A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: 2MnO4 (aq) + 16H* (aq) +5Pb (s) → 2Mn* (aq) +8H₂0 (1) +5Pb²+ (aq) Suppose the cell is prepared with 5.05 M MnO4 and 1.25 M H in one half-cell and 5.98 M Mn²+ Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. 0 0.0 X μ pla 5 and 2.24 M Pb in the other.

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### Understanding a Redox Reaction in a Galvanic Cell

A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction:

\[ 2\text{MnO}_4^{-} (\text{aq}) + 16\text{H}^{+} (\text{aq}) + 5\text{Pb} (s) \rightarrow 2\text{Mn}^{2+} (\text{aq}) + 8\text{H}_2\text{O} (l) + 5\text{Pb}^{2+} (\text{aq}) \]

Suppose the cell is prepared with 5.05 M \(\text{MnO}_4^{-}\) and 1.25 M \(\text{H}^{+}\) in one half-cell, and 5.98 M \(\text{Mn}^{2+}\) and 2.24 M \(\text{Pb}^{2+}) in the other.

Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

### Diagram Explanation:

There is a blank input box followed by a set of options to select units, which could possibly include voltage or concentration units such as molarity (M).

The checkbox options include:
- \[ \boxed{\phantom{x}} \mu \]
- \[ \boxed{\phantom{x}} \text{to the power of 10} \]
- \[ \text{other checkbox options} \]
- Refresh button (to reset input)

The exact nature of each checkbox is not clear, but they are likely for unit selection or for powering the final answer appropriately.

### Calculation Steps (Hypothetical Process):

1. Identify the standard electrode potentials (\( E^\circ \)) of the half-reactions involved.
2. Use the Nernst equation to find the electrode potentials under the given conditions.
3. Combine the electrode potentials to calculate the overall cell voltage.

Ensure to refer to your chemistry textbook or instructor for the values of standard electrode potentials required for accurate computation.
Transcribed Image Text:### Understanding a Redox Reaction in a Galvanic Cell A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: \[ 2\text{MnO}_4^{-} (\text{aq}) + 16\text{H}^{+} (\text{aq}) + 5\text{Pb} (s) \rightarrow 2\text{Mn}^{2+} (\text{aq}) + 8\text{H}_2\text{O} (l) + 5\text{Pb}^{2+} (\text{aq}) \] Suppose the cell is prepared with 5.05 M \(\text{MnO}_4^{-}\) and 1.25 M \(\text{H}^{+}\) in one half-cell, and 5.98 M \(\text{Mn}^{2+}\) and 2.24 M \(\text{Pb}^{2+}) in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. ### Diagram Explanation: There is a blank input box followed by a set of options to select units, which could possibly include voltage or concentration units such as molarity (M). The checkbox options include: - \[ \boxed{\phantom{x}} \mu \] - \[ \boxed{\phantom{x}} \text{to the power of 10} \] - \[ \text{other checkbox options} \] - Refresh button (to reset input) The exact nature of each checkbox is not clear, but they are likely for unit selection or for powering the final answer appropriately. ### Calculation Steps (Hypothetical Process): 1. Identify the standard electrode potentials (\( E^\circ \)) of the half-reactions involved. 2. Use the Nernst equation to find the electrode potentials under the given conditions. 3. Combine the electrode potentials to calculate the overall cell voltage. Ensure to refer to your chemistry textbook or instructor for the values of standard electrode potentials required for accurate computation.
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