Use the table of standard reduction potentials (Appendix M ) to calculate Δr G^∘ for the followi

step 1 All right. Hello, everybody. Today we're going to be finding the Gives Tree Energy for a few rea

Use the table of standard reduction potentials (Appendix M )...

Use the table of standard reduction potentials (Appendix $\mathrm{M}$ ) to calculate $\Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ for the following reactions at $298 \mathrm{K}$ (a) $3 \mathrm{Cu}(\mathrm{s})+2 \mathrm{NO}_{3}^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$ $3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{NO}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell)$ (b) $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$ $\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$

Use the table of standard reduction potentials (Appendix $\mathrm{M}$ ) to calculate $\Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ for the following reactions at $298 \mathrm{K}$
(a) $3 \mathrm{Cu}(\mathrm{s})+2 \mathrm{NO}_{3}^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$
$3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{NO}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell)$
(b) $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$
$\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$

Key Concepts

Standard Reduction Potentials

Standard reduction potentials are measures of the tendency of a chemical species to acquire electrons and be reduced, under standard conditions (1 M concentration, 1 atm pressure, and 25°C). These values are essential for determining the overall cell potential of a redox reaction and predicting the direction of electron flow.

Gibbs Free Energy Change

The Gibbs free energy change (?rG°) quantifies the maximum reversible work obtainable from a chemical reaction at constant temperature and pressure. It can be directly related to the cell potential by the equation ?rG° = -nFE°, where n is the number of electrons transferred and F is Faraday’s constant.

Electron Transfer and Stoichiometry

In electrochemical reactions, it is critical to account for the number of electrons transferred, as this stoichiometric factor (n) is used in calculating the cell potential and subsequently the Gibbs free energy change. Balancing half-reactions correctly ensures accurate determination of the overall reaction energetics.

Faraday’s Constant

Faraday’s constant, approximately 96,485 coulombs per mole of electrons, is the proportionality factor that converts the charge involved in electron transfer into energy units (joules). It is a key element in relating the cell potential to the change in Gibbs free energy.

Transcript

0:00 All right.

00:01 Hello, everybody.

00:01 Today we're going to be finding the gives tree energy for a few reactions.

00:05 So let's go right into it.

00:07 Our first reaction is 3 -cu solid plus 2 -n -o3 minus aqueous plus 8h plus aqueous, giving us 3 -c2 -plus aqueous, plus 2 -2 -2 -2 -liquid.

00:20 So i've taken the liberty of writing out the reduction potentials, reduction specifically, for both of our half -reactors.

00:30 Reactions are n -o3 to n -o and our co2 plus to cu.

00:36 Now notice up here we have cu on the left and co2 plus on the right.

00:40 Down here we have 2 plus on the left and cu on the right, which means that our copper is being oxidized instead of reduced.

00:49 So we're going to take this one, we're going to switch it to negative 0 .337 volts, which means that our e not total is going to be equal to 0 .96 volts minus 0 .337 volts, which when accounting for sigfigs is about 0 .62 volts.

01:13 Okay.

01:15 Cool.

01:16 Now we just plug this into our free energy.

01:18 So delta rg is going to be equal to negative nf -e -0, which is.

01:25 So we know that the equation for copper for the oxidation is, for the oxidation is cu gives us cu2 plus plus 2e minus, right? but we also know that we have 3 cu over here.

01:47 So this entire equation has to be multiplied by 3.

01:51 Let me write that on the other side, which means that our total.

01:56 Number of electrons transferred is going to be six, right? so if we have six electrons, then that's our new n value.

02:05 So we're going to say negative six.

02:09 Of course, we have faraday's constant, 9645, and then our e nods 0 .62...

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