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Math  /  Algebra

QuestionConsider the galvanic cell shown below (the contents of each half-cell are written beneath each compartment).
The standard reduction potentials are as follows: Cr3++3e−→Cr(s)E∘=−0.73 VBr2(aq)+2e−→2Br−E∘=+1.09 V\begin{array}{ll} \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(s) & E^{\circ}=-0.73 \mathrm{~V} \\ \mathrm{Br}_{2}(a q)+2 \mathrm{e}^{-} \rightarrow \\ 2 \mathrm{Br}^{-} \end{array} \quad E^{\circ}=+1.09 \mathrm{~V}
What is the value of EE for this cell at 25∘C25^{\circ} \mathrm{C} ? a. 2.21 V b. 2.12 V c. 0.59 V d. 1.76 V e. 1.88 V

Studdy Solution

STEP 1

1. We are dealing with a galvanic cell.
2. The standard reduction potentials for the half-reactions are given.
3. We need to calculate the cell potential E E at standard conditions (25°C).

STEP 2

1. Identify the oxidation and reduction half-reactions.
2. Calculate the standard cell potential Ecell∘ E^\circ_{\text{cell}} .
3. Determine the overall cell potential E E .

STEP 3

Identify the oxidation and reduction half-reactions.
- The reduction half-reaction is: $ \mathrm{Br}_{2}(aq) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad E^{\circ} = +1.09 \mathrm{~V} \]
- The oxidation half-reaction is the reverse of the given reduction reaction for chromium: $ \mathrm{Cr}(s) \rightarrow \mathrm{Cr}^{3+} + 3 \mathrm{e}^{-} \quad E^{\circ} = +0.73 \mathrm{~V} \]

STEP 4

Calculate the standard cell potential Ecell∘ E^\circ_{\text{cell}} .
The standard cell potential is calculated using the formula: Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
Substitute the values: Ecell∘=(+1.09 V)−(−0.73 V)E^\circ_{\text{cell}} = (+1.09 \, \text{V}) - (-0.73 \, \text{V}) Ecell∘=1.09 V+0.73 VE^\circ_{\text{cell}} = 1.09 \, \text{V} + 0.73 \, \text{V} Ecell∘=1.82 VE^\circ_{\text{cell}} = 1.82 \, \text{V}

STEP 5

Determine the overall cell potential E E .
Since the problem asks for the cell potential at standard conditions and we have calculated Ecell∘ E^\circ_{\text{cell}} , the value of E E is the same as Ecell∘ E^\circ_{\text{cell}} .
Thus, the value of E E for this cell at 25∘C 25^{\circ} \mathrm{C} is: E=1.82 VE = 1.82 \, \text{V}
However, the closest option is not listed. Let's verify the calculation:
Recalculate: Ecell∘=1.09 V+0.73 V=1.82 VE^\circ_{\text{cell}} = 1.09 \, \text{V} + 0.73 \, \text{V} = 1.82 \, \text{V}
The closest option to our calculation is not present, indicating a potential miscalculation or misinterpretation of the options. Based on the given options, the closest value is:
d. 1.76 V

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