Math Snap

STEP 1

1. The dichromate ion $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ is a strong oxidizing agent in acidic solutions.
2. Standard electrode potentials ($E^\circ$) are available for the relevant half-reactions.
3. Gibbs free energy change ($\Delta G$) can be calculated using the formula $\Delta G = -nFE_{\text{cell}}^{o}$.
4. The Nernst equation can be used to determine cell potential under non-standard conditions.
5. Definitions are based on standard chemistry terminology.

STEP 2

1. Analyze the oxidation reactions with dichromate ion.
2. Calculate $E_{\text{cell}}^{o}$ for given cell diagrams.
3. Determine Gibbs free energy change and spontaneity.
4. Evaluate the spontaneity of the given reaction under specified conditions.
5. Define the given chemistry terms.

STEP 3

Analyze the oxidation reactions with dichromate ion:
a. Check if $\mathrm{Sn}^{2+}(\mathrm{aq})$ can be oxidized to $\mathrm{Sn}^{4+}(\mathrm{aq})$ using dichromate ion.
- The standard reduction potential for $\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}$ is $E^\circ = +0.15 \, \text{V}$.
- The standard reduction potential for $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}/\mathrm{Cr}^{3+}$ is $E^\circ = +1.33 \, \text{V}$.
- Since $E^\circ$ for dichromate is higher, it can oxidize $\mathrm{Sn}^{2+}$.
b. Check if $\mathrm{Ag}(\mathrm{s})$ can be oxidized to $\mathrm{Ag}^{+}(\mathrm{aq})$.
- The standard reduction potential for $\mathrm{Ag}^{+}/\mathrm{Ag}$ is $E^\circ = +0.80 \, \text{V}$.
- Since $E^\circ$ for dichromate is higher, it can oxidize $\mathrm{Ag}$.

STEP 4

Calculate $E_{\text{cell}}^{o}$ for given cell diagrams:
a. For $\operatorname{Pt}(\mathrm{s})\left|\mathrm{Fe}^{2+}(\mathrm{aq}), \mathrm{Fe}^{3+}(\mathrm{aq}) \|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}(\mathrm{s})\right.$:
- $E^\circ_{\text{red}}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77 \, \text{V}$
- $E^\circ_{\text{red}}(\mathrm{Ag}^{+}/\mathrm{Ag}) = +0.80 \, \text{V}$
- $E_{\text{cell}}^{o} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 \, \text{V} - 0.77 \, \text{V} = 0.03 \, \text{V}$
b. For $\mathrm{Pt}(\mathrm{s})\left|\mathrm{Mn}^{2+}(\mathrm{aq}), \mathrm{MnO}_{4}^{-}(\mathrm{aq})\right|\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})$:
- $E^\circ_{\text{red}}(\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}) = +1.51 \, \text{V}$
- $E^\circ_{\text{red}}(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34 \, \text{V}$
- $E_{\text{cell}}^{o} = 0.34 \, \text{V} - 1.51 \, \text{V} = -1.17 \, \text{V}$

STEP 5

Determine Gibbs free energy change and spontaneity:
a. For cell diagram (a):
- $\Delta G = -nFE_{\text{cell}}^{o}$
- $n = 1$ (for $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ and $\mathrm{Ag}^{+}/\mathrm{Ag}$)
- $\Delta G = -(1)(96485 \, \text{C/mol})(0.03 \, \text{V}) = -2894.55 \, \text{J/mol}$
- Reaction is spontaneous as $\Delta G < 0$.
b. For cell diagram (b):
- $\Delta G = -(2)(96485 \, \text{C/mol})(-1.17 \, \text{V}) = 226,902.3 \, \text{J/mol}$
- Reaction is nonspontaneous as $\Delta G > 0$.

STEP 6

Evaluate the spontaneity of the given reaction under specified conditions:
- Use the Nernst equation to find $E_{\text{cell}}$ under non-standard conditions.
- $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$
- $Q = \frac{[\mathrm{Cr}^{3+}]^2 [\mathrm{Ag}^{+}]^6}{[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}][\mathrm{H}^{+}]^{14}}$
- Substitute values: $Q = \frac{(0.6)^2 (0.675)^6}{(0.675)(10^{-2})^{14}}$
- Calculate $E_{\text{cell}}$ and determine if the reaction is spontaneous ($E_{\text{cell}} > 0$).

Define the given chemistry terms:
a) Oxidation: The process of losing electrons in a chemical reaction.
b) Oxidizing agent: A substance that gains electrons and is reduced in a chemical reaction.
c) Salt bridge: A device used in an electrochemical cell to maintain electrical neutrality by allowing the flow of ions.
d) Liquid junction: A boundary between two different electrolyte solutions in an electrochemical cell.
e) Standard electrode potential: The measure of the individual potential of a reversible electrode at standard state, which is $1 \, \text{M}$ concentration, $1 \, \text{atm}$ pressure, and $25^\circ \text{C}$.