QuestionMay 6, 2025

The overall reaction 2Co^3+(aq)+2Cl^-(aq)otimes 2Co^2+(aq)+Cl_(2)(g) has the standard cell voltage E_(cell)^circ =0.46V Given E^circ =1.36V for the reaction Cl_(2)(g)+2e^-(2)2Cl^-(aq) calculate the standard reduction potential for the following the half reaction at 25^circ C Co^3++e^-otimes Co^2+ Addyouranswer Integer, decimal, or E notation allowed

The overall reaction 2Co^3+(aq)+2Cl^-(aq)otimes 2Co^2+(aq)+Cl_(2)(g) has the standard cell voltage E_(cell)^circ =0.46V Given E^circ =1.36V for the reaction Cl_(2)(g)+2e^-(2)2Cl^-(aq) calculate the standard reduction potential for the following the half reaction at 25^circ C Co^3++e^-otimes Co^2+ Addyouranswer Integer, decimal, or E notation allowed
The overall reaction 2Co^3+(aq)+2Cl^-(aq)otimes 2Co^2+(aq)+Cl_(2)(g) has the standard cell voltage E_(cell)^circ =0.46V Given E^circ =1.36V for the reaction Cl_(2)(g)+2e^-(2)2Cl^-(aq) calculate the standard reduction potential for the following the half reaction at 25^circ C Co^3++e^-otimes Co^2+ Addyouranswer Integer, decimal, or E notation allowed

Solution
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Answer

0.90V Explanation 1. Write the relationship between cell potential and half-reactions The standard cell potential is related to the reduction potentials of the two half-reactions: E_{cell}^{\circ} = E^{\circ}_{\text{reduction (cathode)}} - E^{\circ}_{\text{oxidation (anode)}} 2. Identify cathode and anode reactions Cathode: Cl_2(g) + 2e^- \rightarrow 2Cl^- (E^{\circ} = 1.36V). Anode: Co^{3+} + e^- \rightarrow Co^{2+} (unknown E^{\circ}). 3. Rearrange formula to solve for E^{\circ}_{\text{anode}} Substitute values into the formula: 0.46 = 1.36 - E^{\circ}_{\text{anode}} Rearranging gives: E^{\circ}_{\text{anode}} = 1.36 - 0.46 4. Perform calculation E^{\circ}_{\text{anode}} = 0.90V

Explanation

1. Write the relationship between cell potential and half-reactions<br /> The standard cell potential is related to the reduction potentials of the two half-reactions:<br />$E_{cell}^{\circ} = E^{\circ}_{\text{reduction (cathode)}} - E^{\circ}_{\text{oxidation (anode)}}$<br /><br />2. Identify cathode and anode reactions<br /> Cathode: $Cl_2(g) + 2e^- \rightarrow 2Cl^-$ ($E^{\circ} = 1.36V$). <br />Anode: $Co^{3+} + e^- \rightarrow Co^{2+}$ (unknown $E^{\circ}$).<br /><br />3. Rearrange formula to solve for $E^{\circ}_{\text{anode}}$<br /> Substitute values into the formula:<br />$0.46 = 1.36 - E^{\circ}_{\text{anode}}$<br /><br />Rearranging gives:<br />$E^{\circ}_{\text{anode}} = 1.36 - 0.46$<br /><br />4. Perform calculation<br /> $E^{\circ}_{\text{anode}} = 0.90V$
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