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Math  /  Algebra

Question6. What is an electrode of the first kind? Write a general Nernst equation for this indicator electrode.
7. A solution of 0.25MCr2O72−0.25 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} and 0.35MCr3+0.35 \mathrm{M} \mathrm{Cr}^{3+} has a pH of 2.0 . What is the Ecell \mathrm{E}_{\text {cell }} of this halfcell?

Studdy Solution

STEP 1

1. We need to define an electrode of the first kind.
2. We need to write the general Nernst equation for an electrode of the first kind.
3. We are given concentrations of Cr2O72−\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} and Cr3+\mathrm{Cr}^{3+} and a pH value to find the Ecell\mathrm{E}_{\text{cell}} of the half-cell.
4. The Nernst equation will be used to calculate the cell potential.

STEP 2

1. Define an electrode of the first kind.
2. Write the general Nernst equation for an electrode of the first kind.
3. Write the balanced half-reaction for the given species.
4. Calculate the standard electrode potential.
5. Use the Nernst equation to calculate Ecell\mathrm{E}_{\text{cell}}.

STEP 3

Define an electrode of the first kind.
An electrode of the first kind is a metal electrode that is in direct contact with its cation in solution. The potential of this electrode is directly related to the concentration of the cation.

STEP 4

Write the general Nernst equation for an electrode of the first kind.
The Nernst equation for an electrode of the first kind is given by:
E=E∘−RTnFln⁡Q E = E^{\circ} - \frac{RT}{nF} \ln Q
Where: - E E is the electrode potential. - E∘ E^{\circ} is the standard electrode potential. - R R is the universal gas constant (8.314 J/mol K8.314 \, \text{J/mol K}). - T T is the temperature in Kelvin. - n n is the number of moles of electrons transferred. - F F is Faraday's constant (96485 C/mol96485 \, \text{C/mol}). - Q Q is the reaction quotient.

STEP 5

Write the balanced half-reaction for the given species.
The half-reaction involving Cr2O72−\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} and Cr3+\mathrm{Cr}^{3+} in acidic solution is:
Cr2O72−+14H++6e−→2Cr3++7H2O \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}

STEP 6

Calculate the standard electrode potential.
The standard electrode potential for the reaction can be found in electrochemical series tables. For this reaction, let's assume:
E∘=1.33 V E^{\circ} = 1.33 \, \text{V}

STEP 7

Use the Nernst equation to calculate Ecell\mathrm{E}_{\text{cell}}.
First, calculate the reaction quotient Q Q :
Q=[Cr3+]2[Cr2O72−][H+]14 Q = \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}][\mathrm{H}^{+}]^{14}}
Given: - [Cr2O72−]=0.25 M[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] = 0.25 \, \text{M} - [Cr3+]=0.35 M[\mathrm{Cr}^{3+}] = 0.35 \, \text{M} - pH=2.0⇒[H+]=10−2 M\mathrm{pH} = 2.0 \Rightarrow [\mathrm{H}^{+}] = 10^{-2} \, \text{M}
Q=(0.35)2(0.25)(10−2)14 Q = \frac{(0.35)^2}{(0.25)(10^{-2})^{14}}
Now, substitute into the Nernst equation:
E=1.33−8.314×2986×96485ln⁡((0.35)2(0.25)(10−2)14) E = 1.33 - \frac{8.314 \times 298}{6 \times 96485} \ln \left( \frac{(0.35)^2}{(0.25)(10^{-2})^{14}} \right)
Calculate the value:
E=1.33−0.02576ln⁡(0.12250.25×10−28) E = 1.33 - \frac{0.0257}{6} \ln \left( \frac{0.1225}{0.25 \times 10^{-28}} \right)
E=1.33−0.00428ln⁡(4.9×1026) E = 1.33 - 0.00428 \ln \left( 4.9 \times 10^{26} \right)
E=1.33−0.00428×61.7 E = 1.33 - 0.00428 \times 61.7
E=1.33−0.264 E = 1.33 - 0.264
E=1.066 V E = 1.066 \, \text{V}
The Ecell\mathrm{E}_{\text{cell}} of this half-cell is:
1.066 V \boxed{1.066 \, \text{V}}

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