Answer :
To determine the overall voltage for the given redox reaction, we need to consider the standard reduction potentials for each half-reaction.
The half-reactions are:
1. \(Ag^{+} + e^{-} \rightarrow Ag(s)\)
2. \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\)
The standard reduction potential for the silver reaction is:
[tex]\[ E^{\circ}_{Ag^{+}/Ag} = +0.80 \, \text{V} \][/tex]
The standard reduction potential for the copper reaction is given for the reduction process \(Cu^{2+} + 2e^{-} \rightarrow Cu(s)\), which is:
[tex]\[ E^{\circ}_{Cu^{2+}/Cu} = +0.34 \, \text{V} \][/tex]
However, the copper reaction we are dealing with is the oxidation process \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\), and we must take the opposite sign for this reaction:
[tex]\[ E^{\circ}_{Cu/Cu^{2+}} = -0.34 \, \text{V} \][/tex]
To find the overall cell voltage, we need to subtract the anode potential (oxidation potential) from the cathode potential (reduction potential). The half-reaction with the more positive reduction potential serves as the cathode. Therefore:
- The cathode reaction (reduction) is \(Ag^{+} + e^{-} \rightarrow Ag(s)\) with \( E^{\circ}_{\text{cathode}} = +0.80 \, \text{V} \).
- The anode reaction (oxidation) is \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\) with \( E^{\circ}_{\text{anode}} = -0.34 \, \text{V} \).
The overall cell voltage is given by:
[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \, \text{V} - (-0.34 \, \text{V}) \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \, \text{V} + 0.34 \, \text{V} \][/tex]
[tex]\[ E_{\text{cell}} = 1.14 \, \text{V} \][/tex]
Thus, the overall voltage for the redox reaction is 1.14 V, and the correct answer is:
[tex]\[ C. \, 0.80 - (-0.34) \][/tex]
The half-reactions are:
1. \(Ag^{+} + e^{-} \rightarrow Ag(s)\)
2. \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\)
The standard reduction potential for the silver reaction is:
[tex]\[ E^{\circ}_{Ag^{+}/Ag} = +0.80 \, \text{V} \][/tex]
The standard reduction potential for the copper reaction is given for the reduction process \(Cu^{2+} + 2e^{-} \rightarrow Cu(s)\), which is:
[tex]\[ E^{\circ}_{Cu^{2+}/Cu} = +0.34 \, \text{V} \][/tex]
However, the copper reaction we are dealing with is the oxidation process \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\), and we must take the opposite sign for this reaction:
[tex]\[ E^{\circ}_{Cu/Cu^{2+}} = -0.34 \, \text{V} \][/tex]
To find the overall cell voltage, we need to subtract the anode potential (oxidation potential) from the cathode potential (reduction potential). The half-reaction with the more positive reduction potential serves as the cathode. Therefore:
- The cathode reaction (reduction) is \(Ag^{+} + e^{-} \rightarrow Ag(s)\) with \( E^{\circ}_{\text{cathode}} = +0.80 \, \text{V} \).
- The anode reaction (oxidation) is \(Cu(s) \rightarrow Cu^{2+} + 2e^{-}\) with \( E^{\circ}_{\text{anode}} = -0.34 \, \text{V} \).
The overall cell voltage is given by:
[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \, \text{V} - (-0.34 \, \text{V}) \][/tex]
[tex]\[ E_{\text{cell}} = +0.80 \, \text{V} + 0.34 \, \text{V} \][/tex]
[tex]\[ E_{\text{cell}} = 1.14 \, \text{V} \][/tex]
Thus, the overall voltage for the redox reaction is 1.14 V, and the correct answer is:
[tex]\[ C. \, 0.80 - (-0.34) \][/tex]
