determine $E$ for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With $\left[\mathrm{Ag}^{+}\right]$ equal to $0.15 \mathrm{M}$, at what concentration of $\mathrm{I}^{-}$ would the cell have zero potential?

Key Concepts

Voltaic Cell

A voltaic cell, or galvanic cell, is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions occurring in separated half-cells. The overall cell potential is determined by combining the potentials from the oxidation and reduction reactions, and the flow of electrons from one electrode to the other produces electrical work.

Half-Reactions

In any redox reaction taking place in an electrochemical cell, the overall reaction is split into two half-reactions: one for reduction (gain of electrons) and one for oxidation (loss of electrons). Each half-reaction is associated with a specific standard reduction potential, and proper identification of these reactions is crucial in determining the cell’s behavior.

Standard Reduction Potentials

Standard reduction potentials are values measured under standard conditions (1 M concentration for solutions, 1 atm pressure, and 25°C) that indicate the tendency of a species to be reduced. Comparing these potentials allows one to predict the direction of electron flow and to identify the stronger oxidizing and reducing agents in the cell.

Nernst Equation

The Nernst equation provides a quantitative relationship between the cell potential and the concentrations (or activities) of the reactants and products. It adjusts the standard reduction potential to account for non-standard conditions by incorporating the reaction quotient, temperature, and number of electrons transferred in the half-reaction.

Anode vs Cathode Identification

The anode is the electrode where oxidation occurs, while the cathode is where reduction takes place. Correct identification is essential because the overall cell potential depends on the difference between the cathode and anode potentials. Changes in ion concentrations can influence which electrode acts as the anode under non-standard operating conditions.

Equilibrium Conditions and Reaction Quotient

The reaction quotient (Q) relates to the current concentrations of the reactants and products in the cell. When the cell potential is zero, the system is at equilibrium, meaning the forward and reverse reactions occur at equal rates. Determining the condition for zero cell potential involves setting the Nernst equation to zero and solving for the concentration that satisfies equilibrium.

Transcript

00:03 So before i ask this question, we need to figure out which one is the cathode, which one is the anode.

00:10 So what i put it right here is the reaction for the silver.

00:19 So this is the half reaction, and its redox potential is 0 .80 a volt.

00:41 And this is the half reaction for iodine and the redox potential is 0 .54 volt.

01:04 Because the radius potential for the silver to silver solid is more positive, so that will be the cathode, which means they're going to have a silver solid is more positive, so that will be the cathode, which means they're going to have a reduction reaction, they will act as the reduction reaction.

01:27 And next one will be the anode, so which means the oxid, this will be an oxidation reaction.

01:40 Okay, so now once we figured out cathode and anode, then we can very easy to determine what is the e0, which will be the e0, the e0.

01:51 Of the right here is reduction potential the reduction potential so the reduction potential of the cathode minus the reduction potential of the anode so that will be 0 .80 minus 0 .54 so that is the standard redox potential.

02:25 So now the condition is not standard.

02:31 So in that case the redox potential will be calculated use this formula e0 minus the n 0 .0592 log q.

02:50 And to determine the q and you need to write down the correct reactions right here.

02:59 So because this is the reduction reaction, this is the oxidation reaction, so you can imagine the reaction should go this way.

03:24 So that should be the reaction because you see the silver go to – silver iron go to silver metal is reduced.

03:32 And iodide go to iodine, which is oxidized.

03:37 So that is what we defined from here.

03:42 So the q right here, now from, once you have this equation, now you can see the q will be, we can use the equation to write q values, q expressions.

03:59 So for this particular case, n equals 2, 0 .0591, and log q right here, so that will be 1 divided by a concentration of silver ion to the power of 2, and i died to the power of 2.

04:21 So 0 .26 minus 2 .0592, and log right here, that will be 0 .1592.

04:37 And 0 .035 squared.

04:45 If you plug all those numbers, then you will get the final answer, 0 .15 volt.

05:02 So the next part right here is asking if the reaction under that condition is spontaneous or not...

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