Math Snap

STEP 1

1. We are given a solution of $0.25 \, \text{M} \, \text{Cr}_2\text{O}_7^{2-}$ and $0.35 \, \text{M} \, \text{Cr}^{3+}$.
2. The solution has a pH of 2.0.
3. We need to calculate the $E_{\text{cell}}$ of this half-cell.
4. We assume standard conditions for temperature (25°C or 298 K).
5. The Nernst equation will be used to find the cell potential.

STEP 2

1. Write the balanced half-reaction.
2. Determine the standard reduction potential.
3. Use the Nernst equation to find the cell potential.

STEP 3

Write the balanced half-reaction for the reduction of $\text{Cr}_2\text{O}_7^{2-}$ to $\text{Cr}^{3+}$.
The balanced half-reaction in acidic solution is:
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

STEP 4

Determine the standard reduction potential ($E^\circ$) for the half-reaction.
The standard reduction potential for the reaction is typically found in a table of standard electrode potentials:
$$E^\circ = 1.33 \, \text{V}$$

Use the Nernst equation to find the cell potential ($E_{\text{cell}}$).
The Nernst equation is given by:
$$E = E^\circ - \frac{RT}{nF} \ln Q$$ Where:
- $E$ is the cell potential.
- $E^\circ$ is the standard reduction potential.
- $R$ is the universal gas constant ($8.314 \, \text{J/mol K}$).
- $T$ is the temperature in Kelvin ($298 \, \text{K}$).
- $n$ is the number of moles of electrons transferred in the reaction ($6$ for this reaction).
- $F$ is Faraday's constant ($96485 \, \text{C/mol}$).
- $Q$ is the reaction quotient.
Calculate $Q$:
$$Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}$$ Given:
- $[\text{Cr}^{3+}] = 0.35 \, \text{M}$
- $[\text{Cr}_2\text{O}_7^{2-}] = 0.25 \, \text{M}$
- $[\text{H}^+] = 10^{-2} \, \text{M}$ (from pH = 2)
$$Q = \frac{(0.35)^2}{0.25 \times (10^{-2})^{14}}$$ Calculate $E_{\text{cell}}$:
$$E = 1.33 - \frac{8.314 \times 298}{6 \times 96485} \ln \left( \frac{(0.35)^2}{0.25 \times (10^{-2})^{14}} \right)$$ $$E = 1.33 - \frac{0.0257}{6} \ln \left( \frac{0.1225}{0.25 \times 10^{-28}} \right)$$ $$E = 1.33 - 0.00428 \ln (4.9 \times 10^{26})$$ $$E = 1.33 - 0.00428 \times 61.6$$ $$E = 1.33 - 0.264$$ $$E = 1.066 \, \text{V}$$ The $E_{\text{cell}}$ of this half-cell is approximately $1.066 \, \text{V}$.