What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the Ag+ concentration is 5.29 × 10-4 M and the Cr³+ concentration is 1.41 M? 3Ag+ (aq) + Cr(s) → 3Ag(s) + Cr³+ (aq) Ag+ (aq) + e → Ag(s) Fo = 0.799 V red Cr³+ (aq) + 3e → Cr(s) E = -0.740 V red Ecell = The cell reaction as written above is spontaneous for the concentrations given: O true O false V

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### Calculation of Cell Potential for an Electrochemical Cell

**Problem Statement:**
What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the \( \text{Ag}^+ \) concentration is \( 5.29 \times 10^{-4} \) M and the \( \text{Cr}^{3+} \) concentration is 1.41 M?

\[ 3\text{Ag}^+(aq) + \text{Cr}(s) \rightarrow 3\text{Ag}(s) + \text{Cr}^{3+}(aq) \]

**Standard Reduction Potentials:**
\[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = 0.799 \ \text{V} \]
\[ \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ_{\text{red}} = -0.740 \ \text{V} \]

**Cell Potential Calculation:**

\[ E_{\text{cell}} = \text{V} \]

**Question:**
The cell reaction as written above is spontaneous for the concentrations given:
- [ ] true
- [ ] false

---

**Explanation:**

The problem requires the calculation of the cell potential \( E_{\text{cell}} \) for a galvanic cell at 298 K (standard temperature). The cell reaction provided involves silver and chromium:

\[ 3\text{Ag}^+(aq) + \text{Cr}(s) \rightarrow 3\text{Ag}(s) + \text{Cr}^{3+}(aq) \]

You must use the given standard reduction potentials to determine the overall cell potential. You are also asked to determine whether the reaction is spontaneous given the specific concentrations of \( \text{Ag}^+ \) and \( \text{Cr}^{3+} \).

The standard electrode potentials provided are:

- For the reduction of silver ions: 
\[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = 0.799 \ \text{V} \]

- For the reduction of chromium ions: 
\[ \text{Cr}^{3
Transcribed Image Text:### Calculation of Cell Potential for an Electrochemical Cell **Problem Statement:** What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the \( \text{Ag}^+ \) concentration is \( 5.29 \times 10^{-4} \) M and the \( \text{Cr}^{3+} \) concentration is 1.41 M? \[ 3\text{Ag}^+(aq) + \text{Cr}(s) \rightarrow 3\text{Ag}(s) + \text{Cr}^{3+}(aq) \] **Standard Reduction Potentials:** \[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = 0.799 \ \text{V} \] \[ \text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ_{\text{red}} = -0.740 \ \text{V} \] **Cell Potential Calculation:** \[ E_{\text{cell}} = \text{V} \] **Question:** The cell reaction as written above is spontaneous for the concentrations given: - [ ] true - [ ] false --- **Explanation:** The problem requires the calculation of the cell potential \( E_{\text{cell}} \) for a galvanic cell at 298 K (standard temperature). The cell reaction provided involves silver and chromium: \[ 3\text{Ag}^+(aq) + \text{Cr}(s) \rightarrow 3\text{Ag}(s) + \text{Cr}^{3+}(aq) \] You must use the given standard reduction potentials to determine the overall cell potential. You are also asked to determine whether the reaction is spontaneous given the specific concentrations of \( \text{Ag}^+ \) and \( \text{Cr}^{3+} \). The standard electrode potentials provided are: - For the reduction of silver ions: \[ \text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ_{\text{red}} = 0.799 \ \text{V} \] - For the reduction of chromium ions: \[ \text{Cr}^{3
### Chemical Equilibrium and Thermodynamics

#### For the reaction:

\[ \text{N}_2 (g) + 2\text{O}_2 (g) \rightarrow 2\text{NO}_2 (g) \]

The change in enthalpy (\(\Delta H^\circ\)) and change in entropy (\(\Delta S^\circ\)) for the reaction are given as:
\[ \Delta H^\circ = 66.4 \text{ kJ} \]
\[ \Delta S^\circ = -122 \text{ J/K} \]

The objective is to find the equilibrium constant for the reaction at a temperature of 343.0 K. Assume that both \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent of temperature. 

#### Calculating the Equilibrium Constant (\(K\))

The equilibrium constant for a reaction can be determined using the Gibbs free energy change (\(\Delta G^\circ\)) at a given temperature. The relationship between \(\Delta G^\circ\) and the equilibrium constant \(K\) is given by the equation:
\[ \Delta G^\circ = -RT \ln K \]

Where:
- \(R\) is the universal gas constant (8.314 J/mol·K).
- \(T\) is the temperature in Kelvin (K).

The Gibbs free energy change (\(\Delta G^\circ\)) can be calculated using the relationship:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]

Given:
\[ \Delta H^\circ = 66.4 \text{ kJ} = 66400 \text{ J} \]
\[ \Delta S^\circ = -122 \text{ J/K} \]
\[ T = 343.0 \text{ K} \]

Let's plug in the values:
\[ \Delta G^\circ = 66400 \text{ J} - (343.0 \text{ K} \times -122 \text{ J/K}) \]
\[ \Delta G^\circ = 66400 \text{ J} + 41846 \text{ J} \]
\[ \Delta G^\circ = 108246 \text{ J} \]

Now use the relationship between \(\Delta G^\circ\) and \(K\):
\[ 108246 \text{ J} = - (8.314
Transcribed Image Text:### Chemical Equilibrium and Thermodynamics #### For the reaction: \[ \text{N}_2 (g) + 2\text{O}_2 (g) \rightarrow 2\text{NO}_2 (g) \] The change in enthalpy (\(\Delta H^\circ\)) and change in entropy (\(\Delta S^\circ\)) for the reaction are given as: \[ \Delta H^\circ = 66.4 \text{ kJ} \] \[ \Delta S^\circ = -122 \text{ J/K} \] The objective is to find the equilibrium constant for the reaction at a temperature of 343.0 K. Assume that both \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent of temperature. #### Calculating the Equilibrium Constant (\(K\)) The equilibrium constant for a reaction can be determined using the Gibbs free energy change (\(\Delta G^\circ\)) at a given temperature. The relationship between \(\Delta G^\circ\) and the equilibrium constant \(K\) is given by the equation: \[ \Delta G^\circ = -RT \ln K \] Where: - \(R\) is the universal gas constant (8.314 J/mol·K). - \(T\) is the temperature in Kelvin (K). The Gibbs free energy change (\(\Delta G^\circ\)) can be calculated using the relationship: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] Given: \[ \Delta H^\circ = 66.4 \text{ kJ} = 66400 \text{ J} \] \[ \Delta S^\circ = -122 \text{ J/K} \] \[ T = 343.0 \text{ K} \] Let's plug in the values: \[ \Delta G^\circ = 66400 \text{ J} - (343.0 \text{ K} \times -122 \text{ J/K}) \] \[ \Delta G^\circ = 66400 \text{ J} + 41846 \text{ J} \] \[ \Delta G^\circ = 108246 \text{ J} \] Now use the relationship between \(\Delta G^\circ\) and \(K\): \[ 108246 \text{ J} = - (8.314
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